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Mathematics of accounting

Curtis, Arthur B.

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business mathematics, accounting.

University of Florida Libraries Gainesville, Florida-

Gift of

Potter, Bower & Company Certified Public Accountants

Mathematics

of Accounting

ARTHUR B. CURTIS, B.C.S., C.P.A. JOHN H. COOPER, B. Acds., C.P.A.

Revised by WILLIAM JAMES McCALLION, M.A.

Associate Professor of Mathematics and Director of Extension, Mc Master University, Hamilton, Ontario

FOURTH EDITION

Englewood Cliffs, N. J. PRENTICE-HALL, INC.

PRENTICE-HALL ACCOUNTING SERIES H. A. Finney, Editor

©, 1925, 1934, 1947, 1961, by PRENTICE-HALL, INC. Englewood Cliffs, N. J.

REPRODUCED IN ANY FORM, BY MIMEOGRAPH OR ANY OTHER

MEANS, WITHOUT PERMISSION IN WRITING FROM THE

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Library of Congress Catalog Card No.: 61-7601

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Printed in the United States of America 56390-C "

Preface

This revision was undertaken with three main objectives: to bring the major portion of the content up to date, to introduce a sound mathe- matical basis to business mathematics, and to make the book more useful to Canadian readers.

The first objective is achieved by securing more current statistics and by including new approaches, new examples, and new problems. The second objective is achieved by giving a more mathematical treatment of simple interest, compound interest, and annuities. These subjects are presented as an exploitation of simple formulae, each of which contains four unknowns, with a solution found for the fourth unknown when values are assigned to the other three. The third objective is attained by includ- ing special notes and explanations where the Canadian theory differs sig- nificantly from the theory in the United States.

It is hoped that this revision will be useful as a textbook for courses in the mathematics of finance without losing any of its value as a reference book for accountants. Emphasis has been placed on the reference-book aspect by including completely detailed examples of a wide variety.

The author is indebted to Professor John L. Zimka, Department of Accounting, Fairleigh Dickinson University, Rutherford, New Jersey, who read the manuscript and made many valuable suggestions, to Miss Lillian Margot and the Editorial Department of Prentice-Hall for their helpful encouragement and conscientious assistance, and to the secretarial staff of the Department of University Extension, McMaster University, for the careful typing of the manuscript and proofreading.

Tables III, IV, V, VI, and VII are from Glover's Tables of Applied Mathematics. The author acknowledges his debt to Professor James W. Glover and his publisher, Mr. George Wahr, for their permission to use them.

William J. McCallion

Contents

PART ONE: Commercial Arithmetic — Simple Interest- Applications

1. Fundamental Processes and Short Methods for

the Accountant

Addition; Streamline addition; Combinations whose sum is 10; Adding where the same number is repeated many times; Group addition; Addi- tion of two columns at a time; Recording addition by columns; Practical applications; Subtraction; Avoid errors; Difference between a given minuend and several subtrahends; Balancing an account; Complement method; Subtracting on an adding machine; Practical problems; Multi- plication ; Contractions in multiplication ; To multiply by factors of the multiplier; To multiply when a part of the multiplier is a factor or multiple of another part; To multiply a number of two figures by 11 ; To multiply any number by 11; Multiplying by 25; Multiplying by 15; Multiplying numbers ending with ciphers; Multiplication by numbers near 100, as 98, 97, 96, and so forth, and by numbers near 1,000, as 997, 996, and so forth; Multiplication of two numbers each near 100, 1,000, and so forth; Multiplying by numbers a little larger than 100, as 101, 102, and so forth; Multiplication of two numbers each a little more than 100; Cross multiplication; To cross-multiply a number of three digits by a number of two digits; To cross-multiply a number of three digits by another number of three digits ; Preparation of a table of multi- ples of a number; Division; To divide by 25, 50, or 125; Abbreviated division; Use of tables of division; Reciprocals in division. CHECK- ING COMPUTATIONS: Methods; Rough check; Absolute check; Check numbers obtained by casting out the nines; Verification of addi- tion; Verification of subtraction; Verification of multiplication; Verifica- tion of division; Verification of division where there is a remainder; Check numbers obtained by casting out the elevens; Verification of addi- tion; Verification of subtraction; Verification of multiplication; Veri- fication of division; Check number thirteen.

2. Factors, Multiples, and Common Fractions 35

Factors; Tests of divisibility; Greatest common divisor; Least common multiple; Cancellation. COMMON FRACTIONS: Terms explained; Reduction of fractions; Principle; Mixed numbers; To change a mixed number to an improper fraction; Addition and subtraction of fractions; Multiplication of fractions; Division of fractions; To find the product of

Contents

any two mixed numbers ending in £; To multiply a mixed number by a mixed number; Decimal fractions; Approximate numbers; Calculations with exact numbers; Calculations with approximate numbers; To change a decimal fraction to an equivalent common fraction ; To change a com- mon fraction to a decimal; Aliquot parts; The use of aliquot parts; Multi- plication by aliquot parts; Division by aliquot parts.

3. Percentage and Applications 58

Relation between percentage and common and decimal fractions; Applications; Definitions; Fundamental processes; Computations; Daily record of departmental sales ; Per cent of returned sales by depart- ments; Clerk's per cent of average sales; Per cent of income by source; Per cent of expense; Per cent of increase or decrease; Operating statis- tics; Budgeting; Profits based on sales; Commissions; Cash discount; Trade discount; Single discount equivalent to a series; Transportation charges on discount invoices; Anticipation.

4. Simple Interest 78

Interest; Simple interest formula; Comparison of simple amount and simple present worth. EXACT AND ORDINARY SIMPLE INTER- EST: Method using aliquot parts; The cancellation method; Dollars- times-days method, 6%; Interchanging principal and time; Exact or accurate interest. BANK DISCOUNT: Definition; Counting time; Finding the difference between dates by use of a table. PARTIAL PAYMENTS: Part payments on debts; Methods; United States Rule; Merchants' Rule. AVERAGING DATES OF INVOICES : Definition ; Use; Term of credit; Average due date; Focal date; Methods; Rule for product method. EQUATION OF ACCOUNTS, OR COMPOUND AVERAGE: Definition; Rule for the product method; When to date forward or backward. INSTALLMENT PLANS AND PERSONAL LOANS: Equal monthly installments plan formula; Equal-monthly- installments-plus-an-odd-final-payment plan formula; Personal loans. ACCOUNT CURRENT: Definition; Methods. STORAGE: Defini- tion; Running account.

5. Inventories 111

Valuation of inventories; Cost or market, whichever is lower; Average cost method; "First-in, first-out" method of inventory; "Last-in, first- out" method of inventory; Merchandise turnover; Number of turnovers; Per cent of mark-down to net cost; Computation of inventory by the retail method; Determining the ratio of cost to retail.

vi Contents

6. Insurance 123

Insurance; Policy; Fire insurance; Form of policy; Rates; To find the premium; Agent's commission; Cancellation of policies; Coinsurance; Use and occupancy insurance; Group life insurance; Health insurance; Workmen's compensation insurance.

7. Gross Profit Computations 134

Gross profit; Rate per cent of gross profit; Procedure; Uses; Cost of goods sold; Rate per cent of cost of sales; Fire losses; Use of gross profit test in verification of taxpayer's inventory; Installment sales of personal property; Computation of gross profit; Reserve for unearned gross profit; Bad debts; Deferring income; its effect on tax.

8. Analysis of Statements lJfB

Financial and operating ratios; Costs, expenses, and profits; Ratio of gross profit to net sales; Ratio of operating profit to net sales; Ratio of net profit to net sales; Ratio of operating profit to total capital employed ; Ratio of net profit to net worth; Earnings on common stockholders' investments; Working capital ratio; Sources of capital; Manner in which capital is invested; Turnover of total capital employed; Turnover of inventories; Turnover of accounts receivable ; Turnover of fixed property investment.

9. Goodwill 159

Definition; Basis of valuation; Earning power determined from profit and loss statements; Methods of valuing goodwill; Case illustrations; Valuation by appraisal; Valuation by number of years' purchase price of net profits; Valuation on basis of excess of profits over interest on net assets; Basis of stock allotment; Common stock only; Preferred stock for net assets; Bonds, preferred stock, and common stock; Conclusion.

10. Business Finance 170

Stock rights; Sale of stock and rights, federal income tax; Working capital; Cumulative voting; Book value of shares of stock; Profits distribution.

11. Partnership 179

Definition; Mathematical calculations; Goodwill; Profit-sharing agree- ments; Lack of agreement; Losses; Arbitrary ratio; Ratio of investment;

Contents vii

Division of profits by first deducting interest on capital; Profits insuf- ficient to cover interest on investment; Adjustments of capital con- tribution; Profit sharing in ratio of average investment; Liquidation of partnership; Methods; Total Distribution; Periodic distribution.

12. Public Finance and Taxation 199

Governmental functions; Purposes of taxes; Appropriations; Kinds of taxes. PROPERTY TAX: Determination of tax rate; To find the amount of tax.

PART TWO: Commercial Algebra — Compound Interest — Applications

13. Fundamentals of Algebra 207

Explanation; Symbols and terms; Positive and negative numbers; Addition of positive and negative numbers; The coefficient; Parentheses, brackets, and braces; Subtraction; Multiplication; Division. EQUA- TIONS: Simple equations; Fractions; Clearing of complex fractions; Simultaneous equations with two or more unknowns.

14. Logarithms

Uses of logarithms; The working rules of logarithms; Proof of working rules; Common logarithms; How to use a table of logarithms; To find a number when the logarithm is given; To find a number whose mantissa is not in the table; The use of logarithms in computation; Multiplication by logarithms; Division by logarithms; Powers of numbers; Roots of numbers; Combination of working rules; The slide rule; Use of slide rule; Accuracy of calculations made by the slide rule; Theory of the slide rule; How to learn to use the slide rule; Reading the slide rule; Construction of model slide rule; Multiplication on the slide rule; Division on the slide rule.

15. Graphs and Index Numbers 2J+3

Charts and graphs; Circle chart; Comparison of circles; Bar chart; Line or curve chart; Rules for coordinate charts; Logarithmic chart; Semi- logarithmic or ratio charts. INDEX NUMBERS: Indicators of business conditions; Index numbers; Price indexes; Quantity indexes; Value indexes; Business activity indexes; Other indexes.

viii Contents

16. Series of Numbers — The Progressions 265

Definition; The arithmetic progression; Symbols; Relation of elements; Formulas for an arithmetical progression; The geometric progression; Elements; Formulas for a geometric progression; Progression problems solved by the use of logarithms; The infinite geometric series; Sums of infinite geometric series.

17. Compound Interest

Compound interest; Compound interest and simple interest compared; Actuarial science; Conversion period and the periodic rate; Compound interest formula; Computation of compound amount of 1 by Table 3; Time beyond table limit; Computation of compound amount of 1 by the use of logarithms; Compound amount of 1 for fractional part of com- pounding period; Present value; Compound discount; Computation of present value of 1 by Table 4; Time beyond table limit; Computation of present value of 1 by the use of logarithms; Nominal and effective rates; Interest equation; Compound amount of 1 and present value of 1 at changing rates; Analysis of the compound interest formula.

18. Ordinary Annuities

Definition. ORDINARY ANNUITIES FORMULAS: Amount of an ordinary annuity formula; Computation of compound amount of an ordinary annuity of 1 per period by Table 5; Time beyond table limit; Present value of an ordinary annuity formula; Computation of present value of an ordinary annuity of 1 per period by Table 6; Time beyond table limit; Analysis of the ordinary annuities formulas. PRESENT VALUE AND COMPOUND AMOUNT OF AN ORDINARY ANNUITY. AMORTIZATION AND SINKING FUNDS: Repay- ment of a debt by the amortization method ; Repayment of a debt by the sinking fund method. TERMS OF AN ORDINARY ANNUITY. RATE OF AN ORDINARY ANNUITY.

19. Special Annuities 312

ANNUITY DUE: Amount of an annuity due formula; Present value of an annuity due formula; Analysis of the annuity due formulas. PRES- ENT VALUE AND COMPOUND AMOUNT OF AN ANNUITY DUE. RENT OF AN ANNUITY DUE. TERM OF AN ANNUITY DUE. RATE OF AN ANNUITY DUE. DEFERRED ANNUI- TIES: Present value of a deferred annuity formulas; Analysis of the present value of a deferred annuity formulas. PERPETUITIES: Present value of a perpetuity formula. CAPITALIZED COST.

Contents ix

20. Bond and Bond Interest Valuation 836

Definitions; Kinds of bonds; Guaranteed bond; Collateral trust bond; Mortgage bond ; Debenture bond ; Coupon bond ; Registered bond ; Fully Registered bond; Registered as to principal only bond; Straight term bond; Serial bond; Annuity bond; Bonds sold at par; Bonds purchased at a discount or at a premium ; Price and rate of yield ; Premium and dis- count formulas; Accumulation of discount; Amortization of premium; Annuity bonds; Bonds redeemable from a fund; Values of bonds between interest dates; Flat price and "and interest" price of a bond; Approxi- mate yield rate of interest on bonds. USE OF BOND TABLE AND INTERPOLATION METHOD: Computation when bond table is not available; Amortization of discount, premium, or discount and expense on serial redemption bonds; Bonds outstanding method; Scientific method.

21. Asset Valuation Accounts 858

Asset valuation; Depreciation; Depletion; Depreciation methods; Straight-line method; Working-hours or unit-product method; Sum-of- digits method; Sinking-fund method; Annuity method of depreciation; Constant-percentage-of-book- value method; Composite life; Depletion; Calculation of depletion; Valuation of wasting assets.

22. Permutations and Combinations 874

Fundamental principles; Permutation; Combinations.

23. Probability 880

Probability ; Permutations and combinations in probabilit}' ; Compound events; Independent events; Mutually exclusive events; Empirical probability.

24. Probability and Mortality 387

Life insurance; Mortality table; Probability of living: Probability of dying; Joint life probabilities.

25. Life Annuities

Factors involved; Pure endowment; Life annuity; Commutation columns; Life annuities due; Use of commutation table; Deferred annuity; Deferred life annuity due; Temporary life annuities ; Temporary annuities due; Life annuities with payments m times a year; Forborne temporary annuity due.

x Contents

26. Net Premiums 403

Net single premium; Annual premiums; Term insurance; Annual premium for term insurance; Net single premium for endowment insurance ; Annual premium for endowment insurance.

27. Valuation of Life Insurance Policies Jf.07

Mortality and the level premium; Policy reserves; Interest and the premium; Loading; Dividends and net cost; Terminal reserves; Retro- spective method; Transformation; Reserve valuation for limited pay- ment life insurance; Preliminary term valuation.

APPENDIXES

1. Practical Business Measurements J/.15

2. Tables of Weights, Measures, and Values

3. Financial Tables

I. American Experience Table of Mortality II. Commutation Columns, 3^- Per Cent 435

III. Amount of 1 at Compound Interest: s — (1 + i)n 437

IV. Present Value of 1 at Compound Interest: vn

(1 + i)» or vn = (1 + i)~n 457

n _j_ j\n _ i

V. Amount of Annuity of 1 per Period: s-i, = : 477

VI. Present Value of Annuity of 1 per Period:

VII. Periodic Rent of Annuity Whose Present Value is 1 :

— = i n\ -w 517

an\i 1 - (1 + 0 n

VIII. Values of (1 + i)"* 537

(I _J_ {)1!p _ 1

IX. Values of sf| = — — 538

X. Values of J- -

sil

XI. Values of a-, = v^ .' - 540

Hi *

XII. Logarithms 54 1

Index 556

(1 +

iy/p

- 1

(t' +

iyiv

- 1

1 -

(1 +

iynp

PART ONE

Commercial Arithmetic Simple Interest Applications

Fundamental Processes and Short Methods for the Accountant

Addition. Addition is the process of combining numbers of the same denomination. Quantities of such unlike measures as dollars and yards cannot be added; but quantities like yards, feet, and inches can be changed to like numbers and then added. Like numbers are numbers that express the same kind of units. The sum is the number resulting from adding two or more like numbers, and the addends are the different numbers to be added.

Addition is the most fundamental of all numerical operations. It is essential that the clerk, the businessman, and the accountant be able to add with precision and rapidity. The ability to recognize the sums of numbers instantly is acquired by constant practice and careful study.

Drill tables. Practice adding the columns of numbers in the follow- ing table until you can complete the operation in 25 seconds, without error. State sums only; that is, do not repeat the numbers to be added.

Practice stating the sums of the following columns of numbers until you can do all of them correctly in less than 2\ minutes.

Fundamental Processes and Short Methods for the Accountant

34234335576523538 22233323323322232

Streamline addition. Omit unnecessary words: that is, do not name the number to be added; name only the sum.

5 In the example at the left, a common way of adding would be (com-

7 mencing at the top): 5 and 7 are 12, 12 and 8 are 20, 20 and 4 are 24,

8 and so forth. Instead of adding in this manner, proceed to the answer 4 by saying (mentally), " 12, 20, 24, 27, 29, 38."

2 _9 38

Drill table.

3426751835 2598934728 7377268671 4669586963 5933872499 6842625746 1584397537

Combinations whose sum is 10. Combinations of two or more numbers whose sum is 10 are of frequent occurrence. When these com-

Fundamental Processes and Short Methods for the Accountant 5

binations are recognized, addition may be shortened by adding such com- binations as 10.

4 In this example, the addition may be performed as follows: (com- 7v mencing at the top) 14, 16, 21, 30, 40.

3 Or, it may be added in this manner: 11, 21, 30, 40.

2 Do not try to form combinations. Unless they are instantly recog-

5 nized, add the numbers in the regular manner. 9

3 2>

Drill table. / / J J

7353264192 25688 9, 2587 1622 554974 5477447835 . 6 9 16 3 15 3 6 1 4843773737 9281582615 8728226425 2457398259

Adding where the same number is repeated many times. In obtaining averages, in adding statistics, and in other work involving addition, often the same number is repeated many times. Use multi- plication to save time in adding.

724 In this example, 7 occurs four times and 6 occurs three times in the 785 third column. The sum of the third column may be found as follows:

773 Carried 4

748 4X7 28

696 3X618

687 50 679 5092

The work is actually performed mentally thus: 4 (28) 32, (18) 50. Where the columns are long, a side calculation may be necessary.

Drill table.

284

34.86

23.56

48.34

71.53

273

33.75

23.95

47.56

72.37

281

32.86

24.72

39.85

72.48

311

31.29

25.31

38.64

69.95

314

34.36

26.54

45.58

68.83

321

32.75

31.72

39.95

68.44

318

33.95

32.69

42.74

67.93

319

36.87

33.47

38.56

71.59

6 Fundamental Processes and Short Methods for the Accountant

Group addition. The most practical method of adding is to group or combine two or more figures mentally, and to name results only.

Mental Steps

Instead of saying, "5 and

4 are 9, and 7 are

16, and 3

are 19, and 8 are

27, and 6

are 33, and 1

are

34,

and 2

are

36," simply

think

"9,

19,

6 1 2

33, 36."

Drill tab]

le.

6 4 5

7 4

2 5 7

3 8

4 5 7

6 9

5 9 7

6 8

4 9 7

8 7

5 8 6

6 2

7 6 3

2 3

7 6 4

5 8

7 7 7

2 1

Addition of two columns at a time. Two columns of figures may be added at the same time, as shown in the following illustration :

Mental Steps Explanation

56 Tens Units Tens Units

28 (1) 76 (2) 84 (1) 56 and 20 = 76 (2) 76 and 8 = 84

43 (3) 124 U) 127 (3) 84 and 40 = 124 (4) 124 and 3 = 127

_21 (5) 147 (6) 148 (5) 127 and 20 = 147 (6) 147 and 1 = 148

148

Drill table.

79 82 24 37 65 39 28 28

48 84 33 44 81 58 39 59

81 95 46 53 42 48 23 86

15 83 52 66 73 73 37 63

Recording addition by columns. Accountants are subject to interruptions, but the time required to re-add a column of figures for the purpose of picking up the carrying figure may be saved if the total of each column is recorded separately. The separate column totals are also convenient to use in checking the work; for instance, if in a final summary of additions there is an error of $100.00, the hundreds' columns of the subtotals may be verified quickly without the necessity of re-adding all the columns.

Fundamental Processes and Short Methods for the Accountant 7

Example — Method 1 Example — Method 2 Example — Method 3

4572 4572 4572

3986 3986 3986

2173 2173 2173

5911 5911 5911

2765 2765 2765

4937 4937 4937

24 24 24

32 34 34

40 43 43

20 24 24

24344 24344

Explanation 1. Add each column separately, setting the sums one place to the left, as in the example. After the last column has been added, add the individual sums in regular order; that is, from right to left.

Explanation 2. In Method 2, a little time is saved by adding to each column the number carried from the column at the right.

Explanation 3. Method 3 differs from Method 2 in the writing of the columns' sums. It is somewhat easier to write the sums one below the other. This cannot be done in Method 1 because carrying figures are not used, and another step is required to complete the answer: that is, finding the grand total of the units, tens, hundreds, and so forth.

A modification of the third method is useful in adding columns of dollars and cents.

$ 644.22 The total, $4,062.08, is obtained by adding each column sepa-

821.94 rately as explained under Method 3. The computation will 314.26 appear as follows, the purpose of the horizontal lines being to 712.84 separate cents from dollars, and hundreds from thousands. 976.54 592.28

$4,062.08

Sum of the first column 28

Sum of the second column, 28 plus 2, carrying number 30

Sum of the third column, 19 plus 3, carrying number 22

Sum of the fourth column, 24 plus 2, carrying number 26

Sum of the fifth column, 38 plus 2, carrying number 40

As there are no more columns, write the carrying number 4

The total, $4,062.08, is obtained by reading the numbers at the right, com- mencing at the bottom.

PROBLEMS

1. 2. 3. 4. 5. 6. 7. 8.

5273 5126 7952 1395 3688 $367.98 $786.42 $498.57

2191 8497 2975 2764 4932 421.74 518.49 822.56

8437 7934 8675 8351 7563 281.34 946.72 753.86

3426 9783 8437 6248 2898 633.46 881.92 629.75

7139 9126 2975 5347 6598 855.91 542.37 367.43

7895 8751 3826 4586 8877 769.25 787.66 521.54

Fundamental Processes and Short Methods for the Accountant

PRACTICE PROBLEMS

Average weekly earnings from payroll reports.

33.20 25.13 37.41 31.65 31.40 22.93 32 16 26.37 36.52 32.05 32.58 23.60 23.44 36.37

35.72

29.88 39.24 33.47 35.29 23.22 33.49 28.34 33.64 34.60 26.49 28.81 37.92 41.54

30.85 21.99 35.31 28.89 27.89 19.11 19.15 18.96 30.73 30.70 18.95 33.45 31.60 30.41

33.20 35.72

42.28 31.56 29.82 37.33 28.97 54.61 39.06 28.29 26.87 27.01 39.52 41.86

28.28 28.24 22.61 21.46 23.91 22.41 17.99 25.21 33.26 28.72 19.49 18.64 37.12 31.04

Tabulation of advertising lineage.

26,228 13,818 27,122 17,077 32,094 32,936 21,499 20,655 22,338 13,412

10.

29,207 17,588 28,267 15,095 36,072 32,835 18,116 24,094 10,365 60,475

11.

22,107 15,977 39,082

9,644 23,449 18,930 46,520 25,140

8,015 38,795

12.

14,849 11,966 36,021 7,888 19,634 15,033 43,778 19,271 13,412 93,323

13.

10,049

14,745

8,562

5,575

12,376

2,175

7,531

8,650

15,530

23,680

37.61 29.97 31.36 21.34 33.34 31.78 34.73 26.74 22.88 28.09 28.20 20.80 37.53 28.94

14.

57,104 71,075 119,035 28,857 39,190 16,085 15,484 28,192 14,711 22,865

7. 22.53 35.62 16.22 31.91 14.16 27.14 31.62 33.84 30.89 39.04 15.76 20.87 38.72 37.16

23.25 35.25

32.18 37.17 35.99 36.66 35.96 37.37 18.17 29.64 15.82 29.87 27.84 36.83

15.

13,022 15,223

17,058 18,048 26,174 28,169 14,949 24,478 22,175 23,680

16.

10,755 16,850 15,573 10,259 19,635 24,572 15,057 19,445 24,493 37,335

Addition of dollars and cents, irregular items.

17.

139.10

12.65

10.57

50.05

1,275.48

260.73

7.81

78.13

2.50

111.82

29.53

54.53

15.36

147.62

5.27

18.

86.35

52.67

44.00

208.33

394.68

64.72

6.29

27.33

.62

27.65

7.33

16.29

4.59

111.52

59.68

19.

209.80 44.82 37.45

127.29

4,151.36

24.94

.72

71.97

12.09

.35

160.31 45.15

128.60 41.51 46.43

20.

45.40

34.20

28.66

135.68

945.21

2.72

118.61

.75

32.49

8.58

33.55

8.45

24.85

138.34

92.54

21.

86.35

52.67

42.57

208.33

878.52

111.56

6.46

44.77

.69

3.07

70.63

5.37

35.15

9.98

214.34

22.

1,955.05 531.03 442.85

2,148.74

30,149.39

112.64

509.74

27.02

153.07

1,512.34

1,002.90

282.51

66.66

146.43

641.51

Fundamental Processes and Short Methods for the Accountant 9

Practical applications. In the following problems will be found examples of business records requiring addition for the completion of the record.

PROBLEM 1

In this problem, cash register tapes provided the source of the entries on Form 1. As the sales were registered, the classification was imprinted on the tape. At the end of the day, the classified items appearing on the tape were accumulated on Form 1, and the totals transferred to Form 2. At the end of the week, Form 2 was added; at the end of the month, the weekly totals were accumulated to monthly totals. Thus, sales for the month were analyzed by departments or classes.

Add the columns on Form 1 (Saturday's sales), transfer the totals to Form 2, and find the total sales for the week.

Form 1

Candy

Cigars

Soda

Drugs

Own Remedies

Patent Medicines

Toilet Articles

.35 1.25 .80 .45 .75 .90

.10 .25 .15 .25 .50 .25

.10 .15 .20 .45 .10 .50

.45

1.64

.10

.75

1.50

1.75

.75 .45 .15

1.25 .90

1.65

1.25 .50

1.50 .89 .33

2.35

.35 1.15

.80 2.65

.75 1.85

Form 2

Day

Candy

Cigars

Soda

Drugs

Own Reme- dies

Patent Medi- cines

Toilet Articles

Totals

Mon.

Tues.

Wed.

Thurs.

Fri.

Sat.

12.65 8.50

11.25 9.65

10.35

19.15

16.10

8.75

4.25

5.55

3.95 6.80 4.50 2.55 3.75

27.63 33.98 15.20 7.65 12 84

4.18 2.47 1.75 2.85 3.68

9.85

12.20

2.55

4.86 5.49

5.00 3.65 10.45 4.63 3.85

Form 2 is self-proving — that is, the sum of the daily totals must equal the sum of the departmental or classification totals.

Fundamental Processes and Short Methods for the Accountant

PROBLEM 2

The "peg board" is used for accumulating numbers having to do with many kinds of information. The numbers are entered on narrow forms which are attached to the "peg board." The forms are held in place and cross extension as well as "footings" are thus permitted.

fj/.a

V-3C ¥?

In the following example, this arrangement is used to accumulate total departmental sales made by a salesman.

Salesman R. F.

Salesman R . F.

Total Sales

Dept

Date 4/2 Date 4/3 Date 4/4 Date 4/5 Date 4/6

128.57 645.39 362.45 472.31 45.97 273.14 928.63

587.23 321.69 847.86 123.45 671.17 372.45 436.49

347.58 123.63 219.23 547.81 359.34 135.67 569.81

237.51 563.85 149.27 462.38 326.49 857.62 318.48

637.82 495.71 826.45 718.26 534.58 149.17 529.32

(a) Find the total of each day's sales.

(b) Find the total sales for each department.

The answer in the lower right corner proves the work.

Subtraction. Subtraction is the process of finding the difference between two like numbers. The minuend is the number to be diminished, and the subtrahend is the number to be taken from the minuend.

Fundamental Processes and Short Methods for the Accountant 1 1

Addition and subtraction are closely related. Subtraction by adding is the method used by the expert cashier and by money changers. The "making change" method of subtraction consists in adding to the amount of the purchase enough to make the sum equal to the amount tendered in payment.

Example. Y buys groceries to the value of $1.34 and gives the cashier two one- dollar bills in payment. How much change should he receive?

Solution. The cashier in making change may return to Fa penny, a nickel, a dime, and a half-dollar, saying: "$1.34, 35, 40, 50, $2.00," which means $1.34 + .01 = $1.35; $1.35 + .05 = $1.40; $1.40 + .10 = $1.50; and $1.50 + .50 = $2.00. Other coins than those mentioned may be returned by the cashier, but it is customary to make change in the largest coins possible.

Exercise. As the cashier, make change, using the largest denominations pos- sible, assuming that each of the following purchases were made and two one-dollar bills were offered in payment.

1. $1.44 5. $1.64 9. $1.17 13. $1.43

2. 1.67 - 6. 1.32 10. 1.29 14. 1.38

3. 1.27 7. 1.82 11. 1.54 15. 1.49

4. 1.41 8. 1.11 12. 1.56 16. 1.05

Avoid errors. Many errors in subtraction are made in borrowing from the next higher order. When that order is reached, it is not uncom- mon to overlook the fact that borrowing has taken place. Errors of this kind can be avoided by changing subtraction to the process of addition; that is, by adding to the subtrahend the number required to make the subtrahend equal to the minuend.

Explanation. Instead of thinking, "7 from 16 is 9," think, "7 + 9 = 16." Write the 9. Add 1, the digit carried over, to the 8, making 9. 9 + 8 = 17. Write, 8, and add 1, Example.

the digit carried over, to 1, making 2. 2 + 0 = 2. Minuend 8276

Write 0. 3 + 5 = 8. Write 5. Answer: 5,089. Subtrahend 3187

Difference 5089

PROBLEMS

1. 9574 2. 7436 3. 6175 4. 8147 5. 6328 6. 5317

5886 3569 2897 4368 2549 3428

Difference between a given minuend and several subtrahends.

In instances similar to the following example, the final result can be found in one operation by the application of the foregoing method of subtraction.

Example. From a fund of $3,456, the following disbursements were made : $594, $375, and $286. What was the balance left in the fund?

Explanation. Write the problem as shown in the solution. Begin at the right, and add the units' column of subtrahends, (6 + 5 + 4), adding (and setting

1 2 Fundamental Processes and Short Methods for the Accountant

down) enough (in this instance, 1) to make the units' figure of

the sum the same as the units' figure of the minuend. Add the Solution.

tens' column of the subtrahends, including the carrying figure, $3,456

(1 + 8 + 7 + 9), adding (and setting down) enough (in this in- 594

stance, 0) to make the tens' figure of the sum equal the tens' 375

figure of the minuend. Add the hundreds' column of the 286

subtrahends, including the carrying figure, (2 + 2 + 3 + 5), $2201

adding (and setting down) enough (in this instance, 2) to make

the hundreds' figure of the minuend. To the carrying figure, 1, add enough

(in this case, 2) to make the thousands' figure of the minuend; set down 2.

PROBLEMS

1. $1,562 2. $2,756 28 3. $5,987 4. $4,875 5. $2,975

437 52.70 235 - 365 762

122 7.55 789 1,529 194

254 528.75 1,526 284 275

Balancing an account. In most cases, inspection will tell which side of the account is the greater in amount. Add the larger side, and put the same footing on the smaller side, leaving space for the balance; then add from the top downward, supplying the figures necessary to make the column total equal to the footing previously placed there. Example.

Debits

Credits

$ 1,956.18

$ 134.26

3,452.75

258.19

289.34

764.83

5,726.31

2,375.94

Balance,

7,891.36

$11,424.58

Explanation. The balance, $7,891.36, was found as follows: Inspection showed the debit side to be the larger in amount. It was therefore added, and the footing of the account, $11,424.58, was placed under both debit and credit columns. The first order of the credits — that is, the cents — adds to 22. Insert 6 to make 28. With 2, the digit carried over, the second order, the dimes, adds to 22. Insert 3 to make 25. The third order, the dollars, with the digit carried over, adds to 23. Insert 1 to make 24. The fourth order, the tens of dollars, with the digit carried over, adds to 23. Insert 9 to make 32. The fifth order, the hundreds of dollars, with the digit carried over, adds to 16. Insert 8 to make 24. The sixth order, the thousands of dollars, with the digit carried over, adds to 4. Insert 7 to make 11.

PROBLEMS

1. Debits Credits 2. Debits Credits 3. Debits Credits

$856.73 $298.56 $725.14 $1,356.17 $3,586.28 $ 591.18

345.96 264.39 239.51 691.35 192.75 2,751.26

298.85 6.15 64.28 256.38 384.72 185.35

142.31 ...■■ 75.19 265.54 _._...-

Complement method. The complement of a number is the dif- ference between that number and the unit of a next higher order. Thus,

Fundamental Processes and Short Methods for the Accountant

the complement of 6 is 4 ; the complement of 8 is 2 ; and the complement of 68 is 32.

If, in subtracting a number less than 10 from a given number, its com- plement is added, the result will be 10 too large. If two complements are added, the result will be 20 too large; and if three complements are added, the result will be 30 too large.

To find the sum of a column containing numbers to be subtracted, add the complements of the subtractive items, and from the sum of each order deduct as many tens as there are subtractive items in the order.

Example. A practical application of the complement method of subtraction is that of finding the net increase in a statistical record such as the following:

Sales

Increase

Dept.

This Mo.

Last Mo.

Decrease*

$ 427.95

$ 346.29

$ 81.66

515.86

457.75

58.11

395.57

385.86

9.71

402.75

416.87

14.12*

$1,742.13

$1,606.77

$135.36

Solution. The difference between the sales this month and the sales last month for each department is shown as an increase or a decrease. The difference between the total sales this month and the total sales last month is $135.36. To prove that the departmental increases and decrease are correct, add the third column, beginning at the top and adding downward, using the comple- ment each time on the last number. Thus, 8 and 8 are 16; write 6, and drop the 10, as one complement was added and the answer is 10 too large. 14 and 9 are 23; write 3 and carry 10, dropping one 10. 19 and 6 are 25; write 5 and carry 1, again dropping one 10. 14 and 9 are 23; write 13, dropping one 10 as before.

Example. Find the net increase of the following items:

Increase

Decrease*

15.60

51*

29.00

62*

124

83*

199

113.79*

132.46

122

580.01

Solution. In this problem there are four items showing decreases; therefore, each time a complement is added, the final result will be 10 too large, and in this case, the final result will be 40 too large, so 40 is deducted each time. Begin at the top and add downward: 9 (comp.), 15, 23 (comp. was 8), 30, 31, 37, 46, 51, subtract 40, write 1 and carry 1.

Now the next column. 7 (6 and 1), 12, 14, 23, 27, 29, 33, 35, 38, 41, 45, 54, 60, subtract 40, write 0 and carry 2. Next column, 7, 13, 20, 21, 30, 32, 36, 45, 46, 55, 62, 64, 68, 70, subtract 40, write 0 and carry 3.

Adding the tens: 4 (1 and 3 carried), 5, 11, 13, 15, 20, 22, 31, 40, 43, 46, 48, but subtract 20 as only two complements were used,

14 Fundamental Processes and Short Methods for the Accountant

write 8 and carry 2. The complement 10 may be added each time there is no item, making the answer 68, then subtract 40, leaving 28 as before. Remem- ber, subtract as many 10's as there are complements added.

Finally the hundreds' column. There are but five items in this column; therefore, with the 2 carried, proceed as follows: 3, 4, 13, 15, subtract 10 (only one complement was added) and write 5. Answer: 580.01.

. $58

9.01*

12*

PROBLEM S

tracted are marked (*) in Problems 1 and 2.

2. $122 65

3. $48.75 Gain

$20.25 Gain

175.50

31.25 Gain

4 . 50 Loss

89.88*

3.20 Loss

41.50 Gain

17.20

65.50 Gain

28.45 Gain

1.48

15.25 Loss

38.47 Gain

8.62*

16.38 Gain

12.34 Loss

36.95

26.65 Gain

49.82 Gain

Subtracting on an adding machine. If increase or decrease col- umns are being verified on an adding machine that does not have a direct subtraction device, add the complements of the numbers to be sub- tracted.

To subtract $219.48, set 780.52 on the keyboard and strike all nines to the left of the number; and to subtract $102.79, set 897.21 and strike all nines to the left of the number. Striking of the nines eliminates from the totalizers the number 1 that would otherwise be included in the answer.

Practical problems. In the following problems, both addition and subtraction have to be performed in order to complete the records.

PROBLEM 1

This problem illustrates a section of a twelve-month moving-average schedule used in cost accounting and other cumulative work. Assuming that twelve months covers a cycle of business changes due to seasonal variations, and so forth, the moving twelve months' total provides a fairly reliable amount for comparative purposes.

The earliest month's results are subtracted from the twelve months' total and the current month's results are added, making a current twelve-month accumulation. The record is self-proving.

Total

Total, 12/31/63. . Deduct Jan., 1963

DepL 1

$125,275.93

9,495.79

DepL 2

$56,472.29

4,907.63

Dept. 3 $4,207.23 368.80

DepL 4 $7,200.49 502.50

Add Jan., 1964. . . 12 mos. totals ....

8,805.67 8,933.07

4,480.25

358.79

588.79

Deduct Feb., 1963

4,093.19

293.67

496.68

Add Feb., 1964. . . 12 mos. totals. .

9,033.48 10,854.92

4,123.97

235 . 80

517.90

Deduct Mar., 1963

4,837.07

331.04

480.09

Add Mar., 1964. . . 12 mos. totals. . . .

8,588.37

4,001.18

334.17

521.72

Fundamental Processes and Short Methods for the Accountant

PROBLEM 2

From the following sales record, find the increase or decrease in sales by department.

Comparative Sales Record

Dept. No. February, 2nd Year

1 $ 7,134.95

2 6,225.19

3 7,934.97

4 6,354.76

5 3,695.15

6 9,767.98

7 8,567.39

8 5,607.18

9 11,365.39

10 14,572.86

Total

Increase or

February, 1st Year Decrease^

$ 6,834.79

5,764.87

8,375.16

5,986.35

3,756.89 9,475.18 8,467.57 4,865.84

10,785.65

13,764.16 .

PROBLEM 3

A daily business record may be prepared from cash register totals and other information. With the aid of the amounts given, complete the record for the day. Some of the sections contain items that are needed to complete other sections.

Cash Receipts Sales

Rec'd. on Acc't. . $234.56 Cash Sales $.

Other Receipts . . 59 . 32 Credit Sales ....

Cash Sales 497 . 85

152.35

Total Receipts ___

Cash on Hand Opening Balance . $250 . 75

Receipts

Total

Paid Out

Closing balance

Accounts Payable

Bal. for'd $315.20

Invoices Today . . 262 . 35

Total

Paid Today 136 . 57

Balance

Total Sales

Bank Account

Bal. for'd $2,872.63

Today's Dept

Total

Today's Cks .. . 175.32

Balance . . .

Cash Sales Summary

Total for'd $2,542.75

Today's Cash

Sales _

Total to for- ward

Cash Paid Out

For Stock $ 85.42

For Expenses. . . 19.56

Personal 27.50

Deposit 652 . 80

Total

Accounts Receivable

Bal. for'd $481.52

Credit Sales . . .

Total

Rec'd. on Acc't. Balance

Credit Sales Summary

Total for'd $638.47

Today's Credit

Sales

Total to forward

PROBLEM 4

In the following table of Gross Profits by Departments, add the Goods on Hand, March 1, 1st Year, to the Purchases for the Year, and from this sum subtract the Goods on Hand, March 1, 2nd Year. This gives the Cost of Goods Sold. The operation should be performed without transferring any of the

1 6 Fundamental Processes and Short Methods for the Accountant

figures. Use the complements of the numbers in the column Goods on Hand, March 1, 2nd Year.

The difference between the Cost of Goods Sold and the Sales will give the Profit or Loss.

To verify the work, add all the columns, and deal with the totals in the same way as with the figures for the departments. The difference between the totals of the Cost of Goods Sold column and the Sales column should equal the difference between the totals of the Profit and the Loss columns, showing the Net Profit of the ten departments for the year.

Gross Profits by Departments

Goods Goods

on Hand Purchases on Hand Cost of

March 1, for the March 1, Goods

Dept. 1st Year Year 2nd Year Sold Sales Profit Loss

1 $3,475.86 $ 9,846.37 $2,347.11 $12,678.92

2 1,357.10 6,725.40 1,475.86 ,..,. 6,188.90

3 3,276.84 10,326.85 3,827.84 .. 8,297.63

4 5,475.90 11,176.98 5,874.13 13,586.47

5 4,276.83 9,798.34 4,207.16 10,508.92

6 3,785.47 8,376.41 3,648.10 8,756.13

7 2,986.17 9,386.57 3,014.74 8,964.85

8 3,275.83 8,724.18 2,817.56 9,575.34

9 2,976.95 9,543.34 2,734.15 10,789.18

10 3,532.25 10,217.60 3,375.89 12,756.84

Footings -

Multiplication. Multiplication is a short process of addition; that is, a number is to be taken as an addend a given number of times.

How many bushels of grain are in three bins each containing 146 bu.?

idition 146

Multiplicatu 146

146

438

Multiplication involves three numbers, the multiplicand (the number to be repeated, 146) ; the multiplier (the number showing the number of repetitions, 3); and the product (the number showing the result, 438).

The multiplicand and the product are always like numbers. 146 bushels multiplied by 3 equals 438 bushels.

PROBLEMS

1. What is the cost of 640 acres of land at $42.50 an acre?

2. How many minutes are there in an ordinary year?

3. A barrel of flour contains 196 pounds. What is the weight of flour pro- duced in one day by a mill that produces 375 barrels?

4. Sound travels about 1,120 feet in a second. How far will it travel in 15 seconds?

5. How many peaches are in 12 crates, if there are 84 peaches in each crate?

Fundamental Processes and Short Methods for the Accountant

Accuracy and speed in multiplication depend largely upon a thorough mastery of the multiplication tables. Tables previously learned should be reviewed. Continue with frequent drills on combinations up to 25 X 25. The following table of multiples from 12 X 12 to 25 X 25 is given for reference and drill. Tables of multiples prepared in this manner facilitate the work of pay roll extension, inventory extension, billing, and so forth.

TABLE OF MULTIPLES

144

156

168

180

192

204

216

228

240

252

264

276

288

300

156

169

182

195

208

221

234

247

260

273

286

299

312

325

168

182

196

210

224

238

252

266

280

294

308

322

336

350

180

195

210

225

240

255

270

285

300

315

330

345

360

375

192

208

224

240

256

272

288

304

320

336

352

368

384

400

204

221

238

255

272

289

306

323

340

357

374

391

408

425

216

234

252

270

288

306

324

342

360

378

396

414

432

450

228

247

266

285

304

323

342

361

380

399

418

437

456

475

240

260

280.

300

320

340

360

380

400

420

440

460

480

500

252

273

294

315

336

357

378

399

420

441

462

483

504

525

264

286

308

330

352

374

396

418

440

462

484

506

528

550

276

299

322

345

368

391

414

437

460

483

506

529

552

575

288

312

336

360

384

408

432

456

480

504

528

552

576

600

300

325

350

375

400

425

450

475

500

525

550

575

600

625

Contractions in multiplication. Contractions in multiplication may often be made by observing the peculiarities of the multiplier and the multiplicand and calling into use factors, multiples, complements, supple- ments, reciprocals, aliquots, and the like.

To multiply by factors of the multiplier. The ordinary method and the shorter method of multiplying by factors are shown in the fol- lowing example. Observe that in the ordinary method there are two multiplications and an addition, while in the shorter method there are only two multiplications.

Example. Multiply 567 by 27.

Solution.

Ordinary Method

567

3969

1134

15309

PROBLEMS

Shorter Method

567 27 = 9 X 3

5103

15309

Multiply:

1. 4,584 by 64.

2. 8,359 by 54.

3. 1,459 by 35.

4. 2,684 by 27.

5. 8,756 by 42.

6. 6,123 by 45.

18 Fundamental Processes and Short Methods for the Accountant

To multiply when a part of the multiplier is a factor or mul- tiple of another part.

Example. Multiply 34,768 by 488.

Solution.

* 34768

488

278144 product by 8 16688640 product of 60 times product by 8 16966784

PROBLEMS

Multiply:

1. 45,692 by 549. 3. 21,347 by 497. 5. 84,123 by 248.

2. 49,871 by 648. 4. 33,546 by 355. 6. 13,456 by 153.

To multiply a number of two figures by 11. Observation of the ordinary method shows that, in the answer, the sum of the two digits is written between the two digits.

Ordinary Method Shorter Method 54 54

11 _U

54 594

594

When the sum of the two digits is 10 or more, 1 must be carried to the digit at the left; for example, 64 X 11 = 704, and 93 X 11 = 1,023.

To multiply any number by 11. Observation of the ordinary method shows that, in the answer, the units' digit of the multiplicand is the units' digit of the product; that the tens' digit of the product is the sum of the units' digit and the tens' digit of the multiplicand; that the hundreds' digit of the product is the sum of the tens' digit and the hun- dreds' digit of the multiplicand; and so on. When the sum of two digits is 10 or more, 1 must be carried.

Ordinary Method Shorter Method 8937 8937

11 11

8937 98307

Multiplying by 25. Annex two ciphers to the multiplicand, and divide by 4.

Fundamental Processes and Short Methods for the Accountant 19

Example. Multiply 7,562 by 25. Solution.

4)756200

189050

PROBLEMS

Multiply each of the following by 25:

1. 3,874. 2. 3,948. 3. 7,981. 4. 5,426.

Multiplying by 15. Annex a cipher to the multiplicand, and increase the result by one-half of the multiplicand.

Example. Multiply 8,435 by 15.

Solution.

84350 42175

126525

PROBLEMS

Multiply each of the following by 15:

1. 7,432. 2. 8,397. 3. 3,926. 4. 9,536.

Multiplying numbers ending with ciphers. Multiply the sig- nificant figures in each number, and to the product annex as many ciphers as there are final ciphers in both the multiplier and the multiplicand.

Example. Multiply 756,000 by 4,200.

Solution.

756 42

31752 Annex five ciphers. Answer: 3,175,200,000.

PROBLEMS

Multiply:

1. 325,000 by 2,300. 3. 24,100 by 4,200.

2. 370 by 480. 4. 8,300 by 2,100.

Multiplication by numbers near 100, as 98, 97, 96, and so forth, and by numbers near 1,000, as 997, 996, and so forth. This method is of value in finding the net proceeds of some amount less 2%, 3%, and so forth, and also in many other situations.

20 Fundamental Processes and Short Methods for the Accountant

Example. Multiply 3,247 by 97.

Solution. Multiply the number by 100, and subtract 3 times the number.

324,700 = 3,247 X 100

9,741 = 3,247 X 3 314,959 = 3,247 X 97

Multiplication by a number near 1,000 is accomplished in the same manner by multiplying by 1,000 instead of by 100.

PROBLEMS

Multiply:

1. 2,459 by 98. 2. 7,318 by 97. 3. 5,438 by 96. 4. 8,752 by 95.

Multiplication of two numbers each near 100, 1,000, and so forth. Products of numbers in this class may be calculated mentally.

Example. Multiply 96 by 98.

Explanation. Step 1. Multiply the complements of the two numbers, and if the product occupies units' place only, prefix a cipher. Result, 08. Solution.

Step 2. Subtract the complement of one number Complement

from the other number, and write the result at the left 96 4

of the result in Step 1. The complement of either 98 2

number subtracted from the other number leaves the 9408 same remainder; as, 96 — 2 or 98 — 4 each equals 94. Answer: 9,408.

Example. Multiply 92 by 88.

Solution.

Complement 92 8

88 12

8096

Explanation. The product of the complements is 96, the last two figures of the answer. 88 — 8 or 92 — 12 = 80, the first two figures of the answer. Answer: 8,096.

Example. Multiply 996 by 988. Solution.

Complement 996 4

988 12 ■

984,048

Explanation. When numbers near 1,000 are multiplied, ciphers are prefixed to the product of the complements, so that the product occupies three places.

Fundamental Processes and Short Methods for the Accountant 21

PROBLEMS

Multiply:

1. 97 by 96. 2. 88 by 98. 3. 995 by 992. 4. 997 by 994.

Multiplying by numbers a little larger than 100, as 101, 102, and so forth. Annex two ciphers to the multiplicand, and to this add the product of the multiplicand and the units' figure of the multiplier. Annex three ciphers for multipliers over 1,000.

Example. Multiply 3,475 by 104.

Solution.

347500

13900 (4 X 3,475) 361400

PROBLEMS

Multiply:

1. 2,875 by 102. 2. 3,496 by 105. 3. 2,972 by 1,004. 4. 4,568 by 1,006.

Multiplication of two numbers each a little more than 100. To

the sum of the numbers (omitting one digit in the hundreds' column), annex two ciphers, and add the product of the supplements (excess over 100).

Example. Multiply 112 by 113.

Solution.

112 113 12500 (sum of numbers, with one digit in the hundreds' column omitted)

156 (product of supplements, 12 X 13) 12656

Explanation. In instances similar to the foregoing, a knowledge of the multi-* plication tables to 20 X 20 makes mental results possible, and is invaluable in inventory and other extensions.

PROBLEMS

Multiply:

1. 114 by 112. 2. 106 by 108. 3. 116 by 111. 4. 118 by 115.

Cross multiplication. When the multiplicand and the multiplier are each numbers of two figures, the work may easily be kept in mind and the partial products added without being written down.

22 Fundamental Processes and Short Methods for the Accountant

Example. Multiply 47 by 38.

Solution Graphic Solution

47 4t tf 47

38 3$ £ ^8

1786 ~~

Explanation. 8 X 7 = 56. Write 6, carry 5. (8 X 4) + (3 X 7) + 5 = 58. Write 8, carry 5. (3 X 4) + 5 = 17. Write 17. Answer: 1,786.

PROBLEMS

Multiply:

1. 53 by 29. 2. 48 by 57. 3. 74 by 32. 4. 65 by 28.

To cross-multiply a number of three digits by a number of two digits. A three-digit number may be multiplied by a two-digit number in a manner similar to that of multiplying a two-digit number by a two- digit number.

Example. Multiply 346 by 28.

Solution.

346

28 9688

Explanation. 8 X 6 = 48. Write 8, carry 4. 4 (carried) + (8 X 4) + (6X2)= 48. Write 8, carry 4. 4 (carried) + (8 X 3) + (4 X 2) = 36. Write 6, carry 3. 3 (carried) + (2 X 3) = 9. Write 9. Answer: 9,688. A graphic presentation of the steps required appears as follows:

34fc

2 3 4 5 6 3& $46

S S

PROBLEMS

324 X 28

428 X 34 7. 289 X 85

10.

693 X 42

543 X 42

516 X 26 8. 356 X 48

11.

384 X 56

658 X 56

513 X 76 9. 785 X 34

12.

473 X 65

To cross-multiply a number of three digits by another number of three digits. Comparison of the graphic presentation with that above shows that the first three steps are the same, the next three are new, and the final three are the same.

Example. Multiply 428 by 356.

Solution Graphic Solution

. ~ ~ l1 ?_3 4£5 78 9

428 42* 4\S 4n£^ %% 8 428

356 35* 3A jHX 3*S 6 &56

152,368

Fundamental Processes and Short Methods for the Accountant 23

Explanation. 6 X 8 = 48. Write 8, carry 4. 4 (carried) + (6 X 2) + (8 X 5) = 56. Write 6, carry 5. 5 (carried) + (6 X 4) + (8 X 3) + (2 X 5) = 63. Write 3, carry 6. 6 (carried) + (5 X 4) + (2 X 3) = 32. Write 2, carry 3. 3 (carried) + (3 X 4) = 15. Write 15. Answer: 152,368.

PROBLEMS

1. 124 X 251 6. 832 X 425 11. 436 X 579

2. 262 X 158 7. 639 X 256 12. 832 X 656

3. 328 X 245 8. 819 X 325 13. 295 X 638

4. 638 X 256 9. 677 X 283 14. 767 X 842

5. 784 X 364 10. 518 X 824 15. 698 X 476

Preparation of a table of multiples of a number. It is not uncommon to have to use the same number many times in making calcula- tions, especially in cost accounting. A saving of time and increased accu- racy in the work are achieved if a table of multiples of the number is con- structed. Suppose that you have to perform a number of multiplications in which 326,834 is one of the factors. A table of multiples may be con- structed with an adding machine by locking the repeat key. Sub-total after each pull of the handle. The sub-totals should check with the prod- uct column shown below. If the table is prepared by repeated additions, and not with an adding machine, the 10th product should be computed, as it will verify all, unless there are compensating errors in the work.

TABLE OF MULTIPLES

Multiplier Product

1 326,834

2 (326,834 + 326,834) 653,668

3 (653,668 + 326,834) 980,502

4 (980,502 + 326,834) 1,307,336

5 (1,307,336 + 326,834) 1,634,170

6 (1,634,170 +326,834) 1,961,004

7 (1,961,004 + 326,834) 2,287,838

8 (2,287,838 + 326,834) 2,614,672

9 (2,614,672 + 326,834) 2,941,506

Verification

10 (2,941,506 + 326,834) 3,268,340

Example. Multiply 326,834 by 5,249.

Solution.

2941506 = 9 times 326,834

1307336 = 4 times 326,834

653668 = 2 times 326,834

1634170 = 5 times 326,834

1715551666 = product

If the table is prepared without the use of an adding machine, proceed as follows.

1. Write 326,834 near the bottom of a slip of paper or a card.

2. Start the table by writing 326,834. Place the slip or card just above this number, thus:

24 Fundamental Processes and Short Methods for the Accountant

326,834

1. 326,834

3. .__

3. Add the two numbers, placing the sum, 653,668, on line 2. This is two times the number.

4. Move the slip or card down one line and add again, placing the sum, 980,502, on line 3, forming three times the number.

5. Continue moving the slip or card down one line each time and adding.

6. When 9 times the number is obtained, check the accuracy of the work by repeating the process once more. The result should be ten times the number.

PROBLEMS

Set up a table of multiples of 245,386, and multiply 245,386 by the following numbers:

1. 2,465 2. 3,542 3. 2,498 4. 5,347 5. 6,173

Division. Division is the process of finding how many times one number is contained in another number. The dividend is the number to be divided, the divisor is the number by which we divide, and the quotient is the number showing how many times the dividend contains the divisor.

The remainder is a number less than the divisor, and results when the dividend does not contain the divisor exactly. It is an undivided portion of the dividend.

Short division is the method used when the products of the divisor and the digits of the quotient are omitted.

Example. Divide 3,476 by 2.

Solution.

2)3476 1738

Long division is the method used when the work is written in full.

Example. Divide 5,839 by 24.

Solution.

24)5839(243 48 103 96 79 72 7

To divide by 25, 50, or 125. The work of division can be lessened by making the operation one of multiplication.

Fundamental Processes and Short Methods for the Accountant 25

Example. Divide 1,400 by 25.

Solution. 14X4 = 56.

Explanation. Divide 1,400 by 100 by dropping the zeros. But, 100 is 4 times the actual divisor; therefore, the quotient 14 is -J- of the actual quotient, so 14 X 4 or 56 is the actual quotient.

In a similar manner, 1,400 divided by 50 is 28; and 14,000 divided by 125 is 112. (Note: Further reference to this method is given under the subject of division by aliquot parts of 100.)

Abbreviated division. Instead of writing the product and then sub- tracting, the product of each digit of the divisor is subtracted mentally, using the " making change" method, and only the remainder is written.

3285

234)768756 667 1995 1236 66

Use of tables in division. If a number of divisions are to be made with the same divisor, it is advantageous to set up a table of multiples of the divisor.

Example. Assume that 328 is to be used a number of times as a divisor, and that one of the dividends is 587,954. A table of multiples could be set up thus :

TABLE OF MULTIPLES

Multiplier Product

1 328

2 656

3 984

4 1,312

5 1,640

6 1,968

7 2,296

8 2,624

9 2,952

Explanation. Inspection shows the first digit in the quotient to be 1. The sec- ond partial dividend is 2,599. The table of

multiples shows the largest product contained Solution.

therein to be 2,296, opposite 7. The third 328)587954(1792T%9T partial dividend is 3,035, and the table of 328

multiples shows the largest product contained 9500

therein to be 2,952, opposite 9. The fourth 2296

partial dividend is 834, and the largest prod- ~oH5k

uct contained therein is 656, opposite 2. The 30

remainder is 178. The fraction £|f may be 2952

reduced to T8^, or it may be changed to a 834 H§ _ 89

decimal. 656 328 164

178

26 Fundamental Processes and Short Methods for the Accountant

Division in this manner is rapid, as no time is lost through selection of a quotient so large that when the product is found it exceeds the dividend, necessitating another trial.

PROBLEMS

Divide the following numbers by 144 after setting up a table of multiples of 144:

1. 374,825. 2. 628,256. 3. 496,287.

Reciprocals in division. The reciprocal of any number is found by dividing 1 by the number. The reciprocal of 5 is 1 -s- 5, or .2, and the reciprocal of 25 is 1 -r- 25, or .04.

The quotient in a division may be found by multiplying the dividend by the reciprocal of the divisor. Hence, in instances in which it is neces- sary to find what per cent each item is of the total of the items, the use of the reciprocal of the divisor will save time and provide a check on these computations.

To find what per cent each item is of the total of the items:

(a) Divide 1 by the total of the items to obtain the reciprocal of the total.

(6) Using the result obtained in (a) as a fixed multiplier, multiply each of the individual items, and the respective results obtained will be the per cents which the individual items are of the total sum.

Example. Find the per cent that each department's monthly expense is of the total monthly expense.

Department Expense

A $ 600.00

B 500.00

C 1,200.00

D 700.00

E 1,000.00

Total $4,000.00

Solution. Divide 1 by 4,000 to obtain the reciprocal, .00025. Multiply the expense of each department by this reciprocal, and the product will be the per cent that the department's expense is of the total expense.

Department Expense Reciprocal Per Cent

A $ 600.00 X .00025 = 15 %

B 500.00 X .00025 = 12j-%

C 1,200.00 X .00025 = 30%

D 700.00 X .00025 = 17£%

E 1,000.00 X .00025 = 25 %

Total $4,000.00 100 %

The foregoing method of calculating the rate per cent has a great many applications in an accountant's work. Another illustration is given —

Fundamental Processes and Short Methods for the Accountant 27

that of calculating the per cent that each item in a profit and loss state- ment is of net sales.

Quality Meat Market

Profit and Loss Statement for the Year

Detail Amount Per Cent

Net sales $20,000.00 100.00

Cost of merchandise sold 15,712.00 78.56

Gross profit $ 4,288.00 21.44

Expenses

Salaries and wages $2 , 266 .00 1 1 . 33

Advertising 22 .00 .11

Wrappings 172 .00 .86

Refrigeration 210.00 1 .05

Heat, light, and power 54 .00 .27

Telephone 54.00 .27

Rent 338.00 1.69

Interest 146.00 .73

Depreciation of store equipment. . . . 152.00 .76

Repairs to store equipment 44 . 00 .22

Insurance 10.00 .05

Taxes 42.00 .21

Losses from bad debts 38.00 .19

Other expenses 284.00 1.42

Total expenses 3,832.00 19.16

Net profit $ 456.00 2.28

Explanation. The foregoing is a simple statement, and the per cents can be determined mentally if each item is divided by the amount of net sales. For the purpose of illustration, however, find the reciprocal of $20,000.00, which is .00005 (1 -5- 20,000); then multiply each item by this reciprocal, and the results will be as shown in the per cent column.

PROBLEMS

1. The floor space occupied by Z Manufacturing Company was as follows:

Service Department X 600 sq. ft.

Service Department Y 1 , 100 sq. ft.

Service Department Z 550 sq. ft.

Producing Department A 2 , 000 sq. ft.

Producing Department B 1 , 568 sq. ft.

Producing Department C 2,234 sq. ft.

Sales Department 600 sq. ft.

Administrative Offices 550 sq. ft.

9,202 sq. ft.

The Building and Maintenance Expense account shows a total of $2,982.50. What amount of this expense should be distributed to each of the departments?

2. In the following tabulation, find the per cent that each department's floor space is of the total floor space:

28 Fundamental Processes and Short Methods for the Accountant

Sq. Ft. Per Cent

Floor Space of Total

Dept. 1 2,456

Dept. 2 1,014

Dept. 3 875

Dept. 4., 1,252

Dept. 5 748

6,345 100.00

3. Calculate the per cent that each item is of net sales.

The Food Mart

Profit and Loss Statement

Net Sales $35,600 100.00%

Cost of Merchandise Sold 27,969

Gross Profit $ 7,631

Expenses

Salaries and Wages $4,080

Advertising 28

Wrappings 266

Refrigeration 308

Heat, Light, and Power 106

Telephone 81

Rent 416

Interest 203 ...

Depreciation of Store Equipment 147

Repairs to Store Equipment 45

Insurance 21

Taxes 39

Losses from Bad Debts 119

Other Expenses 490

Total Expenses 6,349

Net Profit $ 1 ,282

CHECKING COMPUTATIONS

Methods. Addition may be checked by adding the second time, adding from the bottom to the top if the first addition was from the top to the bottom. This is preferable to performing the work in the same way the second time, as a mistake once made is likely to be repeated.

Subtraction may be checked by adding the subtrahend and the remainder. The sum should equal the minuend.

Multiplication may be checked by interchanging the multiplier and the multiplicand and multiplying again.

Division may be checked by multiplying the divisor and the quotient, adding to this product any remainder. The answer should equal the dividend.

Rough check. Rough check is an approximate check and is often used to locate large errors. It is also used in determining approximate

Fundamental Processes and Short Methods for the Accountant

results. It is especially useful in checking misplacement of the decimal point in multiplication and division of decimal fractions. A rough check of addition may be made as follows:

Example 54,892 36,071 53,784 21,342 76,854

242,943

Check 55 36 54 21 _77 243

If the required result is thousands, disregard the three columns at the right, except to increase the fourth-column sum by one if the digit in the third column is 5 or more. The check shows the answer to be approxi- mately 243,000.

Absolute check. There is no such thing as an absolute check, because there are always possibilities of offsetting errors, but the use of several methods of checking computations makes the probability of error so slight that one may rely on the result as correct.

Check numbers obtained by casting out the nines. A simple and easily remembered check is that of casting out the nines. Add the digits of the number, divide the sum by nine, and use the remainder, which is called "the excess," as the check number. In the number 4,875, the sum of the digits is 24, and 24 divided by 9 equals 2 with an excess of 6.

Verification of addition.

Explanation. The sum of the digits of 8,342 is 17 (8 + 9 and set down 8. If a number contains a 9, skip it in adding the digits; thus, in 8,967, 8 + 6 + 7 equals 21. Cast out the nines and set down the excess, 3. Find the check number of each line in the same way. Add the check numbers, and cast the nines out of their sum. Find the check number of the sum of the column being verified. The final check number in each case is 5.

+ 4 + 2).

Cast out

Example

8342

8967

8378

9276

8431

43394—5

32—5

PROBLEMS

Add, and verify by casting out the nines:

2487

7452

4501

1231

3156

8129

2765

4567

2982

5758

4567

1085

4756

2253

8256

3426

8928

7685

2435

7531

30 Fundamental Processes and Short Methods for the Accountant

Verification of subtraction. Example.

7856

2138

5718

Explanation. 7,856 checks 8, and 2,138 checks 5. 8 — 5 = 3, and 5,718 checks 3.

PROBLEMS

Subtract, and verify by casting out the

nines:

1. 2.

7496 7428 2831 1956

4751 3286

Verification of multiplication.

Example.

482 376

181232—8

35—8

8237 5129

Explanation. 482 checks 5, and 376 checks 7. 7X5 = 35. 35 checks 8, and the product, 181,232, also checks 8.

PROBLEMS

Multiply, and verify by casting out the nines:

1. 2. 3. 4.

456 412 832 765 287 654 254 414

Verification of division. Division may be verified by multipli- cation; that is, the product of the quotient and the divisor should equal the dividend. Apply the same principle in verifying with check numbers.

Example.

Explanation. 76,492 checks

114

1. 13 checks 4. 5,884

104

checks 7. 4 X 7 = 28,

109

and 28 checks 1, which

104

is also the check number

~1>2

of the dividend.

Fundamental Processes and Short Methods for the Accountant 31

PROBLEMS

Divide, and verify by casting out the nines:

1. 11,550 by 42. 2. 60,882 by 73. 3. 11,049 by 127. 4. 9,854 by 26.

Verification of division where there is a remainder. The check number of the remainder added to the product of the check number of the quotient and the check number of the divisor should equal the check number of the dividend.

Example. Explanation. Step 1: The re-

32)75892(2371 mainder, 20, checks 2. The

64 quotient, 2,371, checks 4.

TTo The divisor, 32, checks 5.

9g 2 + (4 X 5) = 22, and 22

— checks 4.

lf Step 2: The dividend,

XI 75,892, checks 4.

52 Step 1 and Step 2 should

<& produce the same check number.

PROBLEMS

Divide, and verify by casting out the nines:

1. 34,765 by 52. 2. 29,878 by 87. 3. 95,763 by 26. 4. 8,476 by 41

Check numbers obtained by casting out the elevens. Because casting out nines does not reveal errors in computations if two digits have been transposed, some persons prefer to use eleven as a check number.

Begin with the left-hand digit of the first number, and subtract it from the digit to its immediate right. If the digit to the right is smaller, add eleven before subtracting. Using the remainder as a new digit, subtract it from the third digit from the left, first adding eleven if necessary. Use this remainder as a new digit, and subtract it from the fourth digit from the left, first adding eleven if necessary. Continue in this manner until all the digits in the number have been used. The final remainder is the check number of the number.

Another method of checking results by means of the number eleven is to use alternate digits. From the sum of the first, third, fifth, etc., digits (beginning at units' place) subtract the sum of the second, fourth, sixth, etc., digits. If the subtraction cannot be performed, eleven is first added to the sum of the odd digits, and the sum of the even digits is subtracted, the remainder being the check number.

Fundamental Processes and Short Methods for the Accountant

Verification of addition.

Explanation. Begin at the left with the number 4,324. 4 from 14 (3 -f 11) = 10. 10 from 13 (2 + 11) = 3. 3 from 4=1, the check number of 4,324.

Take the second number, 8,689. 8 from 17 (6 + 11) = 9. 9 from 19 (8 + 11) = 10. 10 from 20 (9 + 11) = 10, the check number of 8,689.

Check all the numbers in the same manner. Add the check numbers. The sum of the check

numbers checks 1, and the sum of the numbers 31791 1 23 1

checks 1.

Example.

4324

8689

6327

8964

3487

31791—1

3789 5462 9581 3998 5314

Verification of subtraction.

Example.

7453

1289

PROBLEMS

; out the elevens:

2456

9755

1279

8256

2075

3851

2754

8632

9287

6311

6164

8307 7165 2693 2198 5183

Explanation. 7,453 checks 6. 1,289 checks 2. 6 - 2 = 4 and 6,164 checks 4.

PROBLEMS

Subtract, and verify by casting out the elevens:

8795

3465

7985

3079

1560

2134

5698

1002

Verification of multiplication,

Example.

584 256

149504 3

Explanation. 584 checks 1. 256 checks 3. 3X1 = 3, and 149,504 checks 3.

Fundamental Processes and Short Methods for the Accountant 33

PROBLEMS

Multiply, and verify by casting out the elevens:

1. 2. 3. 4.

346 4289 7437 287

275 324 2856 36

Verification of division.

Example 1 Example 2

24)89784(3741 31)75893(2448

72 62

177 138

168 124

98 149

96 124

"24 "253

24 248

Explanation 1. 89,784 checks 2. 24 checks 2. 3,741 checks 1. 2X1 = 2, the check number of the dividend.

Explanation 2. 75,893 checks 4. 31 checks 9. 2,448 checks 6. The remainder checks 5. 5 + (9 X 6) = 59. 59 checks 4, the same check number as that of the dividend.

PROBLEMS

Divide, and verify by casting out the elevens:

1. 80,925 by 83. 2. 124,392 by 142. 3. 25,874 by 49. 4. 28,769 by 135.

Check number thirteen. If thirteen is used as a check number, transpositions and shiftings of figures are readily detected. However, in checking by 13, it is necessary actually to divide by 13.

TABLE OF MULTIPLES

1 13 6 78

2 26 7 91

3 39 8 104

4 52 9 117

5 65 10 130

All the dividing is done mentally.

Example. Cast out 13 from 247,563.

Explanation. Begin with the two left-hand digits. 24 checks 11. 11, with the next digit, 7, is 117, and 117 checks 0. Use the next two digits. 56 checks 4. 4 with the next digit is 43, and 43 checks 4.

34 Fundamental Processes and Short Methods for the Accountant

The verification of addition, subtraction, multiplication, and division is performed in the same manner as with 9 and 11. The difference is in the method of arriving at the check number, as has been outlined.

PROBLEMS

1. Add, and verify by check number 13 :

24875 32986 79840 80475 13048 93476

2. Subtract, and verify by check number 13:

84756 21348

3. Multiply, and verify by check number 13:

4875 259

4. Divide, and verify by check number 13:

975,648 348

Factors, Multiples, and Common Fractions

Factors. The factors of a number are the integers whose product is the number. Thus, the factors of 6 are 2 and 3, and the factors of 18 are 3 and 6, or 2 and 9. A prime factor is a prime number, that is, a number not exactly divisible by any number except itself and 1.

Factoring is the process of separating a number into its factors.

Example.

Solution.

What are the prime factors of 315?

3)315 3)105 5) 35

The prime factors of 315 are, therefore, 3X3X5X7.

Example.

Solution.

What are the factors of 315?

9)315

7) 35

The factors of 315 are, therefore, 9X7X5.

Factoring is important for its assistance in the solution of problems in fractions, practical measurements, percentage, and all problems in which cancellation is used. One use of factors was given on page 17, "to multiply by factors of the multiplier," and another on page 18, "to multiply when a part of the multiplier is a factor or multiple of another part."

36 Factors, Multiples, and Common Fractions

Tests of divisibility. To be able to factor a number quickly, one must become thoroughly familiar with the tests of divisibility. A number is divisible by:

1. Two, if it is an even number or if it ends in zero.

2. Three, if the sum of its digits is divisible by 3. Thus, 41754 is divisible by 3 because the sum of the digits is 21, and 21 is divisible by 3.

3. Four, if the two right-hand figures are zeros, or if they express a number divisible by 4. Thus, 13724 is divisible by 4 because 24 is divis- ible by 4.

4. Five, if the units' figure is either a zero or a 5.

5. Six, if it is an even number the sum of whose digits is divisible by 3. Thus, 846, 918, and 54252 are divisible by 6.

6. Eight, if the three right-hand digits are zeros, or if they express a number divisible by 8. Thus, 2000 and 5624 are divisible by 8.

7. Nine, if the sum of its digits is divisible by 9.

8. Ten, if the right-hand figure is zero.

(There is no simple method of testing divisibility by 7.) Greatest common divisor. A common divisor of two or more num- bers is a number that evenly divides each of them. Thus, a common divisor of 16 and 24 is 4.

The greatest common divisor of two or more numbers is the greatest number that will evenly divide each of them. It is the product of all their common factors.

Example. Find the greatest common divisor of 36, 63, and 54.

Solution.

3)36 63 54

3)12 21 18

4 7 6

Since 4, 7, and 6 have no common factors, the G. C. D. is 3 X 3 = 9.

A practical application of the principles involved in finding the G. C. D. is in reducing common fractions to their lowest terms.

PROBLEMS

Find the G. C. D. of the following:

1. 64, 160, 320, 640 3. 32, 48, 128

2. 36, 54, 90 4. 81, 729, 2187

5. X, Y, and Z own land on a new street. X has 600 feet frontage, Y has 720 feet, and Z has 900 feet. If they wish to cut this land into lots of equal width, how wide will the lots be, and how many will each have?

Factors, Multiples, and Common Fractions 37

6. If you have three coils of steel cable measuring, respectively, 2205, 2940, and 4704 feet, and wish to cut the whole quantity into pieces of the greatest equal length possible without waste or splices, what will be the length of each piece? How many lengths will be cut from each coil?

Least common multiple. A common multiple of two or more num- bers is a number that is evenly divisible by each of them. Thus, 24 is a common multiple of 3 and 8.

The least common multiple of two or more numbers is the least number that is evenly divisible by each of them. Thus, 12 is the L. C. M. of 4 and 6.

Example. What is the L. C. M. of 12, 28, 30, 42, and 64?

Solution.

2)12 28 30 42 64

2) 6

15 21

3) 3

15 21

7) 1

5 7

1 1 5 1 16 2X2X3X7X5X16 = 6,720

Explanation. Notice that any number not divisible by the factor is brought down, and the process is repeated as long as at least two of the numbers have a common factor. Finally, the L. C. M. is the product of the factors and the numbers having no common factor.

PROBLEMS

Find the L. C. M. of the following:

1. 6, 18, 30, 42 3. 45, 63, 72, 99

2. 16, 24, 64, 96 4. 14, 35, 42, 28

Cancellation. Certain computations involving division can be shortened by removing or cancelling equal factors from both dividend and divisor.

Example. If 32 units of product sell for $57.60, what will 18 units of the same product sell for at the same rate?

Solution.

3.60 9 UM

nxuM _ 32 40

Z0 &

38 Factors, Multiples, and Common Fractions

PROBLEMS

Using cancellation, divide:

1. 27 X 48 X 96 X 38 2. 8 X 12 X 15 X 6

19X16X9X2 5X4X3X18

3. If 15 tons of coal cost $258.00, how much will 25 tons cost at the same rate?

4. A ship's provisions will last 36 men for 216 days. How long will they last 124 men?

COMMON FRACTIONS

Terms explained. A unit is a single quantity by which another quantity of the same kind is measured : 1 foot is the unit of 5 feet; 1 barrel is the unit of 10 barrels ; 1 acre is the unit of 40 acres, and so forth.

These integral units are often divided into equal parts known as frac- tional units, as \ ft., \ bbl., \ A., and so forth.

k fraction is an expression for one or more of the equal parts of a unit, as \ ft., f ft., f bbl., f A., and so forth.

The number above the line in the expression of a fraction is called the numerator; the number below the line is called the denominator.

The denominator indicates the number (and hence the size) of parts into which the unit is divided.

The numerator indicates the number of these parts taken.

A proper fraction expresses less than a unit, or its numerator is less than its denominator; as, f, f, f, and so forth.

An improper fraction is a fraction whose numerator is equal to or greater than its denominator; as, ■§, f, f, and so forth.

A mixed number is a number expressed by a whole number and a frac- tion; as, 2^, 3|-, 16f, and so forth.

Reduction of fractions. Reduction is the process of changing the numerator and the denominator of a fraction without changing the value of the fraction.

A fraction is reduced to higher terms when the numerator and the denominator are expressed in larger numbers.

A fraction is reduced to lower terms when the numerator and the denom- inator are expressed in smaller numbers, and it is reduced to its lowest terms when there is no common divisor of its numerator and denominator.

Principle. Multiplying or dividing both numerator and denominator of a fraction by the same number does not change the value of the fraction. Thus, Jf may be reduced to the equivalent fraction £ by dividing both terms by 4. The fraction £f has been reduced to lower terms. Again, H may be reduced to the equivalent fraction § by dividing both terms by 8. Here the fraction |f has been reduced to lowest terms, since 2 and 3 do not have a common divisor.

Factors, Multiples, and Common Fractions 39

Conversely, f may be changed to an equivalent fraction whose denom- inator is 24 by multiplying both terms by 8 (obtained by dividing 24 by 3), or if. Thus, the fraction f has been reduced to a higher given denom- inator.

Mixed numbers. It is sometimes desirable to change a mixed num- ber to an improper fraction, or, conversely, to change an improper fraction to a mixed number.

To change a mixed number to an improper fraction. Multiply the whole number by the denominator of the fraction, add the numerator, and place the sum over the denominator; thus, 3^ is -1/, 4f is ¥, and 6iis^.

To change an improper fraction to a whole or a mixed number, divide the numerator by the denominator; thus, -^ is 4, £ is l£, *$■ is If or 1^, and *£■ is 4-f .

PROBLEMS

1. Reduce to lowest terms: T% &, T% ££, £§, ||, ff, %%, £f, |f

2. Change to equivalent fractions having denominators as indicated:

i- to 8ths i to 15ths f to 25ths

| to 6ths i to 24ths T% to 48ths

I- to 20ths f to 24ths § to 32nds

I to 8ths | to 36ths T\ to 36ths.

3. Reduce to equivalent fractions whose denominators are 24: y1^, f, f, f,

7 7 ¥» S'

4. Change to improper fractions: 4J, 3J, 1^, 7J, 8f, 6£, 3|, 5|, 5f, 9f

5. Change to whole or mixed numbers: -\8-, V2-, -3/, -^ Tf, -1/, f, -6t4> ff i ¥"•

6. Is the number of fractional units increased or decreased when we reduce A ^° f? Is the s*ze °f the fractional unit increased or decreased when we reduce

A to |?

Addition and subtraction of fractions. Similar fractions are frac- tions that have a common denominator. Only similar fractions can be added or subtracted.

To add fractions, reduce the fractions to similar fractions having a common denominator and add the numerators.

To subtract fractions, reduce the fractions to similar fractions having a common denominator and subtract the numerators.

Example.

Add: ±-, f, andj.

Solution.

1 4

6 8 3

1 7

Factors, Multiples, and Common Fractions

Explanation. Inspection shows that 12 is the least common denominator, i is T%, f is -£%, and \ is T3^. Adding the numerators of the similar fractions gives 17, and \% is 1T%.

Example. Subtract: -f — T5g-.

Solution.

3 — 12

I~T6

5 — 7

T6 _ T-g-

Multiplication of fractions, (a) To multiply a fraction by a whole number, multiply the numerator or divide the denominator of the fraction by the whole number.

Example. Multiply 6 X TV

Solution.

6 X

_ 3 0 — 91

~ 12 ~ %

or 6 = 2, and f

(6) To multiply a whole number by a fraction, multiply the whole number by the numerator of the fraction and write the product over the denominator. Cancel when possible.

Example. Find f of 35.

Solution.

i X 35 = -V-

7 33

X 1

= 14

(c) To multiply a fraction by a fraction, multiply the numerators to obtain the numerator of the answer, and multiply the denominators to obtain the denominator of the answer. Cancel when possible.

Example.

Solution.

Find | of «.

15 —

T6 ~

30 —

¥8 _

2 X

3 X 10

(rf) To multiply a mixed number by a mixed number, reduce each mixed number to an improper fraction and proceed as in (c).

Example. Find the product of: Z\ X 4^.

Solution.

7 v 33 _ 231 - 147

Factors, Multiples, and Common Fractions

Example. Find the product of : 6f X 5£. Solution.

Find:

32X41 = 164 = 4 5X8 5 5

PROBLEMS

9XT\

24 X!

f of 35 A of 16

4 nf 25

7. 3i X 4£

8. 12f X Si-

Division of fractions, (a) To divide a fraction by a whole number, divide the numerator or multiply the denominator by the whole number.

Example.

Solution.

Divide § f by 5.

25-4-5 5

n x i

28 X $

Answer: /¥

(b) To divide any quantity — a whole number, a mixed number, or a fraction, by a fraction, invert the divisor and multiply.

Example. Divide 8 by f .

Solution.

% X3

= 12

Example. Divide 16^- by f .

Solution.

Example.

Solution.

Divide f by J.

4 X3 2

3X2

4 X 1 2

^ = l9i 2 y2

5-ii

2 2

Divide :

PROBLEMS

^5 36 4^

by 9

8byf 9byf

I6f by i 18* by i

by If

h. 9f by 3i

42 Factors, Multiples, and Common Fractions

1. How many pieces of wire each 8-f inches long can be cut from 40 feet of wire?

2. If f of a ton of coal costs $12.75, what is the cost of one ton?

3. How many sash weights each weighing 2\ pounds can be cast from 120 pounds of pig iron, if \ of the quantity of pig iron is wasted in the casting operation?

4. A room is 18f feet long and 144 feet wide. The width of the room is what part of the length of the room?

5. A carpenter has a board that is 20 feet long, but it is \ longer than he needs. How long a board does he need?

6. What is the cost of 1\ tons of coal at $141- a ton?

7. A house and lot are valued at $13,200. If the lot is worth f as much as the house, what is the value of each?

8. If a man can earn $10f a day, how long will it take him to earn $247^?

9. A table is 20 feet long. How many people can be seated on the two sides if you allow 1^ feet for each person?

10. Henry's time book shows that his working time for one week was as follows: Monday, 74 hours; Tuesday, 8 J hours; Wednesday, 8 hours; Thursday, 9 \ hours; Friday, 8 \ hours; Saturday, 6f hours.

He is paid straight time for 8 hours or less and time and a half for hours in excess of 8 each day other than Saturday, when he receives double-time pay for hours worked. How much did he earn at $lf an hour?

11. The shipping clerk reported that he dispatched 320 packages averaging 28f pounds each. What was the total weight of packages dispatched?

12. A cubic foot of water weighs 62J- pounds, and there are approximately 74 gallons to the cubic foot. Estimate the weight of water that a 10-gallon keg will contain.

To find the product of any two mixed numbers ending in \.

(a) When the sum of the whole numbers is an even number. To the product of the whole numbers, add one-half of their sum, and annex \.

Example. Multiply 244 by 8J.

Solution.

24i

Multiply:

1. 8J by 44.

2. 121- by 81,

192 16

(8 X 24)

(4 of the sum of 24 and 8)

2081

(i annexed) PROBLEMS

3. 281- by 12i

4. 16^ by 14£.

5. 18| by 184

6. 101- by 344

Factors, Multiples, and Common Fractions 43

(b) When the sum of the whole numbers is an odd number. To the product of the whole numbers, add one-half of their sum, less 1, and annex f .

Example. Multiply 15J- by 6^-.

Solution.

15-;

90 (6 X 15)

10 (i of 15 + 6 -

lOOf (| annexed)

PROBLEMS

3. 38i by 5i

4. 13i by 8£.

5. 23^ by \\.

6. 19i by 6±,

Multiply:

1. 18i by 5£.

2. 14i by 7f

To multiply a mixed number by a mixed number.

Example. Multiply 524£ by 27£.

Solution.

524i 27i

14148 6 = common denominator of fractions

174f 4

13^ 3 \ = numerators of changed fractions

h 1

14336^ t = H

Explanation. Multiply 524 by 27, obtaining the first part of the answer, 14,148. Next, take £ of 524, obtaining 174§. Then take \ of 27, obtaining 13£. Finally, take \ of -J, obtaining \. Add the four partial products, and the complete product is 14,336^.

PROBLEMS

Multiply:

1. 247| by 39-i. 3. 59^ by \h\. 5. 181f by 6f.

2. 849^ by 28A, 4. 176| by 34f 6. 56J- by 12|.

Decimal fractions. A decimal fraction is a fraction whose denom- inator is some power of ten, indicated by a decimal point placed just to the right of the units' place. Thus, .1 is -jV, .05 is xfo, and .25 is ^ or \.

Approximate numbers. Since many of the numbers we work with are approximate numbers, it is important that the student learn to recog-

44 Factors, Multiples, and Common Fractions

nize approximate numbers and to appreciate the limitations of results obtained by using approximate numbers.

All measurements are approximate numbers. If a surveyor measures a distance and finds it to be 124.7 feet, it would not be correct to say that the distance is exactly 124.7 feet. We would say that the .distance, correct to four significant figures, is 124.7 feet. By more accurate methods the distance might be found to be 124.73 feet, which would be correct to five significant figures.

A measurement of 20.006 has five significant figures. There are two significant figures in .00043, the three zeros being merely "space-fillers" put in to locate the decimal point.

In 53,000 the number of significant figures is uncertain. When we say that the population of a city is 53,000, we do not usually mean that the population is exactly 53,000, but rather that the population is closer to 53,000 than it is to 52,000 or 54,000. In this case, only the 5 and the 4 are significant. On the other hand, the population may actually be 53,000 in which case the number has five significant figures. It follows that the number might also have three or four significant figures.

The number .05600 has four significant figures. The initial zero is a " space-filler," but the last two zeros show that the number is correct to the nearest one-hundred-thousandth.

The following table gives further examples.

Approximate Number of

Number Significant Digits

325 3

127,000 3 to 6

2.73 3

630 2 or 3

.0005 1

350.0 4

.00370 3

50,000 1 to 5

50,001 5

To round off an approximate number to a number of less accuracy, the following rule is used :

When the digit immediately to the right of the last retained digit is 5 or more, the last retained digit should be increased by one; when the digit immediately to the right of the last retained digit is less than 5, the last retained digit is left unchanged.

For example,

4.01738 rounded off to three decimal places would be 4.017. 3.78 rounded off to one decimal place would become 3.8. 3.065 rounded off to two places would become 3.07.

Factors, Multiples, and Common Fractions 45

Measurements are not the only approximate numbers. Almost all numbers which appear in mathematical tables are approximate. A table of square roots shows that

7 is 2.64575 and a table of logarithms shows that

log 35.4 is 1.54900. These are approximations, both to six significant figures.

PROBLEMS

1. For each of the following approximate numbers, state the number of significant figures and round off to one less significant figure.

(a) 14.76 (/) .00079

(6) .0393 (a) 20.0

(d) .010007 (i) 12.04

(e) 2.400 U) 1-999

2. Round off each of the following numbers to two places of decimal.

(a) 24.768 (6) .033 (c) 6.5439 (d) .006 (e) .255

Calculations with exact numbers. If it is found by counting that there are twelve people in a room, then 12 is an exact number. If a man writes a check for $500.00, then $500.00 is an exact number. If a calculation is performed with exact numbers, the answer will be exactly correct.

Addition and subtraction. To add or to subtract decimals, write the numbers so that the decimal points fall vertically in a line and proceed as in whole numbers.

Example. Add: .01, 4.72, 78.25, and .005.

Solution.

.01

4.72

78.25

.005

82.985

Example.

Subtract: 47.02 -

.92.

Solution.

47.02 .92

46.10

Factors, Multiples, and Common Fractions

PROBLEMS

1. Add: 25.679, .0356, 2.78, and .017.

2. Add: 136.2, 28.348, .004 and 1.356.

3. Subtract: 13.48 from 27.049.

4. Subtract: .003 from .47

Multiplication. To multiply decimal fractions, multiply as in whole numbers and point off as many decimal places in the product as there are places in both multiplicand and multiplier.

Example.

Solution.

Multiply 3.06 X .8.

3.06

2.448

Explanation. Since there are three decimal places in both the multiplicand and the multiplier, point off three decimal places in the product.

Example. Multiply: 23.8564 by 6.72.

Solution.

23.8564

6.72

477128

1669948

1431384

160315008

Explanation. As there are six decimal places in the multiplicand and the multi- plier, point off six decimal places in the product. The answer is 160.315008. Rough check: 24 X 7 = 168.

Division. Proceed as with whole numbers, annexing zeros to the divi- dend if necessary. The number of decimal places in the quotient must equal the number in the dividend minus the number in the divisor.

Example. Divide: 54.864 by .24.

Solution.

.24)54.864(228.6 6 8 2 06 144 0

Explanation. Divide by writing the remainders only. The quotient is 2286. As there are three decimal places in the dividend and two decimal places in the divisor, point off one decimal place in the quotient. The answer is therefore, 228.6.

Factors, Multiples, and Common Fractions 47

Example. Divide: 256.7894 by 5.23. Solution.

49.099

5.23)256.78940 47 58 5194 4870 163

Explanation. Predetermine the placing of the decimal. As there are two decimals in the divisor, place the decimal point over the third decimal place in the dividend. Place the first figure of the quotient over the last figure of the partial dividend. One zero has been annexed to the dividend in order to obtain a quotient to three decimals. Rough check: 49 X 5 = 245.

PROBLEMS

Multiply:

Divide:

1. 34.278 X 1.45"

2. 395.264 X .035

3. 74.26 by .00423

4. .056 by .083

5. 18.42 X .045

6. 5.8769 by 1.34

7. .0084 by 1.5

8. 45.87 by .0056

9. 8.45 by 25.3 10. 956 by 4.87

Calculations with approximate numbers. In calculations with approximate numbers, the results are, of course, approximate numbers. To avoid giving a false appearance of accuracy, such results should be rounded off according to the following rules:

Addition and subtraction. The sum or difference obtained in adding or subtracting approximate numbers cannot be accurate to more decimal places than the least accurate of the numbers. Hence, before adding two or more approximate numbers, the numbers should be rounded off to the number of decimal places in the least accurate of the numbers.

Multiplication and division. The product or quotient obtained in multiplication or division with approximate numbers should not contain more significant figures than the least accurate of the numbers.

Example. Find the sum of 3.875, 24.6, 13.45, and 45.30.

Solution. Here all numbers should be rounded off to one decimal place, since the least accurate number (24.6) has one decimal place.

3.9 24.6 13.5 45.3

87.3

Example. Subtract 34.706 - 17.30522.

48 Factors, Multiples, and Common Fractions

Solution. Here 17.30522 should be rounded off to three decimal places.

34.706 17.305

17.401

Example. A firm which employs eight men has a monthly payroll of $2226.50. What is the average pay for the employees of this company?

Solution. $2226.50 v8 = $278.3125.

Since these numbers are exact, it is correct to say that the average pay is exactly $278.3125.

Example. Find the area of a hall which measures 83.5 ft. by 9.3 ft.

Solution. Multiplying length by width we get:

83.5 9.3

25 05 751 5 776.55

Since the factors are approximate numbers, this product must be rounded off to two significant figures, giving 780 sq. ft., where the zero is not significant.

Example. If there are 16 people in a room of 1041 cu ft., how much space per person is there?

Solution. 1041 + 16 = 65.0625

Since the number 1041 is approximate, the answer must be rounded off to 65.06 cu. ft.

Note : In calculations involving both approximate and exact numbers, the exact number may be regarded as having an infinite number of sig- nificant figures. In the above problem, 16 should be thought of as 16.0000000000 ...

PROBLEMS

1. In the following, all numbers are approximate. Perform the operations indicated, retaining the proper number of significant figures.

(a) 20 .4 X 1.13 (g) .9 X .05

(6) 13.94 + 2.8 + 7.092 + .65 (h) 2.976 - .8437

(d) .05614 ^ 7 0') 110.0 X 50001

(e) 176.30 + 2.47 + .98765 (k) 3 LS

(/) 24300 X 70, where none of (I)

the zeros is significant.

136

9.008 X 3.3

4.6

Factors, Multiples, and Common Fractions 49

2. In each of the following, assume the left-hand number to be exact and all others to be approximate. Perform the operations indicated, retaining the proper number of significant figures.

1 037

(a) 92.41 X 40.1

(b) 2 +

.03 + .0

(c)

3012

4- 1.2

(d)

2.7 -

- 1.04

13 X 486.9

In the following problems, be careful to give answers to the proper number of significant digits.

3. A silver collection taken at a Sunday concert amounted to $252.75. It was estimated that 700 people attended. What was the average contribution?

4. A racing car is clocked at 5.6 seconds for a distance of 1535 feet. Find its average speed in feet per second.

5. What is the area of a table top which measures 2.3 ft. by 5.7 ft.?

Contracted multiplication. For students performing multiplications and divisions of decimal numbers as read from logarithm tables, interest tables, annuity tables, and so forth, which are correct to a designated number of decimal places only, another time-saving method of multipli- cation is presented.

Let us multiply 4.7892 X 3.1765 using the regular method:

4.7892 3.1765 239460 287352 335244 47892 14 3676 15.21289380

Recall that, if two numbers are correct to four decimal places each, the product cannot be assumed to be correct beyond four decimal places. The product above is written to eight decimal places, whereas only four decimal places can be considered correct. We have, therefore, performed unnecessary work on the last four decimal places.

The above multiplication could be written in the following manner:

4.7892

= multiplicand

3.1765

= multiplier

.0023

9460

= 4.7892 X .0005

.0287

352

= 4.7892 X .006

.3352

= 4.7892 X .07

.4789

= 4.7892 X .1

14.3676

= 4.7892 X 3

15.2128

9380

50 Factors, Multiples, and Common Fractions

Obviously, everything to the right of the vertical line is unnecessary. This multiplication could be performed in the reverse order, as below.

4.7892

= multiplicand

5.6713

= multiplier reversed

14.3676

= 4.7892 X 3

.4789

= 4.7892 X .1

.3352

= 4.7892 X .07

.0287

352

= 4.7892 X .006

.0023

9460

= 4.7892 X .0005

15.2128

9380

If we discard everything to the right of the vertical line, the solution is unaffected.

4.7892 5.6713

14 3676 4789 3352

287 24

15.2129

It should be noted that the decimal point in the answer lies directly below the decimal point in the multiplicand.

The steps involved in this solution can now be outlined:

(a) Prepare the multiplier. This is done by moving the decimal point either to the left or to the right so that the multiplier has one non-zero digit to the left of the decimal place.

(b) Adjust the multiplicand. This is done by moving the decimal point either to the right or to the left so that the problem remains iden- tical with the original.

30.657 X 20.342 = 306.57 X 2.0342 365.2422 X 364.31 = 36524.22 X 3.6431 .003657 X 19.57 = .03657 X 1.957 567.93 X .00213 = .56793 X 2.13

It should be noted that if the decimal place is moved x places to the right in the multiplier, then it is moved x places to the left in the multi- plicand; if it is moved x places to the left in the multiplier, then it is moved x places to the right in the multiplicand.

(c) Write the multiplier down in reverse order under the multiplicand, with the right-hand digit under the zth decimal place in the multiplicand if x decimal places are desired in the answer.

(d) Multiply through normally by the first digit on the right.

(e) Cancel the digit you have multiplied by and the digit directly above it.

Factors, Multiples, and Common Fractions 51

(/) Multiply through by the next digit 1, saying, 1X2 = 2, which gives 0 to carry since 2 < 5 but we write nothing. Then, 1X9 = 9, which is written directly under the first digit on the right of the first line in the solution.

(g) Cancel the digit you have multiplied by and the digit directly above it.

(h) Multiply through by the next digit 7, saying, 7 X 9 = 63 so we carry 6; 7 X 8 = 56 and 6 to carry give 62. Write the 2 down directly under the first digit on the right of the first line in the solution and carry 6, and so forth, getting 3352.

(i) Carry through the multiplication by each digit in like manner.

(/) Total the result, getting 152129.

(k) Place the decimal point in the answer directly under the decimal place in the multiplicand (adjusting the multiplier permits this), getting 15.2129 as the answer.

Example. Multiply 0.47869347 by 72.5 and obtain the product correct to three decimal places.

Solution. 0.47869347 X 72.5 = 4.7869347 X 7.25.

4.7309347

33 508 957 239

34.704

PROBLEMS

1. Evaluate 5.987654 X 3.147, correct to four decimals.

2. Evaluate 3.596 X 14.57, correct to three decimals.

3. Evaluate 44.187542 X 6.2434 correct to four decimals.

4. Calculate to the nearest cent the value of each of the following : (a) $2,376,205 X 3.53710872.

(6) $2,811,362 X 2.69159903.

(d) $30,265. X 15.67362495.

Contracted division. Many times an accountant has to do a rather complicated division, and so a time-saving method of division is presented.

Let us divide 3768.943 by 57.68429 using the regular method of long division. For the purpose of comparing this with contracted division, the decimal points are adjusted so that the divisor has one figure to the left of its decimal point.

Factors, Multiples, and Common Fractions

65.337

5.768429 |376.894 346.105

3 74

30.788 28.842

560 145

1.946 1.730

4150

5287

.215 .173

88630 05287

.042 .040

833430 379003

.002

454427

It should be noted immediately that the digits to the right of the ver- tical line do not contribute anything to the required answer. The pro- posed contracted division permits the carrying out of a long division to the desired degree of accuracy without unnecessary labor.

The procedure is:

(a) Prepare the divisor. This is done by moving the decimal point either to the left or to the right so that the divisor has one non-zero digit to the left of the decimal place.

(6) Adjust the dividend. This is done by moving the decimal point either to the left or to the right so that the problem remains identical with the original.

30.657 -r- 20.342 = 3.0657 + 2.0342 .003657 -f- 19.57 = .0003657 -^ 1.957 567.93 -r- .00213 = 567930. -r- 2.13

It should be noted that if the decimal place is moved x places to the right in the divisor, then it is moved x places to the right in the dividend ; and if it is moved x places to the left in the divisor, then it is moved x places to the left in the dividend.

5768429 376.8943

The decimal point in the divisor may be omitted, since the decimal point in the answer will lie directly above the decimal point in the divi- dend. (Adjusting the divisor permits this.)

(d) Since the answer is required correct to two decimal places, the problem should be worked to three decimal places and the answer rounded off to two decimal places. Since we are working to three decimal places, we shall use only the first six digits of the dividend and shall cross off the 3. We shall, therefore, use only the first five digits of the divisor, since any more digits would give us a number which would not be contained in the six-digit dividend.

5768429376.8949-

Factors, Multiples, and Common Fractions 53

(e) Divide 57684 into 376894. Since it goes six times, place a 6 over the 7 in the dividend and a 6 under the 4 in the divisor.

5768429376.8943-

(/) Multiply the divisor by 6 and include the carryover from 6X2 (the 2 in the divisor which was crossed off). Subtract:

5768429376.8943-

6 346 105

30 789

(g) Cross off the 4 in the divisor and divide 5768 into the new divi- dend 30789. Since it goes five times, place 5 next to the 6 in the quotient and under the 8 in the divisor. Multiply by 5, remembering to add the carryover from 5X4.

5768£29|376.8943- 56 346 105

30 789 28 842

1 947

(h) Continue this process until the quotient is evaluated to three decimal places. By this time, all the digits in the divisor will have been crossed out.

65.337

070S42& 376.8940

73356

346 105

30 789

28 842

1 947

1 730

217

173

(t) Round the answer off to two decimal places. The answer is, therefore, 65.34.

Example 1. Divide 7.24464613 by 5.38287878 and obtain the answer correct to 8 decimal places.

54 Factors, Multiples, and Common Fractions

Solution.

1.34586834

5. 3B2B787B 7.24464613

4 38685431 5 38287878

1 86176735 1 61486363

24690372 21531515

3158857 2691439

467418 430630

36788 32297

4491 4306

185

161

42.3 Example 2. Evaluate ■ rn correct to four decimal places. 673.58

Solution. Here we shall work to five places of decimals, since the answer is required to four places. It is necessary to add two zeros to the end of the dividend, since the dividend has only three digits in it.

.06280

6735^.42300

0826 40415 1885 1347

538

This rounds off to .0628. Example 3. Divide .76839 by .234 correct to two decimal places. Solution.

3.283

gg40|7.6839- 7 020 663

468

195

187

Here it was necessary to add a zero to the end of the divisor so that the divisor would have as many digits as the dividend.

Factors, Multiples, and Common Fractions

PROBLEMS

Evaluate correctly to the nearest cent:

1. $2,394,291 -^ 3.497572

2. $9,093,255 + 2.57284313

3. $14,300.4671 + 9.231049

To change a decimal fraction to an equivalent common frac- tion. Write the denominator of the decimal, omit the decimal point, and reduce to lowest terms. Thus, to reduce to common fractions in lowest terms or to mixed numbers :

7C _ 75 — 3 /0 — TOO ~~ T

6.25 = 6T2^

.025 = TM

1000

4.125 = 4TuyV

JL_

= 4i

To change a common fraction to a decimal. A common frac- tion may be regarded as an indicated division. Thus: § may be regarded as 2 v 5; therefore, § expressed as a decimal is .4; similarly, \ is .14y, | is .375, and A is .4375.

Aliquot parts. An aliquot part of any number is a number that is contained in it an integral number of times. Thus, 5, 10, 20, and 50 are aliquot parts of 100; that is, 5 = aV of 100, 10 = to- of 100, and so forth.

The use of aliquot parts. As a means of saving time in multi- plication and in division, it is useful to know the decimal equivalents of common fractions, or, conversely, to know the common-fraction equiva- lents of decimal fractions. Aliquot parts are of value in addition and subtraction if an adding machine or a calculating machine is used, because machines are not adapted for general work involving common fractions.

TABLE OF ALIQUOT PARTS OF 1

Common

Decimal

Common

Decimal

Fraction

Equivalent

Fraction

Equivalent

.50

.11*

•33^

1 10

.66|

1 TT

09TV

1 T

.25

1 T2"

08i

.75

41|

1 5

.20

58i

1 6

•16f

1 1 1 2

91|

•83^

1 15

06|

i T

.14f

06i

.28-f

3 T6

18f

.42f

311

.57|

43|

.71*

56J

.85f

1 1 T6

68f

1 8

.12^

1 5 T6

93f

• 371

1 ^5

• 62i-

03i

.87J

09|

Factors, Multiples, and Common Fractions

The fractions in the above table can be extended as decimals as far as the work demands.

PROBLEMS

Express the following as decimal fractions; non-terminating fractions should be carried to the sixth decimal place and the common fraction annexed:

1 1

^2~

1 5

1 4

Te"

"24

2-0

If 2"

Multiplication by aliquot parts. Example. Find 16| % of $475.34.

Solution.

6) $475 .34

$79.22

Explanation. Since .16| equals i find 1 of $475.34.

Example. Find the cost of 256 units at 37-^ each.

Solution.

256 X f X $1 = $96

Explanation. 37^ is f of $1. Therefore, 256 X | X $1

$96.

PROBLEMS

Extend the following items mentally

1. [email protected]

9. 18 @ .33^

17. 64 @ .25

25.

72 @ .83^

2. 45 @ .11^

10. 39 @ .66|

18. 27 @ .22f

26.

32 @ .87^

3. 24 @ .08^

11. 55 @ .09^

19. 32 @ .18|

27.

36 @ .41f

4. 36 @ .50

12. [email protected]

20. 96 @ .03i

28.

27 @ .44$

5. 15 @ .06f

13. 49 @ .28|

21. 48 @ .56i

29.

12 @ .75

6. 75 @ .93^

14. 32 @ .43f

22. 60 @ .58i

30.

14 @ .07i

7. 48 @ .16f

15. 28 @ .57|

23. 48 @ .37i-

31.

18 @ .16|

8. 32 @ .06i

16. 24 @ .62i

24. 35 @ .14f

32.

16 @ .87|-

Division by aliquot parts. It is difficult to divide a number by a mixed number. If the divisor is an aliquot part, the quotient may be found by multiplication, as follows:

Example. Divide 4,875 by 16f .

Solution.

48.75

292.50

Factors, Multiples, and Common Fractions 57

Explanation. Since 16f is i of 100, divide 4,875 by £ of 100, or ±%£. This is the same as multiplying by t£q. Therefore, divide by 100 by pointing off two decimal places from the right, and multiply the result by 6. The answer is 292.50, or 292£.

Example. The production cost of 1,250 units is $3,170. Find the cost per unit.

Solution.

.3170

2.5360

Explanation. 1,250 is £ of 10,000. Divide $3,170 by 10,000 by pointing off 4 decimal places from the right; then multiply the result by 8. The cost per unit is found to be $2,536.

PROBLEMS

Divide

1. 1,342 by 11-1. 3. 3,126 by 33£. 5. 158 by 6£.

2. 2,578 by 12J. ; 4. 384 by 25. 6. 4,275 by 14f .

PROBLEMS

1. A manufacturer pays dividends amounting to T3g- of his capital. If the dividends amount to $37,500, what is the capital?

2. A fuel dealer had 36 cords of wood and sold f of it. How many cords did he sell?

3. If a merchant buys an article for $12J and sells it for $16, the profit is what fraction of the selling price? What fraction of the cost price?

4. A crate containing 10 dozen oranges cost $4.50. If they are sold at the rate of 65 cents a dozen, but \ dozen are spoiled, the profit is what fraction of the selling price?

5. A man has S37-J- and spends %\2\. What fraction of his money does he keep?

6. A factory normally employed 48 men. During a dull period 16 received temporary layoffs. What fraction of the force continued to work?

7. The last reading of a gas meter was 67,324 cu. ft.; the previous reading was 64,815 cu. ft. At $1.45 a thousand cubic feet, find the amount of the gas bill.

8. An investment of $18,000 produces an annual income of $720. At the same rate, what should an investment of $25,000 produce?

9. Tires costing $18.75 were installed when the speedometer registered 18,985 miles. The four tires were replaced when the speedometer registered 34,652 miles. $1.00 was allowed for each old tire. What was the average tire cost per mile, correct to the nearest tenth of a mill?

10. An excavation 8 feet in depth required the removal of 5,328 cu. ft. of earth and rock. The average depth of earth was 5 ft., and the cost of earth removal was $1-^ a cu. yd. The remainder was rock and cost $4-| a cu. yd. for removal. What was the cost of making the excavation?

Percentage and Applications

Relation between percentage and common and decimal frac- tions. Percentage is a continuation of the subject of fractions. It is the process of computing by hundredths, but instead of the term hun- dredths, the Latin expression per cent is used. The sign (%) generally replaces the words per cent, thus, 5%, 10%, and so forth.

Any per cent may be expressed either as a common fraction or as a decimal, thus:

12*%

100%

300%

.05%

100 10,000

Care should be taken in writing per cents. Do not write both the sign and the decimal point; thus, 2% and .02 are the same, but 2% and .02% are widely different, since the first is equivalent to ^ and the second to

Common Fraction

Decimally

Too

Too 12* 125

ioo or 1000

.01 .05

. 12* or . 15

100 TOO

300 TOO

\ 5 ioo or 1000

1. 3.

. 00* or . 0(

T*oor_!_

.0005

5000-

Applications. Percentage admits of applications in many fields. Business operations are guided by carefully prepared statistics, and the relationships of items in statistics are often more clearly reflected when they are expressed in terms of percentage. There are numerous problems involving percentage besides those having to do with financial consid-

Percentage and Applications 59

erations, such as finding the per cent of increase or decrease in volume; per cent of shrinkage of material; per cent of waste in manufacturing operations; per cent of yield of crops.

Definitions. The base is the number or quantity represented by 100%. The base may be, for example, total sales, total expenses, the face value of a note, the par value of a bond, pounds of material used, capacity, and so forth.

The rate is the number of hundredths, or the per cent. The rate may be, for example, 6% or 25%, which are written decimally as .06 and .25.

The percentage is the product of the base and the rate. The percent- age may be, for example, the interest cost of a sum of money, the depart- mental portions of an expense item, the increase in pounds of material used, and the like.

Fundamental processes. In percentage and its application, three fundamental mathematical principles are involved, namely: (1) to find a given per cent of a number; (2) to find what per cent one number is of another; and (3) to find a number when a certain per cent of it is known.

Computations. Computations in percentage are based on these principles.

Principle 1. The percentage is the product of the base and the rate.

Base X Rate = Percentage. Example. 6% interest on $500 is $30. (500 X .06 = 30).

PROBLEMS

In the following, convert the per cent either to a common fraction or to a decimal fraction, whichever is the easier. Find:

1. 25% of 5,280 ft. 6. 2f % of 180 lbs.

2. 10% of 846 lbs. 7. f % of 240 gal.

3. 16f % of 24 bu. 8. f % of $5,000.

4. 37i% of $60. 9. 20% of 95 yds.

5. 80% of 120 pp. 10. 14f % of 42 in.

11. If an expense item of $16.00 is reduced 6±-%, what will be the amount of this item after the reduction?

12. A commission of 12i% was earned on a $240 sale. What was the commission?

13. A sample of grain showed 2f % weed seed. How many bushels of weed seed are in 600 bushels of this grain?

14. An item sells for 40 cents. What will be the selling price after a reduction of 15%?

15. Anticipated requirements for copper will exceed the manufacturer's stock by 35%. If 185 pounds are on hand, how many pounds will have to be purchased?

60 Percentage and Applications

Principle 2. The rate may be found by dividing the percentage by the base.

Percentage -v- Base = Rate.

Example. $30 + $500 = .06 or 6%.

PROBLEMS

In the following find what per cent of:

1. 72 is 24 6. 12 is 20

2. 60 is 50 7. 64 is 8

3. 180 is 120 8. 90 is 10

4. 360 is 90 9. 150 is 25

5. 50 is 20 10. 125 is 25

11. Last year's taxes on a house were $520. This year's taxes were $640. What per cent were this year's taxes of last year's taxes?

12. A pile of lumber contained 4,500 feet, and 3,300 feet were used. What per cent remained?

13. Wages are increased from $1.50 an hour to $1.75 an hour. Find the per cent of increase.

14. A new style of packaging reduced the shipping weight from 130 lbs. to 121 lbs. What was the per cent of saving in shipping weight?

15. The inspector rejected 5 items out of 140 produced. What was the per cent of rejects?

Principle 3. The base may be found by dividing the percentage by the rate.

Percentage -r- Rate = Base.

Example. 30 4- .06 = 500.

PROBLEMS

Find the number of which:

25 is 20%

125 is 16|%

240 is 75%

48isi%

72isl2^%

86 is 43%

374 is 17%

375 is f%

4iisf%

26 is 40%

11. The fire insurance premium on a house was $22.50. The house was insured for 80% of its value at f %. Find the value of the house.

12. Sales increased each year over the preceding year as follows: 15% the second year, 20 % the third year, and 25 % the fourth year. If the fourth year's sales were $21,562.50, what were the first year's sales?

Percentage and Applications 61

13. A bankrupt can pay his creditors 72 cents on the dollar. If his assets are $13,475.28, what are his liabilities?

14. The gross income of a rental property is $1,800 a year. Expenses are $500. If the net income is a return of 6-J-% on the investment, find the value of the property.

15. One workman completes a unit in 7-J hours. Another workman com- pletes a similar unit in 5f hours. The first workman took what per cent more time than the second workman to complete the unit?

MISCELLANEOUS PROBLEMS

1. A machine that cost $50 was marked up 30%. What was the marked

price

2. After a clerk's salary was increased 6^ %, he received $850 a year. What was his former salary?

3. A 4-apartment building cost $18,000. Repairs average \\% of the cost; taxes, 2-|%; insurance on 90% valuable, f %; other expenses amount to $114.25. What should the annual rental income be in order to return the owner 8 % on his investment? What should be the average monthly rental of each apartment?

4. A product shrinks 16% in processing. How many pounds of raw material will be required to process 252 pounds of finished product?

5. A creditor received $637.73 from a bankrupt estate paying 68 cents on the dollar. What was the creditor's loss on the account?

6. In a certain school 1250 pupils are enrolled. If 725 are boys, what per cent of the pupils are girls?

7. A bushel of wheat weighs 60 pounds and contains 0.6% potash, 1.1% phosphoric acid, and 2.1% nitrogen. How many pounds of each constituent are removed from the farm with 200 bushels of wheat?

8. In a year a businessman's sales amount to $57,325.19. If salaries amount to $15,829.40, wages to $6,324.29, machinery to $10,000, and incidentals to $3,272.17, what per cent of sales is required to meet the overhead expenses?

9. How many pounds of cream containing 30% butter fat can be produced from 625 pounds of milk containing 3.9% butter fat?

10. An apartment house of 11 suites was bought for $200,000.00. Taxes, repairs, insurance, and heating cost $8,023.00 per year. Six suites are rented at $125.00 per month, 4 at $110.00 per month, and 1 suite at $220.00 per month. What per cent is realized on the investment per year?

Daily record of departmental sales. The following tabulation is designed to show the total daily sales by department and the total sales for the week, both by department and for the business as a whole. After Saturday's sales have been entered, the total departmental sales for the week may be found and also the per cent that each department's sales is of total sales. The per cent that each day's sales is of total sales for the week is also obtainable.

62 Percentage and Applications

Daily Record of Departmental Sales

Dept. Mon. Tues. Wed. Thurs. Fri. Sat. Total Per Cent

A $475.86 $275.83 $329.86 $424.83 $387.92 $412.15

B 324.18 174.82 274.19 285.27 304.14 319.28

C 456.19 259.80 179.86 258.24 286.39 305.74

D 421.40 268.75 142.56 280.22 178.90 260.57

E 175.60 125.34 156.85 210.05 162.50 187.50 .^^^ II_=:::::::

Total ::^::^. ". ... -~~~™-4 .^^^ 100.00%

Per

Cent 100.00%

PROBLEM

Prepare a form similar to the above, enter the sales in the proper columns, and find: (a) the total sales for each day in all departments (add downward); (6) the total sales for each department for the week (add across); (c) in two ways, the total sales in all departments for the entire week; (d) the per cent of grand total sales made each day (total for each day divided by the grand total) ; (e) the per cent of grand total sales made in each department (total of each department divided by the grand total).

Per cent of returned sales by departments. In some lines of business it is important to keep a close check on the volume of returned sales. This may be done advantageously by means of per cents derived from tabulated results.

PROBLEM

Prepare a form similar to the following, enter the data, and find: (a) the net sales for each department and the net sales for all the departments; (6) the per cent of returned sales in each department and the total per cent of returned sales.

Sales and Returned Sales by Department

Per Cent

Returned Net of Sales

Dept. Sales Sales Sales Returned

A $ 24,863.95 $ 756.82

B 110,356.80 1,328.95

C 53,768.21 975.32 .....:.....

D 16,135.40 628.74

E 9,356.24 256.48

Total -

Clerk's per cent of average sales. As a measure of efficiency, the following tabulation may be made for a department, and each clerk's weekly or monthly sales compared with the average weekly or monthly sales.

Percentage and Applications 63

Monthly Sales of Clerks — Dept. A

Clerk's Monthly Per Cent of

Number Sales Average

1 $2,756.80

2 1,954.36

3 2,075.83 .

4 2,634.87

5 2,315.62

Total 100.00%

Average

PROBLEM

Prepare a form similar to the above, enter the data, and find: (a) the total monthly sales; (6) the average monthly sales per clerk; and (c) what per cent each clerk's sales are of the average sales per clerk.

Per cent of income by source. In accounting for the income of a public service enterprise, it is desirable to show the per cent of income from each source when the company's activities are varied.

PROBLEMS

1. In the following tabulation of gross earnings of a public utility corporation, find what per cent the earnings from each source are of the total gross earnings.

Source Gross Earnings Per Cent

Electric light and power $15,817,324.00

Electric and steam railroads 6,763,656.00

City railways and bus lines 4,248,824.00

Gas 3,191,720.00

Heat 672,394.00

Bridges 589,691.00

Ice 254,670.00

Water 88,303.00

Miscellaneous 21,816.00

$31,648,398.00 100.00%

2. In the following tabulation of the revenue from transportation of an inter- urban railway, find what per cent each item of revenue is of the total revenue.

Revenue from Transportation

Source Amount Per Cent

Passengers $657,855.00

Baggage 550.00

Parlor and chair cars 9,894.00

Special cars 25.00

Mail 1,500.00

Express 21,962.00

Milk 1,666.00

Freight 264,214.00

Miscellaneous 269.00

$957,935.00 100.00%

64 Percentage and Applications

Per cent of expense. Items of operating expenses and their relation to total expenses are more easily compared if expressed in terms of per- centage.

PROBLEMS

1. In the following report of an interurban railway company, find what per cent each group of expenses is of total operating expenses.

Operating Expenses

Item Amount Per Cent

Way and structures $228,690.00

Equipment 98,979.00

Power 105,890.00

Conducting transportation 249,427.00

Traffic 52,823.00

General and miscellaneous 141,560.00

Transportation for investment (credit) 8,403.00

$868,966.00 100.00%

2. In the following statement of the operating expenses of a restaurant for a period of one month, find what per cent each item of expense is of total oper- ating expenses.

Operating Expenses

Item Amount Per Cent

Superintendent's labor $ 75.00

General labor 1,776.00

Extra labor 160.00

Supplies 200.00

Electricity 58.00

Fuel 75.00

Laundry 103 .00

Ice 22.00

Repairs and renewals — equipment 110.00

Meals to employees 340.00

Music 75.00

Miscellaneous 66.00

Total $3,060.00 100.00%

Per cent of increase or decrease. Percentage is often employed to find the relation between numbers; that is, to find how much larger or smaller one number is than another.

PROBLEMS

1. In the following departmental sales tabulation, find: (a) the increase or the decrease in monthly sales by departments; (6) each department's per cent of increase or decrease (divide increase or decrease in each department by that department's monthly sales for This Month Last Year).

Percentage and Applications

Dept.

E Total

This Month This Year $2,973.69 1,426.85 3,752.89 2,581.28 2,076.82

This Month

Last Year Increase

$2,795.84

1,852.18

3,565.62

2,678.15

1,825.38

Decrease

Per Cent Increase

Per Cent Decrease

2. In the following condensed balance sheet of a municipal railway, find the increase or decrease for each item, and also the per cent of increase or decrease.

Assets This Year Last Year

Capital Assets $ 7,912,526 $7,610,139

Current Assets 2,174,925 2,241,395

Deferred Assets 132,124 132,125

Total Assets $10,219,575 $9,983,659

Increase, Per Cent Decrease^ Inc., Dec.\

Liabilities, Reserves, and Surplus

Funded Debt $ 3,992,000

Current Liabilities 269 , 720

Reserves 1,568,469

Surplus 4,389,386

Total Liabilities, etc '... $10,219,575

$4,192,000

343,126

1,615,743

3,832,790

$9,983,659

3. In the following tabulation of advertising expenditures and direct sales resulting therefrom, compute the totals, the increase, and the per cent of increase.

This A advertising

Jan $2,238.00

Feb 2,154.00

Mar 2,435.86

Apr 2,425.46

May 2,293.12

June 2,035.76

July none

Aug none

Sept none

Oct 2,212.56

Nov 785.24

Dec none

Total

Year ago

Increase

% Increase

fear

Last

Year

Sales

Advertising

Sales

$4,251.44

$ 1,769.64

$ 3,762.00

7,461.60

1,787.96

5,067.16

8,773.84

1,769.53

5,232.48

7,292.12

1,840.26

7,818.00

8,709.04

1,831.70

4,867.20

8,412.28

1,825.49

4,673.12

7,383.46

none

5,083.20

7,656.80

none

4,454.56

8,227.84

none

4,650.88

4,298.70

1,142.04

4,976.40

5,260.84

1,306.26

2,682.00

5,683.96

none

3,542.80

$10,607.52

$37,650.77

66 Percentage and Applications

4. In the following statement, find the increase or decrease of revenues and

expenses and the per cent of increase or decrease:

Increase,

This Year Last Year Decrease^ Per Cent

Railway operating revenue $866, 197 $970,060

Other operating revenue 8,218 7,820

Total operating revenue $874,415 $977,880 '. 7.

Railway operating expense:

Way and structures $ 91,380 $ 85,569 ..

Equipment 64,249 61,866

Power 108,313 114,906

Conducting transportation 196,259 211,144

Traffic 9,496 10,157

General and miscellaneous 128,849 128,887

Depreciation 17,324 44,645

Taxes (except income taxes) 26,185 29,840

Total $642,055 $687,014 ZZZZ

Operating income $232,360 $290,866

Non-operating income:

Interest funded securities 2,579 5,105

Interest unfunded securities 7,765 6,328 .

Total $ 10,344 $ 11,433 ZZZZ

Gross income $242,704 $302,299 ZZZZT .^:z:=:::::::.

Deductions from gross income :

Interest $160,318 $161,402

Miscellaneous 3,216 3,257

Total $163,534 $164,659

Net income $ 79,170 $137,640

Operating statistics. The operations of a public utility engaged in transportation afford an excellent opportunity for the presentation of

statistics for managerial control. The following problem has been derived from the report of such an enterprise.

PROBLEM

From the following data, ascertain the required answers.

Section of Income Statement

Income This Year Last Year

Operating revenue:

Railway operating revenue $ 22,413,689 $ 21,678,906

Coach operating revenue 818,328 51,282

Total operating revenue $ 23,232,017 $ 21,730,188

Non-operating income 184,273 141,767

Total revenue from all sources $ 23,416,290 $ 21,871,955

Operating expenses:

Railway operating expenses. $ 16,572,497 $ 15,383,494

Coach operating expenses 786 , 558 41,701

Total operating expenses $ 17,359,055 $ 15,425,195

Net revenue from all sources $ 6,057,235 $ 6,446,760

Percentage and Applications 67

Statistics This Year Last Year

Railway revenue car-miles 52,863, 111 48,248,330

Coach revenue coach-miles 3,529,795 157,540

Railway revenue car-hours 5,692, 190 5,267, 176

Railway revenue passengers 357,926, 168 346, 116,298

Railway transfer passengers 123,310,526 111,445,912

Railway total passengers 481,236,694 457,562,210

Coach revenue passengers 10,564,723 978,782

Coach transfer passengers 387,228

Coach total passengers 10,951 ,951 978,782

Total revenue and transfer passengers 492, 188,645 458,540,992

Railway operating revenue per car-mile (cents)

Coach operating revenue per coach-mile (cents)

Railway operating expenses per car-mile (cents) _ _

Coach operating expenses per coach-mile (cents) __

Railway operating revenue per car-hour ($ and cents) .._ _

Railway operating expenses per car-hour ($ and cents) _

Ratio of transfer passengers to revenue passengers —

railway (per cent) _

Ratio of transfer passengers to revenue passengers —

coach (per cent)

Railway revenue passengers per car-mile operated _

Railway transfer passengers per car-mile operated _

Total railway passengers per car-mile operated _

Coach revenue passengers per coach-mile operated

Coach transfer passengers per coach-mile operated

Total coach passengers per coach-mile operated :....

Ratio of railway operating expenses to railway oper- ating revenue (per cent)

Ratio of coach operating expenses to coach operating

revenue (per cent)

Budgeting. Percentage is also applied in budgeting, as shown by the following example from hotel accounting.

Example. Among the several items of the budget is China and Glassware, $3,500, to be distributed to four departments on the basis of the previous year's expense for this item in the four departments, as follows:

Department Per Cent

Rooms 11 .29

Restaurant 55 . 29

Coffee Shop 14.86

Beverages 18 . 56

Total 100.00%

Solution.

Department Per Cent Budget

Rooms 11.29 $ 395.00

Restaurant 55.29 1,935.00

Coffee Shop 14.86 520.00

Beverages 18.56 650 .00

Total 100.00% $3,500.00

Percentage and Applications

PROBLEMS

1. The following year it was found that the actual disbursements for China and Glassware amounted to $2,280.74, and other facts were as given in the tabulation below. Compute the per cent for the distribution of the budgeted amount for the next year, and the per cent that the expense of China and Glass- ware is of the income for each department.

Gross Department Income

Rooms $141,857.50

Restaurant 59,626.90

Coffee Shop 33,587.45

Beverages 9,061.65

$244,133.50

China and Per Cent Per Cent Glassware of of

Expense Expense Income

$ 269.53

1,252.16

335.87

423.18

$2,280.74 100.00%

2. The following budget is that of an estimated operating statement.

Per Cent of Total Sales Net sales :

Class A $2,000,000

Class B 200,000

Class C 250,000

Class D _ 50,000

$2,500,000 100.00%

Per Cent of Sales

Production costs:

Class A $1,200,000

Class B 140,000

Class C 162,500

Class D _ 40,000

$1,542,500

Gross margin $ 957 , 500

Selling :

Sales administration $ 50,000

General sales department expense 12 , 500

Special promotion, etc 12 , 500

District operating expense 400 , 000

Advertising A 100,000

Advertising B 12,500

Advertising C _ 25,000

Selling cost $ 612,500

Net margin $ 345,000

Per Cent

Calculate: (a) the per cent of net sales in each class, as compared with total net sales; (6) the per cent of production cost in each class, based on sales of each class; (c) the per cent that selling cost is of total net sales; (d) the per cent that the net margin is of total net sales.

Percentage and Applications 69

3. The following is the budget for the Water Department of a municipality. Find the per cent that each budget expenditure is of the total for the department.

Amount Per Cent

Pump station and filter plant salaries $17,300.00

Office salaries and expenses 4,600.00

Chemicals, filter plant 1,000.00

Power — pump station and filter plant 15,000.00

Light, heat, and supplies 3,000.00

Water service 3,000.00

Meters and installation 6,000.00

Water main extensions and fire hydrants 3,000.00

Motor truck repairs 150.00

Interest on outstanding warrants 4,246.00

Total $57,296.00 100.00%

4. Compute the increase or decrease and the per cent of increase or decrease in the following comparative budget.

Public Buildings

and Utilities

City hall engineers and janitors. .

This

Year

$ 5,060

4,000

1,000

300

900

1,200

600

100

1,000

14,500

2,284

5,000

Last % %

Year Inc. Dec. Inc. Dec.

$ 4,284

City hall fuel and supplies

City hall maintenance and repairs

2,216

850

Detention hospital repairs

400

Detention hospital light and fuel . Park light and fuel

700

1,050 .

Septic tank electric power

Septic tank repairs

600

100

Incinerator fuel and light

1,000

Electric lighting — streets, alleys .

14,000

Interest on warrants

2,170

Contingent fund

4,036

Detention hospital insurance ....

433

Library insurance

281

Park insurance

104

$32,224

$35,944

Profits based on sales. In the income statement, it is customary to base all percentage calculations on sales. With sales equalling 100%, cost of sales, overhead, and net profit are expressed as per cents of sales. Overhead expenses are those incurred in operating a business — such as salaries and wages, rent, heat, light and power, depreciation, taxes, insurance, advertising, telephone, postage, and so forth. In marking goods bought for resale, these expenses must be taken into consideration. A few items of overhead expense do not fluctuate, but many of them have a fairly constant ratio to gross sales. The merchant determines the ratio of overhead expenses to sales from his own experience and that of others engaged in similar businesses. This per cent of cost of doing busi- ness plus the per cent of profit decided upon deducted from 100% deter- mines the per cent which the cost of goods plus freight and drayage bears to the selling price.

Percentage and Applications

Sales = 100%

Invoice Price plus Freight and Cartage

75%

Overhead 15%

Profit 10%

Cost of Sales

75%

Gross Profit on Sales

25%

Example. If overhead charges amount to 15% of sales, and a profit of 10% on sales is desired, what is the selling price of an article with an invoice cost of $21.00 and freight and cartage of $1.50?

Solution.

15% + 10% = 25% 100% - 25% = 75% $21.00 + $1.50 = $22.50, the cost. $22.50 -j- 75% = $30.00, the selling price.

Verification

25% of $30.00 = $7.50, the overhead and profit. $30.00 - $7.50 = $22.50, the cost.

PROBLEMS

1. An article that cost $15.00 was sold for $20.00. What is the profit per cent on the selling price?

2. With an overhead expense of 20%, what per cent of profit on sales is made by selling for $1.50 articles that have an invoice cost of $1.00?

3. What is the per cent of gross profit on sales in Problem 2?

4. How much must the article in Problem 2 be reduced to sell at cost? What per cent is this of the marked price?

5. A merchant sold an article for $12.00 and made a profit of 12^-% on the selling price. What was his profit in dollars?

6. Find the per cent of reduction of marked price to produce cost.

Cost

Marked Price

$ .20

$ .25

2.50

2.75

1.00

1.20

.03

.05

3.00

6.00

.40

.50

.09

.12

15.00

25.00

Per Cent Reduction

Percentage and Applications 71

7. Find the per cent of profit on the selling price.

Per Cent Profit on Cost Selling Price Selling Price

a. $ 1.00 $ 1.20

b. 10.00 15.00

c. .60 .75

d. 3.50 7.00

e. 6.00 8.00 /. 150.00 175.00

g. 16.00 24.00

h. 75.00 125.00

8. The factory price of an automobile is $1,300. Freight charges from factory to dealer are $65.00. If the dealer's overhead is 20% and he expects a net profit of 15% on sales, what should be the selling price of the automobile?

9. A furniture dealer bought a shipment of 20 chairs at $30.00 each. He marked them to sell at a profit of 40% on cost. The entire shipment was sold in the fall clearance sale at 25% reduction from marked price. What was the profit or loss?

10. Complete the following:

Cost

Selling Price

$ 4.00

$ 6.00

15.00

25.00

.16

.20

.04

.08

.10

/•

5.00

7.00

500.00

750.00

24.00

32.00

% on % on Cost Selling Price

11. The invoice price of an article is $12.00. Freight is 75 cents. It costs 18% to do business and you desire a net profit of 10% on sales. What is the selling price of the article?

12. If the invoice cost is $28.00, freight $2.00, overhead 25%, net profit on sales 15%, what is the selling price?

13. A stock of merchandise valued at $8,750.00 was damaged by fire and water. The loss was estimated to be 25%. Find the value of the damaged merchandise.

14. A merchant's overhead, or cost of doing business, is 22f %. He desires to make a net profit of 7^%. What will be the selling price of an item that cost this merchant $4.90?

15. Merchandise is bought for $3.50 less 25% and sold at $3.50 net. What is the rate per cent of profit?

16. A tea and coffee merchant blends a 40^ tea with a 70^ tea in the ratio of 2 to 1. If the blend is sold at 65^ a pound, what is the rate per cent of profit on cost?

17. A chair manufacturer finds the cost of material in a certain type chair to be $7.50. Manufacturing cost (labor and overhead) is $14.80. Selling and

72 Percentage and Applications

administrative expenses are 20% of sales. What is the manufacturer's price for this chair if he desires to net 10% on the selling price?

18. A clerk was ordered to mark a lot of suits so as to make a profit of 20% after allowing 5% discount for cash. By mistake he marked the suits $24.75 each, which resulted in a loss to the clothier of 8%. At what price should the suits have been marked?

Commissions. The commission business in this country is largely the result of our industrial and commercial development. Economic conditions demand that there shall be agents who shall represent either the buyer or the seller. The compensation paid the agent for his serv- ices is called a commission. The principles of percentage apply in commission.

The person who transacts business for another is the agent, and the one for whom the business is transacted is the principal. The fee, usually a per cent of the dollar volume of the transaction, is the commission.

PROBLEMS

1. An agent sells oil for $3,475.00 at Z\% commission. What is the amount of the commission?

2. A merchant buys goods through an agent at a cost of $275.00. The agent charges 2\% commission. What is the total cost of the goods to the merchant?

3. An agent sells a consignment of merchandise for $1,824, retaining his commission of 3%. How much does he remit to his principal?

4. If $302.75 was charged for selling $8,650.00 of merchandise, what was the rate of commission?

5. A realtor's fee for selling a house and lot was $375.00. If the rate was 2^-%, what was the amount received by the principal?

6. An agent's commissions for one week were $216.80. If his sales were $10,840.00, what rate did he charge?

7. The invoice price on a shipment of merchandise was $1,283.38, including agent's commission. If the agent's rate was 3%, what was the commission?

8. The proceeds of a sale received by the principal were $828.78. The commission deducted by the agent was $43.62. What was the rate?

9. The manufacturing cost of a certain type machine is $800.00. The manufacturer wishes to catalog this machine at a list price that will net a profit of 25 % on sales after allowing a dealer's discount of 25 % and agent's commission of 16f %. Find the catalog list price.

10. A collector succeeded in collecting 80% of doubtful accounts, which amounted to $3,295.25. If his commission was 5% of the total he collected, how much did the principal realize?

11. If A remitted $5,297.00 including commission to his agent who purchased goods on a 2% commission, what investment in goods was made in A's name?

Percentage and Applications 73

12. An agent sold a consignment of wheat for $5,280.00 and invested this sum less his commission in corn. His total commission on both transactions amounted to $320.00. If his rate was the same in each case, what was this rate?

13. An agent sold wheat on a commission of \% and bought flour with the net proceeds on a commission of \%. If his total commission was $60.00, what was the cost of the flour?

Cash discount. Cash or time discount is a deduction for immediate payment, or for payment within a definite time. The deduction is a certain per cent of the invoice.

The expression "Terms: 2/10, 1/30, n/60" means that 2% of the in- voice price may be deducted by the purchaser if payment is made within 10 days of the date of the invoice, that 1 % may be deducted if the invoice is paid within 30 days from the date of the invoice, and that the invoice is due in 60 days without discount. In some cases notice is given to the effect that interest at a specified rate will be charged after the due date.

The acceptance of a cash discount is usually of advantage to the pur- chaser. The following table indicates the annual interest rates to which the usual cash discounts are equivalent:

\% 10 days, net 30 days = 9 per cent a year

1 % 10 days, net 30 days = 18 per cent a year \\% 10 days, net 30 days = 27 per cent a year

2% 10 days, net 30 days = 36 per cent a year 2% 10 days, net 60 days = 14.4 per cent a year

2 % 30 days, net 4 months = 8 per cent a year

The rate per cent a year is calculated by taking the number of days between the discount date of payment and the end of the credit period, dividing the number of days in a year (360) by this number, and multi- plying the quotient by the rate of discount under consideration.

zz r r^ =-- X Rate of Discount = Equivalent Annual Interest Rate

Number of Days Between

Discount Date and

End of Credit Period

PROBLEMS

1. Find the equivalent annual interest rate for the following terms:

2% 30 days, net 60 days

3% 10 days, net 30 days

3% 30 days, net 60 days

3% 10 days, net 4 months

2. To pay an invoice of $1,500, with terms 2/10, n/30, the purchaser bor- rowed the money at 6% in order to take advantage of the 2% discount. What benefit did he secure by borrowing the money?

7 A Percentage and Applications

3. A merchant was able to obtain 5% discount on an invoice of $720 by borrowing the money at the bank for 90 days at 6% interest. How much was he able to save?

Trade discount. Mercantile or trade discounts are reductions from list prices, or from the amount of the invoice without regard to time of payment. By offering different rates of trade discounts to wholesalers and retailers, the manufacturer can send the same catalog to both classes of customers. Revised discount sheets are issued as prices fluctuate, but the same catalog may be used a year or more because the list prices are fixed.

Rules of percentage are applied in commercial discounts:

Invoice price = base

Per cent of discount = rate Discount = percentage

Several discounts are sometimes given. These are known as chain discounts or a series of discounts.

The order in which the discounts are deducted will not affect the result; thus, a selling price stated as list price "less 10%, 20%, and 5%" is the same as a selling price stated as list price "less 5%, 20%, and 10%." This is shown in the following example, in which $100.00 is used as the base:

Example.

$100.00 X .10 $10.00 $100.00 X .05 $ 5.00

$100.00 - $10.00 $90.00 $100.00 - $5.00 $95.00

$ 90.00 X .20 $18.00 $ 95.00 X .20 $19.00

$ 90.00 - $18.00 $72.00 $ 95.00 - $19.00 $76.00

$ 72.00 X .05 $ 3.60 $ 76.00 X .10 $ 7.60

$ 72.00 - $3.60 $68.40 $ 76.00 - $7.60 $68.40

$100.00 - $68.40 $31.60 $100.00 - $68.40 $31.60

The total discount in each case is $31.60.

The dollar amount of discount determined from a series of rates is not the same as the amount of discount determined from a single rate equal to the sum of the series of rates. The sum of the series of rates is 35%; 35% of $100.00 is $35.00, whereas the correct discount is $31.60.

Single discount equivalent to a series. Since the rates of the discounts cannot be totalled to find the total discount as shown above, it is necessary to devise a means of evaluating the single discount which would produce the same price as a series of discounts.

A simple formula is derived.

Percentage and Applications 75

Let L = list price

di, d2, . . . , dn = 1st, 2nd, . . . , nth discount rates, respec- tively, expressed as decimal fractions.

Pi, P2, . . . , P„ = 1st, 2nd, . . . , nth. reduced prices, respec- tively.

Then

Pi = L - Ldi

= L(l - di) P2 = Pi - Pid2

= L(l - di) - L{\ - d0d2 = L(1 -dx)(l -eZ2)

Pn-l — P«-2 — Pn-idn-l

= L(l - di)(l - d2) . . . (1 - d„_2) - L(l - rfi) . . . (1 - <4-2R_i = L{\ - d,)(l -d2) . . . (1 - dn_i)

Pn = Pn-l — Pn-ldn

= L(l - di) . . . (1 - rfn_i) - L(l - rfi) . . . (1 - dn_i)dB

= L(1 -di)(l - d2) . . . (1 - dn)

Let d be the single discount rate equivalent to the series dh d2, . . . , dn.

.'. Pn=L -Ld = L(l - d) /. L(l - d) = L(l - 4)(1 - d2) . . . (1— d„) /. 1 - d = (1 - 4)(1 - <&) . . . (1 - dn)

.'. d = i - (i - a)(i - d2) ... (i - dB) (i)

Example 1. What single discount is equivalent to a series of discounts of 20%, 10%, and 10%?

Solution. Here d\ = .2, d2 = .1, and d3 = .1. Substituting these values in the formula, d = i - (1 - rfi)(l - d2)(l - dz) = 1 - (.8)(.9)(.9) = 1 - .648 = .352 .'. single discount is 35.2%.

Example 2. What is the net price of an invoice of $600.00 less 30%, 20%, and 10%?

Solution. Here di = .3, d2 = .2, and d3 = .1.

/. d = 1 - (1 - .3)(1 - ,2)(1 - .1)

= 1 - (.7)(.8)(.9)

= 1 - .504 = .496.

,\ net price = 600.00 - (600.00) (.496)

= 600.00 - 297.60 = $302.40.

Percentage and Applications

PROBLEMS

1. In each of the following, calculate by the short method the single dis- count that is equivalent to the series :

(e) 35% and 10%. (/) 30% and 20%.

(a) 10% and 5%. (c) 40% and 5%.

(b) 20% and 5%. (d) 15% and 10%.

2. In each of the following, find the net price:

(a) $350.00, less 10%, 10%, and 5%. (c) $480.00, less 20%, 10%, and 5%. (6) $500.00, less 33£%, 5%, and 2\ %. (d) $1,200.00, less 5%, 2£%, and 1 %. (e) $900.00, less 50%, 20%, and 5%.

3. The list price of an invoice is $750.00, with discounts of 10%, 5%, and 2\%. The terms of the invoice are: 2/10, 1/30, and n/60. What amount will be necessary to pay the invoice: (a) within the 10-day period; (6) within the 30-day period?

4. B purchases merchandise listed at $3,500.00, less 20% and 25%. He sells this merchandise at the same list price, less 15%, 10%, and 5%. Does he gain or lose, and what amount?

5. A dealer offers merchandise at a list price of $5,000.00, less discounts of 25%, 10%, and 10%. Another dealer offers the same merchandise at a list price of $4,800.00, less discounts of 20%, 15%, and 5%. Which is the better offer, and by what amount?

6. Which is the better offer, and by what amount, on an invoice of $425.00; (a) 30%, 20%, and 10%; or (6) a single discount of 50%?

7. The list price of an item is $24.00. If bought at that price less 33^-% and 10%, and then sold at the same list price less 20% and 5%, what is the profit?

8. The net cost of an invoice of merchandise was $1,200.00. What was the list price, if the cost was 25% and 20% off list?

9. If the list price is $400.00, and the net price is $380.00, what is the single rate of discount?

10. What single discount is equivalent to 25%, 20%, and 12^%?

Transportation charges on discount invoices. In some cases transportation charges are paid by the seller; in other cases, by the pur- chaser. If the purchaser is to pay the transportation charge, and as a matter of convenience the seller prepays it, the seller adds the charge to the invoice. The purchaser is not entitled to cash discount on the added charge.

If a shipment is made "freight allowed," the discount should be figured after the deduction for freight; otherwise, it would be equivalent to taking discount on the transportation charge.

PROBLEMS

1. An invoice of books amounts to $4.85, and parcel post charges are 79 cents, a total of $5.64. If terms are 2/10, what is the discount if paid within the 10-day period?

Percentage and Applications 77

2. Complete the following invoice :

6 doz. Items @$6.65

24 doz. Articles @ .45

6^ doz. Items @ .47^

3 only Items @ 2.34

Less 15%

Less freight allowance

525 lbs. @ .45|- cwt.

Net

If the terms of the above invoice are 1/10, n/30, what will be the discount if paid within 10 days?

3. An invoice for paper, freight allowed, amounted to $1,754.50. The freight bill paid by the purchaser was $238.54. If 2% discount was allowed for pay- ment within 10 days, what was the amount of the check sent within the discount period?

Anticipation. In retail business, invoices for purchases often have dating terms. The terms may be 2/10, 90 days extra. If the merchandise is received within 10 days and checked by the receiving department, the purchaser will deduct 2 % cash discount, and an anticipation discount on the balance computed at 6% (usually) for 90 days, which is equivalent to an additional discount of \\%.

Another case is that of spring purchases of fall merchandise invoiced 2/10, November 1 dating. An invoice with these terms may be dis- counted 2% if paid before November 11, and if paid July 15 would be sub- ject to anticipation discount for 119 days at the customary rate, say, 6%.

Example. An invoice billed April 2, for $3,250.75, terms 2/10, Nov. 1 dating, freight allowed, was paid April 28. Freight paid by the purchaser was $132.48. What was the amount of the check?

Solution.

Invoice $3 , 250 . 75

Less freight 132.48

$3,118.27 Less discount, 2% 62 . 37

$3,055.90

Less anticipation, 6% for 197 days 100.34

Amount of check $2,955.56

PROBLEMS

1. An invoice for $21.25 dated Dec. 28, terms 2/10, Feb. 26, was paid Jan. 7. What was the amount of the check?

2. What is the anticipation on an invoice for $475.50, dated June 10, terms 2/10, 90 days extra, if paid June 25?

3. Find the amount earned by paying an invoice for $1,275.25, dated July 12, terms 2/10, Oct. 1 dating, on July 28.

Simple Interest

Interest, as commonly defined, is a payment for the use of borrowed money or credit. This payment depends upon the rate per cent charged and upon the length of time for which interest is calculated. The sum loaned or the amount of credit used or the capital originally invested is the principal. The number of hundredths of the principal that is paid on the loan is the rate of interest and is usually expressed as a per cent. The length of time over which the principal is used is the time, and the principal plus the interest is the amount.

Simple interest formula. A simple formula for calculating simple interest should now be developed.

Let: P = original principal.

n = time expressed as a fraction of a year. i = interest rate expressed as a decimal fraction. S — amount at the end of time n. I = the total interest in dollars.

Then Therefore, since

and therefore

/ = Pni

S = P + I by definition,

S = P + Pni

S = P(l + ni)

(1)

(2)

Analysis of the simple interest formula. An examination of the simple interest formula (2)

S = P(l + ni)

reveals that it contains four variables, #, P, n, and i. If values are assigned to three of these variables, there remains a linear expression for

Simple Interest 79

the fourth variable which can, therefore, be easily evaluated. Since there are four ways of choosing three things at once from four things, there are four basic problems which can be solved by using the formula:

(a) Given P, n, i, solve to find S.

(b) Given S, n, i, solve to find P.

(d) Given P, S, i, solve to find n.

AMOUNT AT SIMPLE INTEREST

(a) Given P, n, i; to solve and find S.

Example 1. What is the amount of $100.00 for five years at 6%?

Solution. In this case P = 100, i = .06, and n = 5. Substituting these values in the formula,

S = P(\ + ni) = 100[1 + 5(.06)] = 100 (1.30) = $130.00. Example 2. Find the simple interest for two months on $700.00 at 8%.

Solution. In this case P = 700, i = .08, and n = \. Substituting these values in the formula,

/ = pni = 700 X i X .08 = $9.33.

PROBLEMS

1. A man borrows $500.00 for nine years at 4%. What amount of interest will he pay during this period? Write the formula and solution, as shown in the example above.

2. What is the amount of interest due on $300.00 at the end of ten years if the rate is 7%? Write the formula and solution, as shown in the example above.

3. What is the total amount of interest realized on two investments: $400.00 invested for five years at 6% and $750.00 invested for seven years at 4f %?

4. What is the interest accumulation on a debt of $4,270.00 for eight years at 5% simple interest?

Write formulas and solutions for the following:

5. The amount of a $200.00 note due in six years; interest, 5%.

6. The amount due in six years on $530.00 at 6%.

7. The amount due on a note for $750.00 with 4% interest. No interest has been paid during a period of four years.

PRESENT VALUE AND SIMPLE DISCOUNT

(b) Given S, n, i; to solve and find P. This is the problem of find- ing what principal invested now will amount to a certain sum at a definite future time.

80 Simple Interest

Example. What sum invested now at 4|% will amount to $500.00 in three

months?

Solution. In this case S = 500, n = \, % — .045. Substituting these values in

the formula P = — v which is derived from formula (2) by a simple

1+ ra

division,

S 500 500 _

1 + ni 1 + i(.045) ~ 1.01125 ~ UmA*'

The result is commonly called the present value or present worth of $500.00 due in 3 months, money being worth 4-J% per annum.

In the theory of investment, money is never thought to be idle but continuously growing at some interest rate. A sum of money held at present is considered to have future worth (the amount at current interest) and a sum of money to be received in the future is considered to have present worth as shown in the example above. For practical purposes the terms present and future are relative only and may be considered as representative of any two given dates. Therefore, to evaluate a sum on a date later than that on which its value is known, multiply the known value by (1 + ni) ; to evaluate a sum on a date earlier than that on which its value is known, divide the known value by (1 + ni).

(i) To move money forward through time multiply by 1 + ni. (ii) To move money backward through time divide by 1 + ni.

Present value of a debt which bears interest. If any debt bears interest then the maturity value of the debt differs from the face value of the

debt.

If the present value of a debt bearing interest is calculated at the same interest rate, then the present value is the face value. If, however, the interest rate used in calculating the present value is different from the interest rate borne by the debt, there are two steps to the solution:

(i) Find the maturity value of the debt.

(ii) Evaluate the present value of the maturity value at the proper interest rate.

Example. A builder owes $1,000.00 to a wholesaler. His debt matures in three months and bears interest at 7% per annum. What is the present value at 6% per annum?

Solution, (i) To find the maturity value, P = 1,000, n = \, Substitute these values in the formula:

S = P (1 + ni) = 1,000 [1 + i(.07)] = $1,017.50.

Simple Interest

(ii) Now, S = 1,017.50, n = \, and i = .06. Substituting these values in the formula:

P =

1,017.50

1 + ni 1 + i(.06)

= $1,002.46.

Comparison of simple amount and simple present worth. The

following comparative chart is presented to illustrate the accumulation of simple interest on a sum and on the present worth of the same sum:

130

124

118

112

106

Amount ioo

100

Present Worth

76.92

80.65

84.74

89.28

94.34

100

Basis op Amount

Basis op Present Worth

Dec. 31. IstYr.

Dec. 31, 2nd Yr

Dec. 31, 3rd Yr.

Dec. 31. 4th Yr.

Dec. 31, 6 th Yr.

Fig. 4-1.

The amount starts at $100.00, and accumulates to $130.00 in 5 years. The present worth starts at $76.92, and accumulates to $100.00. The rate of interest is the same in each case, 6%.

PROBLEMS

1. What is the present value of a six-year note for $650.00 without interest, if money is worth 5%?

2. A note for $3,500.00, without interest, is due in five years. What is its present value, money being worth 6%?

3. Construct a comparative chart showing the difference in value of the amount and the present worth of $400.00 due in eight years, interest at 5%.

Simple discount or true discount. Simple discount or true discount is the difference between a sum and its earlier value or the difference between the sum due and its present worth. The purpose here is to distinguish simple discount from bank discount, which will be studied later.

82 Simple Interest

In the example above, the simple discount on $500.00 is $500.00 — $494.44 = $5.56. It should be noted that $5.56 is the simple interest on $494.44 for three months at 4^% per annum.

Simple discount formula. Since the present value of S due in time n at

interest rate i is * — ; .> then

1 + ni

Simple discount = S — r — ; . = z — ; .• (3)

1 + m 1 + m

PROBLEMS

1. What is the true discount on a debt of $750.00 due in three years, money being worth 4%? Write the formula and solution.

2. Find the difference between the present value and the face value of a non-interest-bearing note for $500.00, due in four years, money being worth 6%.

3. Find the difference between the true discount and the simple interest on $650.00 for eight years at 4%.

Equivalent discount and interest rates. The equivalent discount rate for a period of time is the discount rate which applied to S for that period of time would produce the same discount as the simple discount for the same period of time as calculated by formula (3). -

The discount on S at interest rate i per period for time n is y— j- — . by

l -j— n%

formula (3). Let d be the equivalent discount rate for the same period. In this way, the discount is Snd.

Therefore, -z — : . = Snd

1 + m

and therefore, d = - — = .• (4)

1 + m

Solve (4) for i and find

' = T^Td (5)

Example 1. A dealer's business nets him 35% interest per annum on money invested in it. What cash discount can he afford to quote for payment of a bill one month before it is due?

Solution. In this case, i = .35, n = ^-. Substituting these values in the formula,

i 35

d = TT^i = 1 + AC85) = -340L

.*. d = 34.01 % interest per annum. /. the discount rate for 1 month = (34.01) X TV = 2.83%.

Simple Interest 83

Alternative solution. Let the payment to settle the bill be $100.00.

/. at 35% the interest on $100.00 for 1 month is 100 X TV X .35 = $2.92.

,'. $100.00 settles a bill of $102.92.

/. $2.92 is the discount on $102.92.

2.92 .*. the rate of discount = ■ = 2.83%.

Example 2. A cash discount of 2% is given if a bill is paid a month before it is due. At what interest rate can a person afford to borrow money for one month to take advantage of the discount?

Solution. A discount of 2% is earned in one month. Therefore, the discount rate for 1 year is .02 X 1 2 = .24.

Therefore, d = .24 and n = xg. Substitute these values in the formula :

• - _A_ = -24 =

1 l-nd 1 - TV(.24) ' D'

Therefore, i = 24^-% per annum.

Alternative solution. Let the bill be $100.00. Then the discount if the bill is paid one month before it is due is $2.00. .'. $98.00 is paid, and therefore $98.00 earns $2.00 in 1 month. .'. the interest rate for 1 month is ^-. .*. the interest rate for 1 year = ¥2¥ X 12 = .245. .*. i = 24^-% per annum.

PROBLEMS

1. Money earns 25% per annum when invested in a certain business. What discount can be offered by this firm on an invoice of $600.00 for payment 40 days before it is due?

2. Find the interest rate per annum which is equivalent to a discount rate of 1 % for payment of a bill 60 days before it is due.

3. The A. B. Hughes Co. estimates that money put into their business yields a net return of 18% per annum. Find the highest discount rate they can afford to allow for payment made one month before the net amount of an invoice is due.

RATE

(c) Given P, S, n; to find i . Since P and S are known in a rate prob- lem, / can be calculated directly &s I = S — P. In this way a rate problem can always be solved by means of the formula (1): / = Pni.

Example. At what interest rate per annum will $100.00 amount to $124.00 in 4 years?

Solution. Here / = 8 - P = $124.00 - $100.00 = $24.00, n = 4, andP = 100.

84 Simple Interest

Substituting these values in the formula,

. _ I_ 24

* " Pn ~ 100 X 4 " m

.*. the interest rate is 6% per annum.

TIME

(d) Given P, S, i; to find n. This solution is identical to the rate solution above, except that we use the formula n = -p-.-

Example. How long will it take $325 to earn $2.60 if invested at 4% per annum?

Solution. In this case, P = 325, / = 2.60, and i = .04. Substituting these values in the formula,

= 2-60 9 = 1

n 325 X .04 5*

Therefore, the time is ^ of a year.

PROBLEMS

Write formulas and solutions for each of the following:

Principal Interest Time Rate 1.$ 400.00 $48.00 3 years

2. 2,000.00 500.00 5 years

3. 800.00 336.00 6 years

4. 300.00 126.00 7 years

5. 150.00 40.50 6 years

6. 1,000.00 240.00 6%

7. 750.00 90.00 4%

8. 360.00 81.00 4i%

EXACT AND ORDINARY SIMPLE INTEREST

In practice, most business transactions involving simple interest are confined to periods of time which are shorter than one year. When eval- uating n as a fraction of a year, there are two methods which are used :

(a) Exact simple interest. This is the method which uses a 365-day year and counts the exact number of days in each month.

(b) Ordinary simple interest. This is the method which uses a 360-day year. If we let Ie denote exact simple interest and I0 denote ordinary simple interest, it is easy to determine the relation between Ie and I0 for :

, Pti , T Pti Ie ~ 365 and Io ~ 360;

where t = number of days.

Simple Interest 85

Therefore ^ = ^X^ = ^? = ^.

ineretore, ^ 365 x pfo. 365 ?3

/. /, = f| Z0 or /0 = if 7e. (6)

Note: In Canada, exact interest and discount are used exclusively. In the United States of America, a 366-day year is used for time involving a leap year, whereas in Canada a 365-day year is used at all times. In the United States of America, three days of grace are not allowed at any time, whereas in Canada three days of grace are always allowed on time drafts and " after sight" drafts as well as on promissory notes but not on demand or "at sight" drafts.

Computing the time for simple interest. If the duration of the loan is given between dates such as from January 29 to March 1, proceed as follows :

Time left in January 2 days.

February 28 days (29 if leap year).

March 1 day.

31 days (32 if leap year).

It should be noted that the first day, January 29, was not included but the last day, March 1, was included.

Methods of computing ordinary simple interest. There are a great many methods of computing simple interest, each of them possessing more or less merit. With the accountant, however, the chief consideration is not how many methods there are, but rather how accurately and how quickly he can solve a problem in interest.

The computation of the product of principal, rate, and time is the short- est method when the time is full years or fractional parts of a year, such as i> i> t> any number of 12ths, and so forth.

Sixty-day method. To find the interest at 6% for:

6 days, point off three additional places to the left of the decimal point

in the principal. 60 days, point off two additional places to the left of the decimal point

in the principal. 600 days, point off one additional place to the left of the decimal point

in the principal.

For 6,000 days, the interest will be the same as the principal.

Justification: $ X 6 X '°6 = $ x .001

360

j X 60 X .06 _ ;

360 ~%X M

$ X 600 X .06

360 "" ' X 'l

$ X 6,000 X .06

360 =®X1

Simple Interest

Example. Find the interest on $256.75 for six days at 6%.

Solution. Pointing off three places to the left of the decimal point in the principal gives 25675, or 26^.

Example. Find the interest on $345.65 for 36 days at 6%.

Solution. Point off three additional places to the left of the decimal point in the principal, and multiply by 6. The answer is $2.07.

This method may be used for any rate by adding to or subtracting from the interest computed at 6%, the fractional part thereof that the specified rate is greater or less than the 6% rate.

For 8 %, increase the interest by -^ of the amount computed at 6 %

7 %, increase the interest by ^ of the amount computed at 6 %

5%, decrease the interest by ^ of the amount computed at 6%

4%, decrease the interest by ^ of the amount computed at 6%

4^-%, decrease the interest by ^ of the amount computed at 6%

PROBLEMS

Find the interest at 6% on.

1. $180.00 for 60 days.

2. $150.00 for 54 days.

3. $262.50 for 24 days.

4. $32.75 for 36 days.

5. $65.50 for 12 days.

6. $26.50 for 18 days

7. $752.25 for 6 days

8. $15.80 for 54 days

9. $75.40 for 30 days 10. $12.85 for 24 days

Method using aliquot parts. Example. Find the interest on $275.84 for 124 days at 6%.

Solution.

$2 2

75 . 84 = interest for 60 days 75 . 84 = interest for 60 days 18 . 38 = interest for 4 days 70.06 = interest for 124 days

Explanation. Pointing off two decimals, as indicated by the vertical line, gives the interest for 60 days. Double this to find the interest for 120 days. Four days' interest is y1^ of 60 days' interest. The sum, $5.70, is the interest for 124 days.

Example. Find the interest on $754.90 for 137 days at 6%.

Solution.

54 . 90 = interest for 60 days

$17

54 . 90 = interest for 60 days 50.98 = interest for 12 days 62 . 90 = interest for 5 days

23.68 = interest for 137 days

Simple Interest 87

Explanation. Pointing off two decimal places gives the interest for 60 days. Double this to find the interest for 120 days. Twelve days is ^ of 60 days; therefore, the interest for 12 days is ^ of 60 days' interest, or $1.5098. Five days is ^ of 60 days, and the interest is ^ of $7.5490, or $0,629. The sum, $17.24, is the interest for 137 days.

After a little practice, any number of days can be resolved into 6- or 60-day periods and easy fractions thereof.

Example. Find the interest on $247.64 for eight days at 6%.

Solution.

247 . 64 — interest for 6 days 082 . 54 = interest for 2 days

330 . 18 = interest for 8 days

Explanation. To find the interest for six days, point off three decimals, as indi- cated by the vertical line. Two days' interest is ^ of six days' interest. The answer is, therefore, 33^.

For rates other than 6%, see adjustments under previous method.

PROBLEMS

Find the interest at 6% on:

1. $286.75 for 9 days. 6. $175.82 for 34 days.

2. $189.22 for 8 days. 7. $38.95 for 19 days.

3. $256.35 for 27 days. 8. $47.56 for 17 days.

4. $178.56 for 39 days. 9. $29.10 for 2 days.

5. $38.29 for 40 days. 10. $1,286.75 for 21 days.

The cancellation method. The cancellation method may be used to advantage in many interest calculations, especially in those having fractional rates and rates other than 6%.

Example. Find the interest on $750.00 for 45 days at 5%.

Solution.

125 15 .01

im xigx .m = 18.75

X2 X 30 4

4 0

4.687, or $4.69

Explanation. Writing below the line 12 times 30, instead of 360 days, facilitates cancellation.

Example. Find the interest on $345.75 for 90 days at 4^-%.

Simple Interest

Solution.

345.75 X 00 X .09 31.1175

n X 30 X 2 4

= $3,889, or $3.89

PROBLEMS

Find the interest, by the cancellation method, on:

1. $840.00 for 12 days at 2%.

2. $320.00 for 15 days at 4%.

3. $160.80 for 16 days at 5%.

4. $275.75 for 74 days at 6%.

5. $112.50 for 85 days at 4%.

6. $284.00 for 34 days at 6%.

7. $368.00 for 56 days at 5%.

8. $775.14 for 79 days at 5%.

9. $250.00 for 91 days at 6%.

10. $500.00 for 102 days at 4%.

11. $360.80 for 38 days at 4^%.

12. $312.32 for 45 days at 4|%.

13. $1,000.00 for 40 days at 5±-%.

14. $1,600.00 for 75 days at 4i%.

Dollars-times-days method, 6%. This method is rapid, and is particularly valuable when a calculating machine is used. It is a modi- fication of the cancellation method, where 6% and 360 days are two of the factors. Thus:

$ X Days X .06

300 6,000

Assume that there are no other items that can be cancelled. The number of dollars is multiplied by the number of days, and the product divided by 6,000. Any number may be divided by 6,000 by pointing off three decimals, and dividing the resultant number by 6.

Example. Find the interest on $256.50 for 28 days at 6%.

Solution. Multiply the number of dollars by the number of days, point off three decimal places in addition to the number of decimal places in the principal, then divide by 6.

$256.50

6)7.182 00

1.197 or $1.20, the interest.

For rates other than 6%, see adjustments under " Sixty-day method."

PROBLEMS

Find the interest on the following:

1. $275.12 for 73 days at 5%.

2. $132.86 for 28 days at 8%.

3. $280.60 for 70 days at 4%.

4. $138.42 for 28 days at 4J%.

5. $276.95 for 17 days at 8%.

6. $640.64 for 56 days at 7%.

Simple Interest 89

Interchanging principal and time. Under the 60-day method, the computations may often be shortened by interchanging the principal and the time.

Example. Find the interest on $6,000.00 for 31 days at 6%.

Solution. Interchanging the principal and the time, the problem becomes that of finding the interest on $31.00 for 6,000 days. Apply the 6%, 60-day method, and the interest is found to be $31.00, since the interest is equal to the principal when the rate is 6% and the time is 6,000 days.

PROBLEMS

Find the interest on the following:

1. $2,400.00 for 23 days at 6%. 4. $3,000.00 for 193 days at 6%.

2. $3,600.00 for 7 days at 6%. 5. $4,500.00 for 38 days at 6%.

3. $6,000.00 for 156 days at 6%. 6. $4,200.00 for 41 days at 6%.

Exact or accurate interest. Exact or accurate interest is that which is obtained when a year is taken as 365 days. For full years, all methods of computing interest give the same result — a certain per cent of the principal ; hence the results differ only when fractional parts of a year are used.

Example. Find the exact interest on $1,200.00 for 93 days at 6%.

Solution. The cancellation method previously explained is the method used, as it is probably the most practical.

240 Z200 X 93 X .06 = 1,339.20

300 ~ 73

= $18.35

PROBLEMS

Find the exact interest on :

1. $750.00 for 45 days at 6%.

2. $1,200.00 for 68 days at 7%.

3. $1,600.00 for 73 days at 6±-%.

4. $273.00 at 4% from May 12, 1961 to October 4, 1961.

5. $1,600.00 at 2|-% from June 3, 1961 to December 17, 1961.

6. $1,103.00 at 6% from May 17, 1962 to September 9, 1962.

7. $12,235.00 at 4^% from April 19, 1960 to August 20, 1960.

8. If the ordinary interest is $104.15, what is the exact interest?

9. What is the ordinary interest if the exact interest is $107.23?

Simple Interest

Promissory notes. A promissory note is a written promise to pay on demand or at a stipulated time a sum of money to settle an obligation. It is signed by the debtor, who is called the maker of the note.

! $2,000.00

Hamilton

,Ont., May 17, 1962

Ninety days

after date

promise

to pay

to the order of Two thousand

David

J. Jones

- _ _ _ 100 Dollars

! for value received with interest at 6%

per annum.

Kenneth Roy Brown

Fig. 4-2. A Promissory Note.

Drafts. A draft is a written order, signed by the creditor, called the drawer, requiring the debtor, called the drawee, to pay on demand or at a fixed or specified time in the future, a stipulated amount. The drawer sends it to the drawee for acceptance and the drawee writes "Accepted" across it, along with his signature, and returns it to the drawer. The draft does not bear interest. If the draft is an "after date" draft, the term begins with the date on the draft; if it is an "after sight" draft, the term begins with the date of acceptance.

Note to Canadian readers. Three days' grace is allowed on promissory notes, time drafts, and "after sight" drafts, but not on demand or "at sight" drafts.

$1,000.00

Hamilton, Ont. April 17, 1961

Ninety days after date pay to the order of Myself

One thousand ___ 100 Dollars

Value received

and charge the same to account of Kenneth Roy Brown

To Samuel R. Green Accepted

Toronto, Ontario Payable at Bank of Commerce

Toronto, Ontario

Fig. 4-3. "After Date" Draft.

Simple Interest 91

$5,000.00

Hamilton,

Ont. Nov.

24, 1961

Ninety days <

after si

ght

pay to the order of

Myself

Five thousand -

100 Dollars Fohn Gray

Value received and charge

the same

to account of

Douglas c

> To Samuel R

. Green

Accepted

Toronto,

Ontario

Payable at Bank of Toronto

Commerce , Ontario

Fig. 4-4. "After Sight" Draft.

Trade Acceptance. A trade acceptance is a draft arising from the sale of goods and is drawn by the seller of the goods on the purchaser. A trade acceptance bears no interest.

Buffalo, N.Y., Jan. 29, 1961 $2,317.45

ninety days after date pay to the order of OURSELVES

45 Two thousand three hundred seventeen and 100 Dollars

This obligation of the acceptor hereof is in payment of goods purchased from the drawer. The acceptor may make the accept- ance payable at any bank or trust company in the United States.

To N. L. White Artistic Drapery Co.

Columbus, Ohio by J. E. L. Graham, Manager

Fig. 4-5. Trade Acceptance.

BANK DISCOUNT

Definition. Bank discount is a deduction made from the amount due at maturity on a note or draft, in consideration of its being converted into cash before maturity. If the note does not bear interest, its face value is the amount due at maturity. If the note does bear interest, the amount due at maturity is the face value plus interest on the face value for the period and at the rate specified in the note.

Simple Interest

In bank discount, the time is the period from the date of discount to the date of maturity of the note. The date of maturity of a note is the day on which it is due. Notes due a given number of days after date mature after the exact number of days have elapsed. Notes due a given number of months after date mature on the same date so many months hence, except notes made on the 31st and falling due in a 30-day month, which mature on the 30th, and notes made on the 29th, 30th, or 31st of some month and falling due in February, which mature on the last day of February.

Example. A note due 30 days after January 31 will mature on March 2; but if the note is due in one month, it will mature on the last day of the succeeding month, or February 28. If the year should be a leap year, the maturity dates would be March 1 and February 29.

Counting time. In counting time, the usual method is to count the first succeeding day as one day. To illustrate, if a note is given on Jan- uary 15 for 10 days, the 16th is counted as the first, the 17th as the second, the 18th as the third, and January 25 as the tenth day.

Finding the difference between dates by use of a table. By numbering the days of the year, a calendar may be made for determining the number of days between any two dates. A portion of such a table, and the use made of it, are illustrated now.

May 1 121 Nov. 1

2 122 2

3 123 3

4 124 4

5 125 5

6 126 6

7 127 7

8 128 8

9 . 129 9

10 130 10

305 306 307 308 309 310 311 312 313 314

The number of days between May 4 and November 9 is found as follows :

The table shows that November 9 is the 313th day of the year The table shows that May 4 is the 124th day of the year

Therefore, the difference is 189 days, the time required

Another form of table is one that shows the number of days from any day of any month to the corresponding day of any other month not more than one year later.

Simple Interest 93

Jan. Feb. Mar. Apr. May June July Aug. Sept. Oct. Nov. Dec.

January 365 31 59 90 120 151 181 212 243 273 304 334

February 334 365 28 59 89 120 150 181 212 242 273 303

March 306 337 365 31 61 92 122 153 184 214 245 275

April 275 306 334 365 30 61 91 122 153 183 214 244

May 245 276 304 335 365 31 61 92 123 153 184 214

June 214 245 273 304 334 365 30 61 92 122 153 183

July 184 215 243 274 304 335 365 31 62 92 123 153

August 153 184 212 243 273 304 334 365 31 61 92 122

September 122 153 181 212 242 273 303 334 365 30 61 91

October.... 92 123 151 182 212 243 273 304 335 365 31 61

November 61 92 120 151 181 212 242 273 304 334 365 30

December 31 62 90 121 151 182 212 243 274 304 335 365

Example. A note due August 17 was discounted June 10. What was the term of discount?

Solution.

From the table, June 10 to August 10 61 days

August 10 to August 17 7 days

Total 68 days

Exact bank discount is evaluated when a 365-day year is used and the exact number of days in each month is counted. This is the only method permitted in Canada, and the three days' grace must be included in the calculations.

Ordinary bank discount is evaluated when a 360-day year is used. This is the method which is always used in the United States.

Proceeds. The proceeds of a note is the difference between the amount due at maturity and the bank discount. Bank discount and proceeds formulas. Let S = maturity value.

n = time expressed as a fraction of a year. d = discount rate expressed as a decimal fraction. Db = bank discount. Pb = proceeds. Then Db = Snd (7)

but Pb = S - Snd by def"

:. Pb = 8(1 - nd) (8)

Example 1. Find the bank discount and the proceeds if a non-interest-bearing note for $420.00 is discounted 90 days before it is due at 6%.

Solution (United States). In this case S = 420.00, n = -££$, and d = .06. Sub- stituting these values in formula (7),

Db = Snd = (420.00) (^%%) (.06) = $6.30 :. Pb = $420.00 - $6.30 = $413.70.

Solution (Canada). In this case S = 420.00, n = ^\\, and d = .06. Substitut- ing these values in formula (7),

Db = Snd = (420.00)(^V)(.06) = $6.42, .*. Pb = $420.00 - $6.42 = $413.58.

94 Simple Interest

Example 2. A note for $780.00 dated May 17, payable in six months with interest at 6%, is discounted at 6% on August 15. Find the bank discount and the proceeds.

Solution (United States). Maturity value of note = 780[1 + i(.06)] = $803.40

Db = (803.40) (JfoX.06) = -12.59 /. Pb = $803.40 - $12.59 = $790.81

Solution (Canada). Maturity value of note = 780[1 + |(.06)] = $803.40

Db = (803.40) (^5 ) (.06) = $12.81 Pb = $803.40 - $12.81 = $790.59

PROBLEMS

Students in the United States should find the ordinary bank discount and students in Canada should find the exact bank discount, remembering to add the three days' grace. It would be good practice, however, to try some questions by both methods.

Find the bank discount and the proceeds on :

1. A note for $750.00, due May 17 without interest, and discounted Sept. 9 at 6%.

2. A note for $1,200.00, due December 4 without interest, and discounted Oct. 29 at 6%.

3. A note for $1,500.00 dated October 8 and due in 4 months with interest at 6%, discounted December 1 at 6%.

4. A note for $800.00, dated September 9 and due in 6 months with interest at 7%, discounted November 11 at 6%.

5. A note for $250.00, dated July 11 and due in 90 days with interest at BJ-%, discounted September 1 at 6%.

6. $443.03 was received as the proceeds of a 90-day note discounted at 6%. What was the face of the note?

7. For what sum must a 60-day note be drawn in order that the proceeds will be $600.00 when the note is discounted at 6%?

8. Find the date of maturity, the term of discount, the bank discount, and the proceeds of a 60-day note for $750.00, dated July 8 and discounted July 17 at 5%.

9. Find the date of maturity and the term of discount of a 90-day sight draft, dated May 14, accepted May 17, and discounted June 10.

10. Find the date of maturity, the term of discount, the bank discount, and the proceeds of a note for $650.00, dated November 30, due in three months, and discounted January 5 at 6%.

PARTIAL PAYMENTS

Part payments on debts. A debtor may by agreement make equal or unequal payments on the principal at regular or irregular intervals. Any partial payment of a note or draft should be recorded on the back of the note or draft.

Methods. There are two methods of applying payments of principal and interest to the reduction of an interest-bearing debt. The method

Simple Interest 95

adopted by the Supreme Court of the United States is termed the " United States Rule"; the other method, which has been widely adopted by busi- nessmen, is termed the " Merchants' Rule."

United States Rule. The United States Rule is now a law in many of the states, having been made so either by statute or by court decision.

The court holds that when a part payment is made on an interest- bearing debt, the payment must first be used to discharge the accumu- lated interest, and what remains of the payment is then applied in cancel- lation of the principal. If the payment is smaller than the accumulated interest, no cancellation takes place, and the previous principal continues to draw interest until the accumulated payments exceed the accumulated interest.

Example. An interest-bearing note for $1,800.00 dated March 1, 1944, had the following indorsements :

September 27, 1959 $500.00

March 15, 1960 25.00

June 1, 1960. .-. $700.00

How much was due September 1, 1960?

Solution.

Yr.

Date of note 1959

First payment, $500.00 1959

Second payment, $25.00 1960

Third payment, $700.00 1960

Settlement 1960

1 _6 0

Explanation. The time is found by successive subtractions of the first date from the second, the second from the third, and so on. The sum of the different times is equal to the time between the date of the note and the date of settlement.

Face of note, March 1, 1959 $1 , 800 . 00

Interest on $1,800.00 at 6% from March 1 to Sept. 27, 6 months and

26 days 61.80

Amount due Sept. 27, 1959 $1,861 .80

Deduct payment 500 . 00

Balance due Sept. 27, 1959 $1,361.80

Interest on $1,361.80 at 6% from Sept. 27 to March 15, 5 months and

18 days, $38.13. As this interest is larger in amount than the payment

made at March 15, the interest is not added and the payment is not

deducted. Interest on $1,361.80 at 6% from Sept. 27 to June 1, 1960, 8 months and

4 days 55.38

Amount due June 1, 1960 $1 , 417 . 18

Deduct sum of payments : March 15 $ 25 . 00

June 1 700.00 725.00

Balance due June 1, 1960 $ 692 . 18

Interest on $692.18 at 6% from June 1 to Sept. 1, 1960, 3 months 10.38

Balance due September 1, 1960 $ 702 . 56

lo.

Da.

Yrs.

Mos.

Days

Simple Interest

Merchants' Rule. Find the amount of the debt (principal and interest) to the date of final settlement, or, if the debt runs for more than one year, find the amount to the end of the first year. Deduct from this the sum of all the payments and interest on the same to the date of settle- ment or to the end of the year. The remainder will be the amount due at the date of settlement or at the beginning of the next year.

Example. For purposes of comparison, the same problem will be used here as was used to illustrate the United States Rule.

Solution.

Face of note, March 1, 1959 $1 , 800 . 00

Interest, 1 year at 6% to March 1, 1960 108.00

$1,908.00 Deduct:

First payment, Sept. 27, 1959 $500 .00

Interest at 6% to March 1, 1960, 5 months and 4 days 12.83 512.83

Balance due at beginning of second year $1 ,395 . 17

Interest on $1,395.17 at 6%, March 1 to Sept. 1, 1960, 6 months _ 41.86

$1,437.03 Deduct:

Second payment, March 15, 1960 $ 25.00

Interest at 6% from March 15 to Sept. 1, 1960, 5 months and

16 days .69

Third payment, June 1, 1960 700.00

Interest at 6% from June 1 to Sept. 1, 1960, 3 months 10.50 736.19

Balance due $ 700 . 84

The difference of $1.72 between the balance as computed by the Mer- chants' Rule and the balance as computed by the United States Rule is small, but a much greater difference will occur when the time is long and the amount large.

It is usual to compute the balance due on notes of one year or less by the Merchants' Rule.

Note to Canadian readers. In Canada both the Merchants' Rule and the United States Rule would be applicable using exact simple interest rather than ordinary simple interest as used in the examples.

PROBLEMS

(Readers in the United States should use ordinary simple interest for com- putations and readers in Canada should use exact simple interest for their computations.)

Solve each of the following problems by both the United States Rule and the Merchants' Rule and compare the results.

1. A note for $1,650.00 was dated May 20, 1959. The interest was 6% from date, and the following payments were indorsed:

Simple Interest 97

Sept. 8, 1959 $ 45.00

Dec. 14, 1959 20.00

Feb. 26, 1960 50.00

July 5, 1960 90.00

Nov. 14, 1960 250.00

What amount was due December 17, 1960?

2. A note for $1,200.00 is dated June 20, 1961. The interest was 6% from date, and the following payments are indorsed:

Oct. 2, 1961 $120 . 40

February 8, 1962 29 . 50

May 23, 1962 56.40

December 11, 1962 388.75

What amount is due January 23, 1963?

3. A note for $1,000.00 was dated April 10, 1958. The interest is 7% from date, and the following payments are made:

November 10, 1959 $ 80.50

July 5, 1960 100.00

January 10, 1961 450 . 80

October 1, 1963 500.00

What amount is due January 1, 1964?

4. A note for $950.00 with interest at 6% was dated Feb. 3, 1960, and has the following indorsements:

March 1, 1960 $150.00

June 3, 1960 96.00

July 8, 1960 300.00

December 20, 1960 250.00

What amount is due January 17, 1961?

5. A note for $791.84 with interest at 6% was dated December 14, 1960, and bore the following indorsements:

January 3, 1961 $100.00

March 16, 1961 240.00

July 29, 1961 324.00

August 3, 1961 20.00

What amount is due November 14, 1961?

6. A note for $1,200.00 dated April 1, 1960, bore interest at 7% and had the following indorsements:

April 12, 1960 $161.08

July 19, 1960 224.14

July 28, 1960 17.90

January 29, 1961 100 . 25

What amount is due April 1, 1961?

AVERAGING DATES OF INVOICES

Definition. Averaging dates of invoices is the process of finding the date when several invoices due at different dates may be paid in one

98 Simple Interest

amount, without loss of interest to either debtor or creditor. This date is called the equated date of payment.

Use. The process of averaging the dates of invoices is most fre- quently used in bankruptcy settlements, where claims when filed with a trustee must show the average due date of the items if interest is to be obtained on overdue amounts. In general, the equated date is important in the settlement of bills of long standing and in the fixing of the date of a note in settlement of invoices.

Term of credit. A term of credit is the time elapsing between the date of a bill and the date on which it becomes due; as, "Bill purchased January 10, Term of Credit 10 days." The due date would be January 20.

Average due date. The average due date is the date on which settle- ment of the complete account should be made by payment of the amount of the invoices, without charge for interest on overdue items or allowance for discount on prepaid items.

Focal date. The focal date is an assumed date of settlement with which the due dates of the several items may be compared, to determine the equated date of payment.

Any date may be used as the focal date, and the final result will be the same. In the interest method, any rate per cent may be used, and the result will be the same. However, 6% is usually used, as the computa- tions are then less complicated.

In all calculations, use the nearest dollar. For example, for $115.29, use $115.00; and for $161.84, use $162.00.

When several bills are sold, some of which have a term of credit, first find the due date of those with a term of credit, and then find the equated date of the several bills.

With respect to bills with terms of credit, the due date of such bills, rather than the invoice date, is used in computing the equated date.

Do not use fractions of a day in determining the average date.

Methods. There are two methods in common use: the Product Method, and the Interest Method.

Rule for product method. Use as the focal date the last day of the month preceding the first item. Multiply each item by the number of days intervening between the assumed date and the due date of the item, and divide the sum of the several products by the sum of the account. Count forward from the assumed date the number of days obtained in the quotient. The result will be the average due date.

Example. Find the date at which the following bills of merchandise may be paid in one amount without loss to either party: Due January 1, $150.00; February 14, $200.00; April 20, $155.00; June 15, $200.00.

Simple Interest 99

Solution by Product Method

(Focal date, Dec. 31)

Due January 1 $150 X 1 = 150

Due February 14 200 X 45 = 9,000

Due April 20 155 X 110 = 17,050

Due June 15 200 X 166 = 33,200

705 59,400

59,400 -*- 705 = 84 days. 84 days after December 31 is March 25.

Explanation. For convenience, assume December 31 as the date of settlement. On the first bill, which is due January 1, there would be interest for 1 day. On the second bill there would be interest from December 31 to February 14, or 45 days, which is equivalent to interest on $9,000.00 for 1 day. On the third bill there would be interest from December 31 to April 20, or 110 days, which is equivalent to interest on $17,050.00 for 1 day. On the fourth bill there would be interest from December 31 to June 15, or 166 days, which is equivalent to interest on $33,200.00 for 1 day.

If all the bills were paid December 31, the debtor would be entitled to interest on $59,400.00 for 1 day, or interest on $705.00, the amount of the account, for 84 days. It is evident that the bills could be paid at a time 84 days later than December 31, or March 25, without loss to either party.

Verification

The interest on $150.00 for 83 days is $2 .08

The interest on $200.00 for 39 days is 1.30

Total gain of interest to debtor $3.38

The interest on $155.00 for 26 days is $ .67

The interest on $200.00 for 82 days is 2.72

Total gain of interest to creditor $3 . 39

The gain of interest to the debtor is on all bills paid after they are due.

The gain of interest to the creditor is on all bills paid before they are due. These two results should be equal, or within a few cents of the same amount. The reason for a little discrepancy is the fraction of a day which is disregarded in determining the due date of the account.

Interest Method

The solution by this method is as follows:

January 1 $150.00 for 1 day = $0.03, interest

February 14 200.00 for 45 days = 1 .50, interest

April 20 155 .00 for 110 days = 2 .84, interest

June 15 200 .00 for 166 days = 5.53, interest

Total interest $9 . 90

Interest on $705.00 for 1 day is $0.1175.

$9.90 -=- $0.1175 = 84, or 84 days.

Explanation. Assume December 31, the last day of the month preceding the first item, to be the date of settlement. If the amount of the account, $705.00, is paid December 31, there will be a loss of interest to the debtor

100 Simple Interest

of $9.90. The interest on $705.00 for 1 day at 6% is $0.1175. It will take a principal of $705.00 as many days to produce $9.90 as the number of times that $0.1175 is contained in $9.90, or 84 days, the same result as was obtained by the product method.

Note to Canadian readers. It is to be remembered that ordinary simple interest is used in all calculations in this section, but in Canada exact simple interest would be used. Otherwise, the methods are iden- tical.

PROBLEMS

1. Several invoices mature as follows:

April 12 $260.00 August 18 $120.00

May 25 500.00 September 2 300.00

At what date may the foregoing invoices be paid in one amount without loss to either party?

2. Calculate the average due date of the following invoices:

June 10 $400.00 August 15 $250.00

July 27 100.00 September 22 300.00

3. Calculate the average due date of the following invoices:

May 8 $275.00 on 60 days' credit

May 24 150.00 on 2 months' credit

June 10 300.00 on 90 days' credit

July 1 250 . 00 on 30 days' credit

EQUATION OF ACCOUNTS, OR COMPOUND AVERAGE

Definition. Equation of accounts, or compound average, is the process of finding the date when the balance of an account having both debits and credits can be paid without loss to either debtor or creditor. With respect to bills with terms of credit, the due date of the bill rather than the invoice date is used in computing the equated date. Credits other than for cash (such as noninterest-bearing notes) are extended to the due date thereof. Summarizing briefly, the date to be used for each debit and credit is the date when the item has a cash value of the amount shown in the entry.

Rule for the product method. After finding the date that each item has a cash value, use the last day of the month preceding the earliest date as the focal date for both sides of the account. Find the number of days between the focal date and the due date of each item ; multiply each

Simple Interest 101

item by the number of days intervening between the focal date and the due date of the item. Find the sum of the products on both the debit and the credit sides of the account. Divide the difference between the sums of the debit and the credit products by the balance of the account. The quotient will be the number of days between the focal date and the average date of the account.

When to date forward or backward. The average date is forward from the focal date when the balance of the account and the excess of the products are on the same side (both debits or both credits) ; if they are on opposite sides, the average date is backward from the focal date.

Example. At what date may the balance of the following account be paid without loss of interest to either party?

Debits Credits

July 1. Mdse. 30 days . $250 . 00 Aug. 15, Cash $400 . 00

July 26, Mdse. 30 days 425.00 Sept. 10, Cash 300.00

Aug. 15, Mdse. 60 days 320.00 Sept. 20, Cash 150.00

Aug. 30, Mdse. 60 days 500 . 00

Solution by Product Method. Since the earliest date is July 31, the assumed focal date would be June 30. However, since July 31 is an end-of-month date, this date is used, as each multiplier is 31 less than it would be if June 30 were used.

July 31, $ 250.00 X 0 = $00,000 Aug. 15, $400.00 X 15 = $ 6,000

Aug. 25, 425.00 X 25 = 10,625 Sept. 10, 300.00 X 41 = 12,300

Oct. 14, 320.00X75= 24,000 Sept. 20, 150.00X51= 7,650

Oct. 29, 500.00 X 90 = 45,000

$1,495.00 $79,625 $850.00 $25,950

Debit side $1,495.00 $79,625.00

Credit side 850 . 00 25,950.00

$ 645.00 $53,675.00

$53,675.00 -J- $645.00 = 83.

The equated date is, therefore, 83 days after July 31, or October 22.

Explanation. First find the due date of each item. For convenience, assume July 31, the earliest due date, as the day of settlement for all the items on each side of the account. Proceed as in the process of averaging dates, which was described in the preceding chapter. With July 31 used as the focal date, there is a loss of interest on the total debits equivalent to the interest on $79,625.00 for 1 day, and a gain of interest on the total credits equivalent to the interest on $25,950.00 for 1 day; or a net loss of interest equivalent to the interest on $53,675.00 for 1 day, which is equal to the interest on $645.00 for 83 days. It is evident that the date when there would be no loss of interest to either party must be 83 days after July 31, or October 22.

102

Simple Interest

Solution by Interest Method.

Debits

July 31, 0 days' interest on $ 250.00 = $ .00

Aug. 25, 25 days' interest on 425.00 = 1.77

Oct. 14, 75 days' interest on 320.00= 4.00

Oct. 29, 90 days' interest on 500.00 = 7.50

$1,495.00 $13.27

Credits Aug. 15, 15 days' interest on $ Sept. 10, 41 days' interest on Sept. 20, 51 days' interest on

400.00 = $ 1.00 300.00 = 2.05 150.00 = 1.28

Dr. $1,495.00 Cr. 850.00

6000) 645.00

$ 850.00 $ 4.33

Interest, $13.27

Interest, 4 . 33

$ 8.94

$ . 1075, interest for one day. $8.94 ^ .1075 = 83.

The equated date of payment is, therefore, 83 days after July 31, or October 22.

Explanation. The explanation of the product method is applicable to the interest method. The variance is in finding interest on each item and dividing the interest balance by the interest for 1 day on the balance of the account.

Note to Canadian readers. Perform the same operations, using exact simple interest instead of ordinary simple interest.

PROBLEMS

Find the equated date in each of the following:

Debits

June 10, Mdse $500 . 00

Aug. 20, Mdse 100.00

Oct. 30, Mdse 250.00

Debits

May 3. Mdse. 60 days $300.00

June 15, Mdse. 60 days 250.00

July 20, Mdse. 30 days 175 . 00

Aug. 27, Mdse. 60 days 225 . 00

Debits

Mar. 1, Mdse. 30 days $225.00

Mar. 20, Mdse. 2 mos 300 . 00

Apr. 5, Mdse. 60 days 150 . 00

Credits

July 5, Cash $300.00

Aug. 10, Cash 150.00

Sept. 25, Cash 200.00

Credits

June 20, Cash $150.00

July 1, Note, 30 days without

int 200 . 00

Aug. 10, Note, 60 days, int., 6% 300 . 00

Credits

Mar. 31, Cash $150.00

Apr. 15, Cash 100.00

May 10, Cash 200.00

INSTALLMENT PLANS AND PERSONAL LOANS

In recent years installment buying has increased rapidly, as has the practice of taking small personal loans from finance companies. Many

Simple Interest

103

purchases of household appliances, furniture, and automobiles are financed by the seller on an installment plan or are financed through an acceptance corporation. It has become important, therefore, to know how ^o calculate the equivalent rate of simple interest that is actually Joeing charged, since it is quite different from the implied rate. Equal monthly installments plan formula.

Let R = the balance owing immediately after the down payment has been made. / = the interest or carrying charges, m = the number of monthly installments. i = the equivalent rate per annum of simple interest. a = the amount of each monthly installment.

1st 2nd 3rd 4th 5th pay- pay- pay- pay- pay- ment ment ment ment ment

(m — l)st mth

payment payment

Obviously, from the chart, the amount of the principal P at rate i at the end of m months must be equal to the sum of the amounts of the installments at rate i at the right-hand end of the time scale.

••.p.(i + 5*)-.-+.(i + ^) + .(i + §) + --"-

d I D m • . i ai i , o ai .

, (m — 2) at . . (m — \)ai + a + ,rt + a +

7Y\ fit

;. P + P — i = ma + -I [1 + 2 + • • • + (m - 2) + (m - 1)]

1 04 Simple Interest

The terms in the square brackets form an arithmetic series whose sum is

m(m — 1)

_ m . . ra(ra — l)<n ••P+Pl2^ma + 24

Since the sum of the installments is equal to the balance P plus the carrying charges 7, we have

P + I

ma = P + I or a =

:.P+P™i = m[ :__l_^ ] 4

m m(m - 1) (l±l)

12 \ m J 24

• P+P™i = P + I + (m~ 1)(p + /^ " 12 24

m . __ , mPz — Pi + (m — l)7i

" 12 * " + 24~

.*. 2Pmi = 247 + Pmi - Pi + (m - l)Ii

/. P(m + l)t - (m - l)/t = 24/

/. {(m + 1)P - (m - 1)7 }* = 247

. .= 247

" l (m + 1)P - (m - 1)7 U;

Example 1. A washing machine which is listed as having a cash price of $120.00 may be purchased for $25.00 down and $20.00 per month for five months. What is the actual interest rate per annum?

Solution. In this case P = 95, 7 = 5, and m = 5. Substituting these values in the formula,

i = 247 . 24X5 = 120 =

(m + 1)P - (m - 1)7 6 X 95 - 4 X 5 550 '

.*. The equivalent simple interest rate is 21.82% per annum.

Example 2. An article listed at $500.00 is sold for $50.00 down, plus a series of 12 monthly payments of $40.00 each, covering the $450.00 unpaid balance and the $30.00 installment charge. Find the equivalent simple interest rate per annum.

Solution. P = $480.00, m = 12, and 7 = $30.00.

247 24 X 30

= .1218

(m + 1)P - (m - 1)7 13 X 480 - 11 X 30

.'. The equivalent simple interest rate is 12.18% per annum.

Example 3. A record player is worth $150.00 but is purchased on the following budget scheme. A down payment of $30.00 and an installment charge of $10.00, both paid at time of purchase, and ten monthly installments of $11.50 each. What is the equivalent simple interest rate?

Simple Interest 105

Solution. P = $110.00

/ = $115.00 - $110 = $5.00 m = 10 ' = 247 = 24X5 =

1 (ro + 1)P - ■ (to - 1)1 11 X 110 - 9 X 5 '

.*. the equivalent simple interest rate is 10.3% per annum.

PROBLEMS

1. An article selling for $60.00 can be purchased on the installment plan with seven monthly installments of $9.00 each. Find the equivalent simple interest rate per annum.

2. A radio can be purchased for $100.00 cash or $6 down and ten monthly payments of $10.00 each. What is the equivalent simple interest rate per annum?

3. A fur coat is listed at $900.00 or $81.00 down and $50.00 per month for one and one-half years. Find the equivalent simple interest rate per annum.

4. A $4,000.00 mortgage is to be repaid over a period of four years through monthly payments of $100.00 each, such payments covering the mortgage and $800.00 carry charges. Find the equivalent simple interest rate per annum.

5. An article priced at $115.00 may be purchased for a cash payment of $20.00 and a $5.00 installment charge, both paid at the time of purchase, and eight monthly payments of $12.00 each. Find the equivalent simple interest rate per annum in this plan.

Equal-monthly-installments-plus-an-odd-final-payment plan formula. In the programming of an installment plan, it is frequently convenient to make all the installments equal except the last one. If, for example, P + I = $76.98, then it could be convenient to have six payments of $12.00 each and an odd final payment of $4.98.

To develop a formula that is applicable for such a plan, the symbols in the previous section will be used with the same meanings, and the amount of the final payment will be denoted by /.

There are, therefore, m — 1 installments of a each and an mth or final payment of /.

Proceeding as in the previous section,

.'. P + P j2 i = / + (m - l)a + 24

Since the sum of the installments is equal to the balance P plus the carrying charges /, we have

106 Simple Interest

f+(m-l)a = P + Ior(m-l)a = P + I-f

.*. 2Pwu = 247 + Pmz + Imi — fmi :. (P - I +f)mi = 24/

24/

m(P - I + /)

(2)

Example. Suppose a $600.00 article can be purchased for nine monthly pay- ments of $70.00 and a tenth final payment of $10.00. Find the equivalent simple interest rate per annum.

Solution. P = $600

/ = $640 - $600 = $40 m = 10 / = $10.00 . = 24X40 =

10(600 - 40 + 10) '

.*. The equivalent simple interest rate is 16.8% per annum.

PROBLEMS

1. A radio is priced at $70.00 cash or $7.00 down, $10.00 per month for six months, and a final payment of $8.00 at the end of the seventh month. Find the equivalent rate of simple interest.

2. A vacuum cleaner is for sale at $109.50 or $10.00 down, $15.00 per month for seven months and $3.50 at the end of the eighth month. Find the equivalent simple interest rate per annum.

3. A gas hot water heater is sold for $125.00 cash or $25.00 down, $10.00 per month for eleven months, and $8.00 at the end of the twelfth month. Find the equivalent simple interest rate per annum.

4. Mr. Smith purchased a washer-and-dryer combination for $592. After paying $100, he found that he could not pay the balance. This balance was financed by an acceptance corporation, with the following terms quoted: 23 monthly installments of $21.00 each plus a final payment of $9.00. Find the equivalent rate of simple interest.

Personal loans.

Formulas (1) and (2) developed in the previous sections apply equally well to personal loans which are repaid by monthly installments.

Example 1. Peter Smith borrows $600.00 from a finance company and con- tracts to repay with eight monthly installments of $78.00 each. What equivalent simple interest rate per annum is he paying?

Solution. P = $600

/ = $624 - $600 = $24 m = 8

Simple Interest

Substituting these values in formula (1),

24/ 24 X 24

107

i =

X 24 = .11.

(m + 1)P - (m - 1)7 9 X 600 - 7 .-. the equivalent simple interest rate is 1 1 % per annum.

Example 2. Miss Harris borrows $75.00 from a personal loan company and agrees to pay $8.00 per month for nine months and $6.75 in the tenth and final month. What simple interest rate per annum is she being charged?

Solution

P = $75.00 / = $78.75 - $75.00 m = 10 / = $6.75

Substituting these values in formula (2),

24/ 24 X 3.75

$3.75

m(P - I + /) 10(75 - 3.75 + 6.75) Miss Harris is* being charged 11.6% per annum.

= .116.

PROBLEMS

1. The following table is taken from a newspaper advertisement of a small loan company. Calculate the equivalent simple interest rate per annum for each loan.

You Receive

Monthly Payment

Number of Payments

$ 147.16

310.78

506 . 94

809.74

1,525.00

$ 9.00 19.00 22.00 34.00 39.00

20 20 30 30 60

2. A borrowed $1,000.00 from B and promised to repay $100.00 per month for ten months and $50.00 in the eleventh month. What interest rate was B charging At

ACCOUNT CURRENT

Definition. An account current is a transcript of the ledger account. It should show the dates on which sales were made, the term of credit for each item, cash payments, and, if settlements were made by note, the date and other details of each note.

Methods. Two methods are used in rinding the amount due: the Interest Method, and the Product Method.

108 Simple Interest

Example. Find the balance due January 1 on the following ledger account, which bears interest at 6%.

J. B. JOHNSON Dr. Cr.

Sept. 1, Balance $1,200.00 Oct. 1, Cash $1,000.00

Sept. 20, Mdse. 30 days 400.00 Nov. 10, Cash 200.00

Oct. 30, Mdse. 30 days 520.00 Dec. 3, Cash 400.00

Nov. 25, Mdse. 30 days 350 . 00 Dec. 15, Note 10 days 300 . 00

Solution by Interest Method.

J. B. JOHNSON

Dr. Cr.

Date Due Amount Days Interest Date Amount Days Interest

Sept. 1 $1,200.00 122 $24.40 Oct. 1 $1,000.00 92 $15.33

Oct. 20 400.00 72 4.87 Nov. 10 200.00 52 1.73

Nov. 29 520.00 33 2.86 Dec. 3 400.00 29 1.93

Dec. 25 350.00 7 .41 Dec. 25 300.00 7 .35

$2,470.00 $32.54 $1,900.00 $19.34

1,900.00 19.34

$ 570.00 + $13.20 = $583.20

Explanation. The number of days opposite each entry is the actual number of days from the date of the item to January 1, the date which is taken as the focal date.

Solution by Product Method.

Dr.

Cr.

Date Due

Amount Days

Product

Date

Amount Days

Product

Sept. 1

$1,200 X 122

= $146,400

Oct. 1

$1,000 X 92

= $ 92,000

Oct. 20

400 X 73

= 29,200

Nov. 10

200 X 52

= 10,400

Nov. 29

520 X 33

= 17,161

Dec. 3

400 X 29

= 11,600

Dec. 25

350 X 7

2,450

Dec. 25

300 X 7

2,100

$2,470

$195,210

$1,900

$116,100

1,900

116,100

$ 570 $ 79,110

The interest on $79,110 for 1 day = $ 13.20 $570.00 + $13.20 = $583.20

In some instances it is more convenient to find the equated due date, and then calculate the interest on the balance of the account from that date to the date of settlement.

PROBLE MS

1. Find the amount that will settle the following account Sept. 10, interest at 6%.

Dr. Cr.

Mar. 15, Mdse. 4 mos $450 . 00 July 5, Cash $400 . 00

Mar. 30, Mdse. 60 days 375.00 July 30, Cash 375.00

Apr. 18, Mdse. 30 days 700 . 00 Aug. 15, Cash 690 . 00

May 15, Mdse. 4 mos 620 . 00 Sept. 5, Cash 615 . 00

May 30, Mdse. 4 mos, 410.00

Simple Interest 109

2. Find the amount that will settle the following account on June 1, interest at 6%.

Dr. Cr.

Jan. 4, Mdse. 30 days $500.00 Feb. 20, Cash $300.00

Jan. 30, Mdse. 30 days 200.00 Feb. 28, Note, 60 days with

Feb. 5, Mdse. 30 days 600 . 00 interest at 6% 300 . 00

Mar. 1, Mdse. 30 days 400.00 Mar. 20, Cash 150.00

3. A borrowed $10,000.00 from a bank on January 2, for building a home, giving a note, secured by a mortgage, due in one year, with interest at 6%. From time to time the bank advanced him money to pay contractors' estimates. Before maturity the bank had actually advanced $9,000.00, as follows:

January 31 $3,000.00

March 15 3,000.00

April 15 1,500.00

May 15 1,500.00

On June 1, the following year, the maker of the note desires to pay it. (a) How should interest be computed? (6) What amount is due June 1?

STORAGE

Definition. Storage is the charge made by a warehouse or depositary for the storing of goods until they are required for use or for transportation to some other point.

Running account. When goods are being received and delivered, the storage company keeps a running account, showing the dates at which goods are received and delivered, together with details of the num- ber of packages, barrels, and so forth. Storage is charged for the average number of days for which one package, barrel, or box has remained in storage. The average number of days is divided by 30 to reduce the average number of days to months, or by 7 to reduce the average number of days to weeks, as the case may be; then the number of months or weeks is multiplied by the price per month or per week.

Example. The following is a memorandum of the quantity of salt stored with a storage company at 4j£ per barrel per term of 30 days' average storage.

Time in Equivalent

Date Receipts Deliveries Balance Storage for 1 Day

June 4 120 bbl. 120 bbl. 28 days 3,360 bbl.

July 2 20 bbl. 100 bbl. 18 days 1,800 bbl.

July 20 100 bbl. 200 bbl. 10 days 2,000 bbl.

July 30 50 bbl. 150 bbl. 11 days 1,650 bbl.

Aug. 10 100 bbl. 50 bbl. 15 days 750 bbl.

Aug. 25 50 bbl. 0 bbl.

9,560 bbl.

Explanation. 9,560 bbl. for 1 day are equivalent to 1 barrel for 9,560 days, and 9,560 divided by 30 (the number of days per term) equals 318f terms. In some cases a full month's storage is charged for any part of a month that goods

1 1 0 Simple Interest

remain in storage; in other cases, 15 days or less are called one-half of a month, and any period of over 15 days is counted as a whole month. 318|^ terms would be charged for as 319 terms, and 319 X .04 = 12.76. Therefore, $12.76 is the storage charge.

PROBLEMS

1. On the following memoranda, compute storage at 4^ per barrel per term of 30 days' average storage:

Received Delivered

Feb. 10 300 bbl. Feb. 20 150 bbl.

Feb. 19 150 bbl. Mar. 5 200 bbl.

Mar. 12 500 bbl. Mar. 15 400 bbl.

Mar. 30 300 bbl. Apr. 14 300 bbl.

2. A grower stored 5,000 bushels of potatoes at 5J^ per cwt., the term being 30 days' average storage. The following is a memorandum of the transactions that occurred. Compute the amount of storage.

Received Delivered

Sept. 1 2,500 bushels Nov. 4 500 bushels

Sept. 10 1 ,500 bushels Dec. 10 600 bushels

Oct. 5 1 ,000 bushels Jan. 15 750 bushels

Feb. 1 1,500 bushels

Mar. 18 750 bushels

Apr. 2 900 bushels

Inventories

Valuation of inventories. The bases of inventory valuation most commonly used by business concerns are: (a) cost; and (b) cost or market, whichever is lower. However, the average cost method is used in some instances — the tobacco industry, for example — and market value as a basis is used in grain and cotton inventories and in inventories of dealers in securities.

Cost or market, whichever is lower. In valuing inventories at cost or market, whichever is lower, a comparison of inventory totals at the two values is not sufficient. It is necessary to consider each item or group of similar items purchased at the same price, and to make the exten- sion at the cost or market price, whichever is lower.

Example. One hundred tons of sugar (200,000 lbs.) were purchased at li a pound, and later 50 tons (100,000 lbs.) were purchased at 6^ a pound. The entire 150 tons were on hand at the close of the year, at which time the market value of sugar was 6^ a pound. Compute the inventory at: (a) cost; and (b) at cost or market, whichever is lower.

Solution.

(a) 200,000 lbs. @ .07 $14,000

100,000 lbs. @ .06 6,000

Inventory at cost $20 , 000

(b) 200,000 lbs. @ .065 $13,000

100,000 lbs. @ .06 6,000

Inventory at cost or market, whichever is lower $19,000

PROBL EM S

1. Given the following inventory of a retail shop for children's clothing, to3Ts, and so forth (correct as to quantities and values), state the amount which should

111

1 1 2 Inventories

be shown on a balance sheet as merchandise inventory, adopting the method of valuing inventory at cost or market, whichever is lower.

Value Per Unit Total Value

Item Cost Market Cost Market

150 knit towels $ 0.38 $ 0.35 $ 57.00 $ 52.50

16 crepe de chine carriage sets .... 10 . 00 12 . 50 160 . 00 200 . 00

125 lingerie and pongee hats 4 . 00 3 . 50 500 . 00 437 . 50

85 rubber bibs with sleeves 1 . 00 1 . 00 85 . 00 85 . 00

240 creepers 3.05 2.98 732.00 715.20

200 spring coats 19.50 20.00 3,900.00 4,000.00

50 spring coats 27.50 29.50 1,375.00 1,475.00

8 play yards 18.00 17.25 144.00 138.00

8 desks used in office 110.00 120.00 880.00 960.00

140 shirts 4.50 5.00 630.00 700.00

200 boys' wash suits 6.20 5.98 1,240.00 1,196.00

125 bloomers 2.85 2.80 356.25 350.00

5 cribs 31.00 29.98 155.00 149.90

12 electric trains 19.00 19.00 228.00 228.00

Total $10,442.25 $10,687.10

2. You are called in by the X. Y. Z. Clothing Company to advise them on the calculation of their inventory. They have always followed the policy of cost or market, whichever is lower. You are informed that the inventory will be used for the tax return, as well as for the annual report to stockholders.

Value per Unit Total Value

Item Cost Market Cost Market

13 suits, grade A $120.00 $110.00 $1,560.00 $1,430.00

12 suits, grade B 80.00 75.00 960.00 900.00

17 suits, grade C 60.00 60.00 1,020.00 1,020.00

7 suits, grade D 40.00 44.00 280.00 308.00

24 overcoats, grade 1 150.00 160.00 3,600.00 3,840.00

5 overcoats, grade 2 80.00 90.00 400.00 450.00

10 overcoats, grade 3 60.00 50.00 600.00 500.00

6 topcoats, grade X 40.00 35.00 240.00 210.00

9 topcoats, grade Y 30.00 25.00 270.00 225.00

18 topcoats, grade Z 20.00 22.00 360.00 396.00

Total $9,290.00 $9,279.00

Which total figure would you advise the company to use for: (a) tax reports; (6) annual report to stockholders?

Average cost method. The general rule that the average cost meth- od of valuing inventories will not be accepted for income tax purposes is subject to certain exceptions. In the tobacco industry, for example, tobacco is bought from the producer, usually in small quantities and at greatly varying prices. Different grades of tobacco are mixed and stored in hogsheads, and it is practically impossible to determine the exact cost of any particular hogshead. The inventory is therefore averaged monthly, according to grades, as follows:

First method. From the inventory of each grade at the beginning of the month is subtracted the amount of tobacco of that grade used, leaving so many pounds costing so many dollars; to this is added the tobacco of

Inventories

113

that grade purchased during the month, and a new average is deter- mined. This is the inventory for the close of the month, and is con- sequently the opening inventory of the next month.

Example.

Stock Card

Received

Issued

Balance

Date

Quan- tity

Rate

Amount

Date

Quan- tity

Rate

Quan- tity

Rate

Amount

6-29

9-30

12-10

100,000

80,000

125,000

$1.00

1.10

.95

$100,000

88,000

118,750

9-1 12-5 12-18

80,000 30,000 20,000

$1.00 1.08 1.08

100,000 20,000

100,000 70,000

195,000

175,000

$1.00 1.00 1.08 1.08

$100,000

20,000

108,000

75,600

194,350

172,750

Inv't 12-31

175,000

$0,987

$172,750

Explanation. It will be noticed that the receipt of 125,000 at .95 on Dec. 10 was extended into the balance column in quantity and amount only, and that the issuance on Dec. 18 was made at the rate established on the first of the month. This is the method that is used when receipts are frequent, as it saves the time that would be required to compute a new rate after each receipt, and establishes a standard rate of issuance for the month.

Second method. When receipts are not frequent and are large in amount, a new average price is computed upon the entry of each receipt.

Example.

Stock Card

Received

Issued

Balance

Date

Quan- tity

Rate

Amount

Date

Quan- tity

Rate

Quan- tity

Rate

Amount

6-29

9-30

12-10

100,000

80,000

125,000

$1. 00

1.10

.95

$100,000

88,000

118,750

9-1 12-5 12-18

80,000 30,000 20,000

$1.00 1.08 • 99±-

100,000 20,000

100,000 70,000

195,000

175,000

$1.00

1.00

1.08

.99^-

•99i

$100,000

20,000

108,000

75,600

194,350

174,450

Inv't 12-31

175,000

$0.99^

$174,450

114

Inventories

PROBLEMS

1. Rule two stock cards as in the preceding example, and enter the following data. Compute the balances: (a) by the first method; and (6) by the second method.

Received

July 5 80,000 units @ $0.90

Aug. 15 20, 000 units® 1.00

Sept. 1 30,000units@ 1.10

Dec. 8 20,000 units® 1.20

Issued

Aug. 1 50,000 units

Dec. 2 20,000 units

Dec. 20 30,000 units

2. Complete the following stock ledger card, using the average-price method.

Stores Ledger

Actual Receipt Price

.20

Average Price

./70f

Name /zJudXt^-A

- 3" yVaM

Part No. SS'67^

Minimum /#? u

Maximum 300

Location /,3

Drawing No.

Unit

REFERENCE

QUANTITY

ON HAND

Date Number

Remarks

Received

Issued

Quantity

Value

e s

£1

OCT 4 4523 OCT 10 34567 OCT 12 35654 OCT 13 38765 OCT 15 39458 OCT 16 4587 OCT 19 40156

100 50

5 10

3 12

5 6

8 9

11 12

"First-in, first-out" method of inventory. Where the same merchandise has been purchased at various prices during the year, and the goods on hand cannot be identified with specific invoices, the amount on hand at the end of the year may be inventoried at the latest purchase price. If, however, the quantity on hand is greater than the amount purchased at the last price, the balance may be inventoried at the next- to-the-last purchase price, and so on. This method is termed " first-in, first-out method" of inventory.

Example.

Inventory, December 31 275,000 units

Invoices :

November 10 125,000 units @ $0.95 per C

September5 80, 000 units© 1 . 10 per C

June 10 70,000 units @ 1 .00 per C

Inventories 115

How should the foregoing inventory be valued? Solution.

125,000 units @ $0.95 per C $1,187.50

80,000 units @ 1 . 10 per C 880.00

70,000 units @ 1.00 per C 700 . 00

275,000 units inventoried $2,767.50

"Last-in, first-out" method of inventory. Under the "last-in, first-out" method, inventories are valued at the cost of goods earliest acquired, and in computing profits from sales the cost of goods last acquired is used. This method will show smaller profits when prices are rising and larger profits when prices are falling than the "first-in, first- out" method. Businesses which use raw materials or other goods includ- able in inventory which are subject to sharp price fluctuations, businesses in which the value of inventory is large compared with other assets and sales, and businesses in which production consumes an extended period are most likely to benefit from the use of this method. (Consult the Internal Revenue Code relative to the requirements incident to adoption and use of this method.)

Example. A has an opening inventory of 10 units at 10 cents a unit, and during the year he makes purchases of 10 units as follows:

January 1 @ .11 = .11

April 2 © .12 = .24

July 3 @ .13 = .39

October 4 @ .14 = .56

10 T730

His closing inventory shows 15 units. What is the value of the closing inventory?

Solution.

10 @ .10 = 1.00

1 @, .11 (Jan.) = .11

2 @ .12 (Apr.) = .24 _2 © .13 (July) = .26

Totals 15 1.61

PROBLEM

Value the closing inventory, using the " last-in, first-out" method.

Opening inventory: 50 units at $1.00 Production :

First quarter: 50 units at $1.50

Second quarter: 100 units at $1.75

Third quarter: 50 units at $2.00

Fourth quarter: 100 units at $2.25 Closing inventory: 150 units

116 Inventories

Merchandise turnover. The number of times that the value of the inventory is contained in the cost of sales is the merchandise turnover.

The final inventory should not be used in computing turnover, unless it represents a normal inventory for the fiscal period or is the first inven- tory that has been taken.

If a perpetual inventory system is in use, the monthly inventories should be added to the inventory at the beginning of the period, and the sum divided by the number of months in the fiscal period plus one. In a year there would thus be 13 inventories — the one at January 1, and the 12 inventories at the ends of the months. When a perpetual inventory is not used, add the inventory at the beginning of the fiscal period to the one at the close of the period; then divide by 2. The quotient will be the estimated average inventory for the period. If semiannual inven- tories are taken, use three inventories and divide by 3. If quarterly inventories are taken, use five inventories and divide by 5.

Formula

Cost of Sales -f- Average Inventory at Cost = Rate of Turnover

Example. A department store found the average inventory of Department A for the fiscal period to be $30,000. The cost of sales for the same period was $120,000.

$120,000 -r- $30,000 = 4, the rate of turnover.

An estimated inventory at the end of any period may be obtained by dividing the sales for the period by 100 % plus the per cent of gross profit based on cost, and deducting the quotient from the total of purchases and first-of-period inventory. A more complete discussion of the gross profit test is given in Chapter 7.

Example. In the above example, assume that in Department A the total cost of merchandise was $150,000, that the sales were $144,000, and that the average profits were 20%. Using the per cent of gross profits to determine the average inventory, the solution would be as follows:

$144,000 (sales) -r- 120% = $120,000, the cost of sales. $150,000 (total cost of goods) - $120,000

(cost of sales) = $30,000, the estimated inventory. $120,000 (cost of sales) -r- $30,000

(inventory) = 4, the rate of turnover.

Number of turnovers. The number of turnovers varies in different lines of business. Records show turnovers varying from 1 to more than 20, depending on the kind of business. It is possible to make a larger profit by several turnovers with a small mark-up* than by 1 or 2 turn- overs with a large mark-up. Limited capital and frequent turnovers

* " Mark-up," as used in this text, refers to the addition made to the cost of mer- chandise to produce the selling price.

Inventories 117

can produce a profit equal to that produced by a greater capital turned fewer times a year. If a merchant turns $1 eight times in the course of a year, he has used i of the capital that would be required if the rate of turnover were 1.

Example 1. A merchant had a rate of mark-up of 50%, with a turnover of 1. He found that by using a rate of mark-up of 30%, he had a turnover of 2. If his former sales were $300,000 annually, how much were his gross profits increased, provided he continued to use the same investment in merchandise?

$300,000 -h 150%, = $200,000, cost of sales. $300,000 - $200,000 = $100,000, gross profits.

Under the new policy he turns the $200,000 twice, the equivalent of $400,000 annually.

$400,000 at 30% = $120,000, profits. $120,000 - $100,000 = $20,000, increased profits due to lowering

the rate of mark-up and increasing the rate of turnover.

Example 2. What investment in merchandise would be required under the new policy to make the same amount of profits that was made under the old policy?

$100,000 -T- 30% = $333,333.33, cost of goods sold to make profits of $100,000.

Since there were two turnovers, the cost of goods sold was twice the amount of the average inventory. Therefore:

$333,333.33 -=- 2 = $166,666.67, the average inventory.

Hence, the merchant could make the same amount of gross profits with an investment $33,333.33 smaller than that required under his old policy.

PROBLEMS

1. A rate of mark-up of 30% results in two turnovers of an average inventory of $30,000. If the expenses of conducting the business are $8,000, what is the net profit?

2. A rate of mark-up of 20% results in three turnovers of an average inventory of $30,000. If expenses remain at $8,000, what is the net profit?

3. The cost of sales in Department B was $42,000. The average inventory was $12,000. What was the number of turnovers?

4. A merchant's sales amounted to $42,000. His average inventory was $10,000, and the average rate of mark-up was 40%. Find the number of turnovers.

5. Commodity X, with a rate of mark-up of 40%, had a turnover of 2. With a rate of mark-up of 30%, it had a turnover of 3. If prior sales were $56,000, find the sales and the increase in gross profit with the 30% rate of mark-up.

6. A rate of mark-up of 35 % results in a turnover of 2 and in sales amounting to $540,000. A rate of mark-up of 20% results in a turnover of 4. How much

1 1 8 Inventories

less capital under the latter plan is required to make as much profit as under the former plan?

7.* On January 1, a concern dealing in a single commodity had an inventory of merchandise which cost $20,000. The goods were marked to sell at 125% of cost, and all subsequent purchases during the six months ending June 30 were marked at the same rate. The selling price of the inventory at June 30 was $24,000. Purchases and sales by months were:

Purchases Sales

(Cost) (Selling Price)

January $8,000 $9,000

February 9,000 9,500

March 14,000 12,000

April 16,000 18,000

May 13,000 22,000

June 10,000 18,000

(a) Compute estimated inventories at cost price at the end of each of the six months.

(6) Compute the rate of turnover for the six months' period, using (1) the January 1 and June 30 inventories; (2) all the inventories.

Per cent of mark-down to net cost. If an item costs $1 and is marked $1.25, in order to sell the item for cost the price must be reduced 25 jzL The marked price is the base when prices are reduced. 25 i is i of $1.25. i = 20%.

An item costs $1 and is marked $1.50. 50 ^ reduction is ^ of $1.50, or 33*%.

PROBLEMS

Calculate the per cent of mark-down for each of the following items:

Marked Per Cent of Mark-down

Item A

Cost

$ 2.00

1.00

15.00

3.50

5.00

2.00

40.00

12.00

Price

$ 2.50

1.25

.40

1.00

25.00

4.00

.30

7.00

.10

4.00

75.00

18.00

to Produce Cost

Computation of inventory by the retail method. The need for frequent inventories has led many department stores to adopt the " retail method" of computing inventories. The accuracy of the inventory by this method depends upon the care exercised in recording the mark-ups

* American Institute Examination.

Inventories

119

and the mark-downs of merchandise prices, and the classification of merchandise into departments and groups and sub-classes within the departments. In addition to the usual records showing sales (at selling price only), records are kept which show the opening inventory and pur- chases at cost and at retail (or selling) prices. An estimated inventory may be prepared from such records in the following manner.

Inventory Computation

Cost

Retail $ 8,000

111,200

Inventory, beginning of period $ 6 , 000

Purchases during the period, including freight and

cartage 74 , 000

Totals $80,000 $119,200

(% Mark-on = $39,200 -f- $119,200 or 32.8859%.)

Sales 104,200

Inventory at retail $ 15,000

Estimated inventory = $15,000 - ($15,000 X 32.8859%) = $10,067.

The foregoing illustration does not take into consideration changes in selling price after the original mark-up. Price changes must be dealt with, and the retail mercantile business has terms for these changes that are not generally understood; therefore, to prevent any misunderstanding, the diagram in Figure 5-1 is presented and the terms explained.

Fig. 5-1.

Original mark-up. The amount by which the original retail price of an article exceeds the cost is the original mark-up.

Additional mark-up. An amount that increases the original retail price is an additional mark-up.

Additional mark-up cancellation. A reduction in the additional mark- up is an additional mark-up cancellation, and the amount cannot exceed the amount of the additional mark-up.

120

Inventories

Net mark-up. The sum of additional mark-ups minus the sum of additional mark-up cancellations is the net mark-up.

Mark-downs. Deductions from the original retail price to establish a new but lower retail price are mark-downs.

Mark-down cancellations. A reduction of the amount of a mark-down is a mark-down cancellation. Mark-down cancellations cannot exceed the total mark-downs. It is evident that the retail price of merchandise is increased when the mark-down is reduced, but such an increase is not to be considered as an additional mark-up.

Net mark-down. The difference between the sum of the mark-downs and the sum of the mark-down cancellations is the net mark-down.

Mark-on. The difference between cost and the original retail plus the net mark-up is the mark-on.

To illustrate the terms, let the following transactions be assumed.

Cost of Article $1.00

Original

Mark-Up

500

Original Retail (or Selling) Price $1.00 + .50 = $1.50

$1.50

Addn'l

Mark-Up

25^

1st Adjusted Retail Price

$1.50 + .25 = $1.75

2nd Adjusted Retail Price

$1.75 - .10 = $1.65

a 100

3rd Adjusted Retail Price

$1.65 - .15 = $1.50

b 150

4th Adjusted Retail Price $1.50 - .15 = $1.35

c 150

5th Adjusted Retail Price $1.35 - .35 = $1.00

d 350

$1.00

e 250

6th Adjusted Retail Price $1.00 + .25 = $1.25

a. Additional mark-up cancellation

Net mark-up = 150 Mark-on =650

b. Additional mark-up cancellation

Net mark-up = 0 Mark-on =500

c. Mark-down

d. Mark-down

e. Mark-down cancellation

Net mark-down =

.15 + .35 - .25 = .25

Inventories

121

Determining the ratio of cost to retail. In determining the ratio of cost to retail, it is customary to include additional mark-ups and addi- tional mark-up cancellations but to exclude mark-downs and mark-down cancellations. To illustrate, let us assume the following facts:

Inventory at beginning of month:

Cost $30,000 00

Retail 43,000.00

Purchases :

Cost 46,000.00

Retail 55,000.00

Returned purchases :

Cost 1,000.00

Retail 1,500.00

Additional mark-ups 5 , 500 . 00

Additional mark-up cancellations 2 , 000 . 00

Mark-downs 6,000.00

Mark-down cancellations 1 , 000 . 00

Sales at retail 71,000.00

Compute the inventory by the retail method.

Solution.

Cost Retail

Inventory $30,000.00 $ 43,000.00

Purchases 46,000.00 55,000.00

$76,000.00 $ 98,000.00

Deduct: Returned purchases 1,000.00 1 , 500 . 00

$75,000.00 $ 96,500.00

Additional mark-ups less cancellations thereof 3,500.00

$100,000.00 $100,000 - $75,000 = $25,000. $25,000 + $100,000 = 25%.

Mark-downs less mark-down cancellations 5 , 000 . 00

$ 95,000.00

Sales at retail 71,000.00

End-of-month inventory at retail value $ 24 , 000 00

$24,000 X 25% = $6,000. $24,000 - $6,000 = $18,000, the cost value of the inventory.

PROBLEMS

1. From the records kept for Department B, the following information is obtained :

Cost Retail

Inventory at Beginning of Month $15,000.00 $25,000.00

Purchases 36,000.00 54,000.00

Returned Purchases 500 .00 700 .00

Additional Mark-Ups 2,000 .00

Additional Mark-Up Cancellations 1 , 000 . 00

Sales 60,000.00

122 Inventories

Calculate by the retail method of inventory the cost of the book inventory at the end of the month.

2.* In a certain department of a large dry-goods house, the purchases for one year were $30,000. They were in the first place marked up for selling pur- poses to $45,000. Later, additional mark-ups amounting to $2,000 were made, and mark-downs aggregating $5,000 were also recorded. At the end of the fiscal period there were found to be on hand goods of a marked selling value of $10,000. State how you would arrive at their inventory value for the purpose of closing the books, and calculate the amount. Explain fully.

* American Institute Examination.

Insurance

Insurance. There are at least 21 kinds of insurance applicable to the ordinary business being done in big cities and as many as 150 kinds of insurance covering all branches of human endeavor.

Policy. An insurance policy is a written contract. The consider- ation given for the protection promised consists of a premium to be paid in money and the fulfillment by the insured of acts of commission and omission according to the terms and conditions set forth in the policy.

Fire insurance. Fire insurance is guaranty of indemnity for loss or damage to property by fire. Insurance companies are liable for loss or damage resulting from the use of water or chemicals used in extinguish- ing the fire and from smoke. A fire loss is predicated on the sound value at the time the loss is sustained and not at the time the insurance is written.

Form of policy. With but a few exceptions, fire insurance companies use a State standard-form policy made mandatory by the State in which they operate. The New York State standard form of policy is the one that is generally used, as it embraces nearly all that is contained in other forms. The form attached to the policy is known as a rider. The rider form directly applies the insurance to fit the facts and conditions of the particular risk. It also amends the standard form, which is not a contract until completed by descriptions and amendments.

Rates. Probably no phase of insurance interests the businessman more than his insurance rate. Independent rating bureaus operate in different parts of the country. Their business is to inspect and to measure the hazards in terms of rates. Rate schedules are compiled for this pur- pose. The charges are in the nature of penalties for hazards.

123

1 24 Insurance

Example. A particular building has been inspected and surveyed by the rater. The degree of municipal and local protection has been measured. This establishes the basic rate. Assume the basic rate to be .40, which is a charge commensurate with the degree of protection and covers all general hazards that cannot be segregated and measured. The better the city protection, the lower will be the basic rate.

Basic rate 40

Area: 15,800 sq. ft 04

(The standard unit area is 1,000 sq. ft., and an additional charge is made for larger areas.)

Parapet wall deficiency 04

Skylights not standard construction .02

Metal stacks through roof 06

Outside wood cornices, loading docks, and wooden conveyor 06

Gallery decks used for storage 03

Occupancy hazard (woodworking mill) 92

Shavings allowed to accumulate 05

No cans for collecting waste 05

No drip pans under machines 05

Floor oil-soaked 05

Total T777

Credit for open finish (inside walls) 08

Building rate unexposed 1 . 69

Exposure :

From buildings No. 2 and No. 5 at 18 ft 34

From building No. 6 at 15 ft 02

From office at 23 ft 05

From buildings No. 9 and No. 10 07

Exposure charge , .48

Total building rate 2.17

If this assured would have the parapet wall brought up to the standard requirements, his rate would be reduced .04. By having the shavings removed daily, and by installing waste cans and drip pans under the machines, the rate could be reduced another .15. As a matter of fact, the owner makes his own rate — the rater simply measures the hazards in terms of rates.

To find the premium. Insurance companies charge a certain num- ber of cents or dollars for insuring each $100.00 worth of property. Thus, the insurance rate in the foregoing example is $2.17 for each hundred dollars of insurance carried. If the building is valued at $65,000.00 and is insured for full value, the amount of the premium would be computed as follows :

$2 . 17 the rate per $100.00 of insurance. X 6 50 the number of hundred dollars of insurance purchased. $1,410.50 the premium, or cost of the insurance for one year.

Agent's commission. Local agents of the fire insurance companies are located in almost every city and town. They act as the represen-

Insurance 1 25

tatives of the companies, soliciting the business and collecting the pre- miums. For this service they receive a certain per cent of the premiums.

Example. A store building valued at $10,000.00 was insured for 80% of its value, the rate being $1.25 a hundred. What was the agent's commission if he received 15% of the premium?

Solution.

80% of $10,000.00 = $8,000.00, the insured value. 80 X $1.25 = $100.00, the premium. 15% of $100.00 = $15.00, the agent's commission.

Cancellation of policies. Both the insurance company and the insured have the right to cancel an insurance policy at any time. When the policy is canceled by the insurance company, the portion of the pre- mium to be repaid to the insured is determined pro rata.

Example. On April 10, the owner of a building insured his property for one year. The premium was $36.00. On October 20, the policy was canceled by the insurance company. What rebate did the insured receive for the unex- pired term?

Solution. The time from April 10 to October 20 is 193 days, expired term of the policy. (See page 93 for table of number of days between dates.)

^§-f of $36.00 is $19.04, amount of premium earned. $36.00 - $19.04 = $16.96, amount of premium returned.

When the policy is cancelled by the insured, the amount of premium returned is determined by the " short rate." The short rate is an arbi- trary per cent fixed by the insurance companies, and is shown by a table like the one at the top of page 126.

Example. On May 2, a one-year policy was written on a shop. The premium was $38.75. On September 26, the policy was canceled at the request of the insured. What rebate did the insured receive?

Solution. From May 2 to September 26 is 147 days. The table shows that 60% of the premium is to be retained when the policy has been in force 150 days, which is the number of days next higher than 147. Then 40% of the premium will be returned.

$38.75 X .40 = $15.50, the return premium.

Coinsurance. This is a form of insurance in which the person who insures his property agrees to carry insurance equal to a certain percentage of the valuation of the property. If he fails to carry that percentage with an insurance company, he (the insured) becomes a coinsurer on the loss,

126

Insurance

1 day.

2 days .

3 days.

4 days .

5 days .

6 days .

7 days .

8 days .

9 days .

SHORT RATE CANCELLATION TABLE

Period exceeding 20 days and not exceeding 25 days, to be the rate of 25 days,

and so on up to one year.

Per Cent Per Cent of Annual of Annual

Policy in Force Prem. Policy in Force Prem.

2% 55 days 29%

4% 60 days or 2 months 30%

5% 65 days 33%

6% 70 days 36%

7% 75 days 37%

8% 80 days 38%

9% 85 days 39%

9% 90 days or 3 months 40%

10% 105 days 46%

120 days or 4 months 50 %

135 days 56%

150 days or 5 months 60%

165 days 66%

180 days or 6 months 70%

195 days 73%

210 days or 7 months 75%

225 days 78%

240 days or 8 months 80%

255 days 83 %

270 days or 9 months 85 %

285 days 88%

300 days or 10 months 90%

315 days 93%

330 days or 11 months 95%

345 days 98%

360 days or 12 months 100%

10 days 10%

11 days 11%

12 days 11%

13 days 12%

14 days 13%

15 days 13%

16 days 14%

17 days 15%

18 days 16%

19 days 16%

20 days 17%

25 days 19%

30 days or 1 month 20 %

35 days 23%

40 days 25%

45 days 27%

50 days 28%

in the ratio which his lack of insurance bears to the amount he should have carried.

Illustration of 80% coinsurance clause:

Case 1. Value of building and contents $75,000

Assured should carry 80 % of value or 60 , 000

Insurance actually carried 45 , 000

Loss by fire 10 , 000

Paid by insurance company, 75 % of loss, or 7 , 500

Assured must bear 25 % of loss, or 2 , 500

Insurance carried was only 75% of what assured should have carried to comply with the 80% clause.

Case 2. Value of property $10,000

Insurance required 8 , 000

Insurance carried 9 , 000

Losses up to $9,000 Paid in full

Case 3. Value of property $10,000

Insurance required 8, 000

Insurance carried 8 , 000

Losses exceeding $8,000

Face of policy, $8,000, is paid.

Insurance 1 27

Case 4. Value of property $10,000

Insurance required 8,000

Insurance carried 5 , 000

Losses exceeding $8,000

Face of policy, $5,000, is paid.

Losses under $8,000

Paid in the proportion that $5,000 bears to $8,000, or f of the loss.

PROBLEMS

1. What premium must be paid on a policy for $3,760 at $1.50 a hundred?

2. A house worth $12,000 is insured for f of its value for three years at $2.35 a hundred. How much is the premium?

3. An agent wrote a policy of $4,500 on a store building at a rate of 85 cents. If the agent's commission was 15%, what was the amount of his commission?

4. Find the amount paid by the insurance company under the 80% coinsur- ance clause in the following:

(a) (b) (c) (d)

Value of property $50,000 $75,000 $100,000 $200,000

Insurance carried 40,000 60,000 80,000 80,000

Loss by fire 10,000 45,000 40,000 40,000

Paid by insurance company

5. You are presented the following tornado insurance plan and are asked to select one of the four policies and to decide upon whether to insure for one or three years. In your opinion, what policy should be taken and for how long a term? The sound value of the property to be insured is $1,242,000.

Amount of One-Year Three-Year

Coinsurance Insurance Rate Premium Rate Premium

(1) None $ 200,000 .20 50

(2) 50% 621,000 .102 255

(3) 80% 993,600 .0749 1872

(4) 90% 1,117,800 .0678 1695

Compute the premiums at the one-year rate and at the three-year rate. Find the average yearly premium on each policy at the three-year rate, and make comparisons in order to determine which policy to accept.

6. A one-year policy on a dwelling was dated June 5. The premium was $42.50. On October 1, the policy was canceled at the request of the insured. Find the amount of return premium.

Use and occupancy insurance. This kind of insurance is protec- tion against loss due to interruption of business by fire or tornado. It is insurance against a loss that is suffered on account of destruction of the property.

1 28 Insurance

The insurance recovery or indemnity is the profit that would have been made if business had not been interrupted and, in addition, the total of expenses that must continue during suspension of business. A busi- ness that is not profitable may be so insured in order to recover the con- tinuing expenses.

Generally speaking, use and occupancy insurance insures gross profits plus the salaries of key employees kept on the payroll account. The policy excepts payroll (other than that of the key employees), heat, light, power, and expenses of maintaining properties not destroyed (such as taxes, depreciation, and maintenance thereof). These items can be picked up by analyzing the running expense accounts. In no event does the policy pay expenses required to be insured unless it is proved that they continue after the fire.

Coinsurance clauses are also applicable in use and occupancy insurance.

A simple procedure to arrive at use and occupancy value for the past twelve months is as follows:

Total Sales

Deduct :

Cost of Merchandise

(Opening Inventory + Purchases — Closing In- ventory)

Ordinary Labor Payroll

Light, Heat, and Power

Total deductions

Actual 100% use and occupancy value for the period

The foregoing procedure is predicated on the assumption that all expenses other than ordinary payroll and light, heat, and power will con- tinue at the same cost as if the business were operating.

A more exact method is one considered in the light of a problem in arithmetic or algebra, as follows:

Let x = Use and Occupancy Insurable Interest Each Day

a = Expenses That Do Not Continue During Suspension of

Business b = Selling Price of Merchandise c = Cost of Merchandise d = Number of Working Days in the Month Then:

b — c — a

Example. The expenses of a business for a given month were determined as follows:

Insurance

129

Item Total

Payroll $45,000

Salaries and Wages of Key Employees

Who Must Be Retained

Salaries and Wages of Employees Not Retained

Power

Heat and Light

Leasehold

Advertising

Taxes

Insurance

Interest

Other Expenses

$54,000

Part of Expense Part of Expense

That Must That Will Not

Continue During Continue During

Suspension Suspension

$20,000

750

—

525

225

1,200

1,725

725

950

1,375

500

525

1,950

875

$25,000

$25,000 750 300

1,000

875

1,075 $29,000

Find the estimated amount of insurance to be carried for each day of the month if sales are estimated to be $181,500 and cost of merchandise sold $109,500. Average number of -working days each month is 25.

Solution.

b - c

181,500 - 109,500 - 29,000 25

1,720

Therefore, on the basis of estimates, $1,720 is the amount of insurance to be carried for each day in the month.

The same result may be obtained in the following manner, using the estimates given:

Sales for the Month $181 ,500

Less: Cost of Sales 109,500

Gross Profit 72,000

Total Expenses 54,000

Net Profit 18,000

Add : Expenses That Must Continue During Suspension .... 25 , 000

Use and Occupancy Value for the Month $ 43 , 000

43,000 -^ 25 = 1,720

PROBL EMS

1. The gross profit of a business was $200,000 after charging raw materials and payroll into manufacturing cost, but excluding light, heat, and power. Continuing payroll of key men was fixed at $20,000. If the policy contained the 80% clause, what was the required amount of insurance?

2. An audit of the expense accounts of the company insured in Problem 1 showed that items that would not have to be continued after the fire totaled $60,000. What amount would be collectible for a twelve-month period, other facts being as stated in Problem 1 ?

3. Assume that it takes 15 months to rebuild the plant. How much insur- ance would be collectible?

1 30 Insurance

4. If a manufacturer on a Sept. 30 fiscal-year basis had a fire on April 1 and it is shown by previous experience that the following six months are the most profitable — in fact, that 66f % of the net earnings are made in that period — would the adjustment take this into consideration, or would it be made on an average for the year?

5. If the conditions in Problem 4 were reversed, what earnings would the adjustment reflect?

6. Compute the use and occupancy value from the following data: Beginning inventory, $170,482.66; ending inventory, $171,721.77; manufacturing — including raw materials, labor, light, heat, and power, maintenance, depreciation, adminis- tration, insurance, taxes, interest, advertising, and all other expenses, $3,409,- 658.42. Fixed charges that are included in the foregoing and that are expected to continue are: administrative salaries, $35,200; interest, $4,800; taxes, $9,961.22; dues and pledges, $6,150; credit information, $235; insurance, $7,918.49; salaries of office, supervisors, and foremen that will have to be retained, $237,075; miscellaneous expenses, $42,395.62. Sales were $3,551,708.81.

Group life insurance. Although this type of insurance is a part of the subject of life insurance, it is presented in this chapter because it is a common form of business insurance. The principles of life insurance are presented in another chapter.

Group life insurance affords employees of a business ordinary life insur- ance at low cost so long as they are employed by the particular employer, as the employer pays a part of the premium. The operation of this type of insurance is best explained by an example of an actual plan.

GROUP LIFE PLAN

1. Eligibility. The following plan of group life insurance is offered to all present employees of the company who will have completed six months or more of continuous service on November 11, 19 — , and to all new employees after they have been with the company for six months.

2. Amounts of insurance. The amount of insurance available to each employee under age 65, nearest birthday, will be based on annual earn- ings as follows:

Class Annual Earnings Life Insurance

1. Less than $1,200 $1 , 000

2. $1,200, but less than $2,200 1 ,500

3. $2,200, but less than $2,800 2,000

4. $2,800, but less than $3,200 2,500

5. $3,200, but less than $3,800 3,000

6. $3,800, but less than $4,200 3,500

7. $4,200, but less than $4,800 4,000

8. $4,800, but less than $5,200 4,500

9. $5,200 and over 5,000

3. Cost of insurance. The monthly cost of the insurance will be based on the employee's insurance age on each anniversary date of the plan, as shown in the following schedule :

Insurance 131

Employee's Monthly Attained Age on Policy Contribution per

Anniversary Each Year $1,000 of Insurance

Age 44 and under $0 . 70

Ages 45 to 54, inclusive 1 . 00

Ages 55 to 59, inclusive 1 . 50

Age 60 and over 1 . 80

PROBLEMS

1. Employee Y is 42 years of age and his earning classification is Class 5. What is the monthly deduction for his insurance?

2. If Y were 14 years older, what would be the monthly deduction?

3. B is 46 years of age and earns $3,000 a year. How much insurance is available to him, and what will be his monthly contribution?

4. Company X insures each of its employees for $1,000. Under age 50 the cost to the employee is 60 cents a month; at age 50 or over, the cost is $1.00 a month. There are 54 employees, classified as follows:

Age Number

18 1

22 6

25 10

29... 4

30 7

45 12

47 8

52 2

56 3

58 1

What is the amount of the monthly payroll deduction?

5. The manual shows the cost of group insurance on a monthly basis to be as follows:

Age Premium

18 $ .51

22 53

25 54

29 55

30 55

45 80

47 90

52 1.26

56 1.71

58 2.00

With the number in each age group being that given in Problem 4, what is the amount of insurance premium that is borne by Company X?

Health insurance. Some plans are contributory and others non- contributory. In either case, the benefits are much the same; but in contributory plans the employee pays a part of the cost in the form of a monthly premium deducted from wages, while in the non-contributory

1 32 Insurance

plans the cost is borne by the employer. Few businesses have their own insurance departments, most of the plans being handled by insurance companies under a group plan.

Incapacities include sickness and non-occupational accidents (occupa- tional accidents being covered by Workmen's Compensation Insurance), but the employer usually reserves the right to withhold benefits if the incapacity is the result of the employee's misconduct or negligence.

The following examples are illustrative of the many ways in which the factor of service is employed to favor the veteran worker.

Example 1.

Amount and Duration of Length of Service Disability Benefits

Under 2 years Such practice as the company may establish

2 but less than 5 years Full pay 4 weeks, half -pay 9 weeks

5 but less than 10 years Full pay 13 weeks, half-pay 13 weeks

10 years and over Full pay 13 weeks, half -pay 39 weeks

Example 2.

Amount and Duration of Length of Service Disability Benefits

1 but less than 10 years 50% of wages

10 but less than 30 years 75 % of wages

30 years and over 100 % of wages

Maximum : 26 weeks in 3 years Example 3.

Amount and Duration of Length of Service Disability Benefits

6 months but less than 1 year 35% of wages; maximum: $35.00

per week, for 6 weeks

1 but less than 2 years 50% of wages; maximum: $50.00

per week, for 13 weeks

2 but less than 3 years 60% of wages; maximum: $60.00

per week, for 13 weeks

3 but less than 4 years 70% of wages; maximum: $70.00

per week, for 26 weeks

4 years and over 75% of wages; maximum: $75.00

per week for 26 weeks.

PROBLEMS

1. A was insured under the plan in Example 1. He was employed for 3 years and became incapacitated for a period of 6 weeks. His average weekly wage was $74.80. What amount of disability benefit was he entitled to receive?

2. B was insured under the plan in Example 2. He had been with the same employer for 12 years. Two years ago he drew compensation for 8 weeks, and last year for 12 weeks. This year he was again incapacitated for a period of 8 weeks. If his average weekly wage was $95.00, what amount of disability benefit was he entitled to this year?

3. C was employed by an employer using the plan in Example 3, and had worked for this employer for a period of 6 years. He became incapacitated

Insurance 1 33

when receiving a weekly salary of $110.00, and was unemployed for 10 weeks. What was the amount of compensation paid?

Workmen's compensation insurance. This type of insurance is financial protection against loss of time for the wage-earning group resulting from accident or occupational sickness while on duty. The cost is levied on the employer in the form of a premium on the payroll classified according to the hazard of occupation. A few states have their own Workmen's Compensation Insurance Departments, but in most states the insurance is carried by the insurance companies specializing in this type of insurance, generally referred to as casualty insurance com- panies.

PROB LEMS

Find the cost of workmen's compensation insurance on payrolls divided into four classifications with respective rates as follows:

1. $239,530.39 @ .611 per C

75,535.62 @ .519 per C

241,327.85 @> .081 per C

99,791.48 @ .586 per C

2. $272,584.07 @, .611 per C

91,856.68 @ .519 per C 292,258.87 @ .081 per C 148,735.42 @, .586 per C

3. $254,248.83 @, .581 per C

79,950.31 @ .548 per C 272,368.08 @, .085 per C 105,553.36 @ .564 per C

4. A deposit of $100.00 was made on a public liability policy. The payroll audit was as follows:

$381,839.77 @ .052 per C 294,212.15 @, .026 per C 138,631.05 @ .026 per C

What amount of additional premium was due on completion of the payroll audit?

Gross Profit Computations

Gross profit. The gross profit represents the margin between the sales and the cost of goods sold, and when expressed as a per cent of sales indicates to one who is familiar with trade practice whether a sufficient margin of profit is being made. Use of the per cent of gross profit to check the correctness of the value set upon the inventory is called the gross profit test of inventory.

Rate per cent of gross profit. The gross profit test is based on the supposition that, in normal times and under normal conditions, any busi- ness will produce approximately the same per cent of gross profit on sales in any one period of time as in any other corresponding period of time.

Procedure. Statements of the gross profit and sales for each of several prior periods should be obtained. The gross profit for any one period divided by the sales for the same period gives the rate of gross profit for that period, based on sales. Disregard any per cent that is abnormal. Add the remaining per cents, and divide by the number added. The quotient is the average per cent of gross profit in prior periods.

Uses. The per cent of gross profit may be used in two ways: first, to prove inventories, and second, to compute the estimated inventory when it is impossible or impracticable to take a physical inventory.

Example. Assume that the average gross profit for the past five years has been 40 % of sales, and that an audit of the books shows that the inventory, taken prior to the beginning of the audit, and valued at $100,000, seems smaller than it should be, while the previous inventory and purchases amounted to $400,000. The sales for the period are $400,000. Show by comparative statement the possibility of error.

134

Gross Profit Computations 135

Solution. In the following set-up, both the average and the current per cents and results are shown. As the profit in prior years has been 40% of sales, the cost of goods sold has been 60% of sales. 60% of $400,000 (sales) = $240,000, cost of sales.

Current Year in Terms

Current Year of an Average Year

Actual Current Test Average

Amounts Per Cent Amounts '■ Per Cent

Sales. .... .' $400,000 100 $400,000 100

Cost of Sales First-of-year inventory

and purchases $400 , 000

Less: Current inv't 100,000 300,000 75 240,000 60

Gross profit $100,000 25% $160,000 40%

If the sales are correct, the cost of sales is $60,000 too high, unless the rate has really changed. This discrepancy may be caused by any of the following: the volume of sales may be incorrectly stated ; the current inventory may be errone- ous, and the cost of sales affected thereby; or there may be an abnormal increase in the cost of merchandise purchased, when compared with the 5-year average. The accountant should determine the reason for the discrepancy.

Cost of goods sold. The average rate per cent of gross profit, applied to the sales for the current period, will give the estimated gross profit for the current period. Deduction of the estimated gross profit from the sales gives the estimated cost of goods sold. This procedure may be reduced to a formula as follows:

Average for Prior Periods

1. Sales — Cost of sales = Gross profit.

2. Gross profit 4- Sales = Per cent of gross profit (based on sales).

Application to Current Period

3. Sales X Per cent of gross profit (prior periods) = Estimated gross profit.

4. Sales — Estimated gross profit (current period) = Estimated cost of sales.

Example.

Cost Gross

Sales of Sales Profit

First period $400,000 $300,000 $100,000

Second period 450,000 340,000 110,000

Third period 350,000 260,000 90,000

Fourth period 100,000

What was the cost of sales during the fourth period

136 Gross Profit Computations

Solution.

Average for Prior Periods

First period, $100,000 -i- $400,000 =25.00% Second period, 110,000 h- 450,000 = 24.44% Third period, 90,000 -^ 350,000 = 25.71%

75.15%

•75% -J- 3 = 25%, the average rate of gross profit.

Application to Current Period

$100,000 X 25% = $25,000, estimated gross profit. $100,000 - $25,000 = $75,000, estimated cost of sales.

Rate per cent of cost of sales. If the rate of profit has been based on cost price instead of on selling price, the cost of sales may be tested by the following computations:

Average for Prior Periods

1. Sales — Cost of sales = Gross profit.

2. Gross profit -J- Cost of sales = Per cent of gross profit (based on cost of sales).

Application to Current Period

3. Sales -s- (100% + Per cent of gross profit) = Cost of sales.

Example.

Sales Cost of Sales

First period $400,000 $300,000

Second period 450,000 340,000

Third period 350,000 260,000

Fourth period 100,000

What was the cost of sales for the last period?

Solution.

Average for Prior Periods

First period, $100,000 4- $300,000 = 33.33% Second period, 110,000 + 340,000= 32.35% Third period, 90,000 + 260,000 = 34.61%

100.29%

100% -s- 3 = 33^%, average per cent of gross profit.

Application to Current Period

$100,000 -J- 1.33£ (1 + .33£) = $75,000, cost of sales. $100,000 - $75,000 = $25,000, gross profit.

It follows that if the cost of sales can be found, any element (inven- tory at beginning of period, purchases, closing inventory, and so forth) which goes to make up the cost of sales can be found, provided the other elements of the costs are given.

Gross Profit Computations 137

Fire losses. Insurance companies are generally willing to settle inventory losses resulting from fire on the basis of values determined by the gross profit method.

Example. The insurance company agrees that the following facts are to be the basis of its reimbursement to the insured for his fire losses :

Average gross profit for 4 years, 40% of sales. Sales for this period to date of fire, $50,000. Cost of goods available for sale, $300,000.

Solution.

$50,000 (sales) X 40% (rate of gross profit) = $20,000, gross profit.

$50,000 (sales) - $20,000 (gross profit) = $30,000, cost of goods sold. $300,000 (goods available for sale) - $30,000

(cost of goods sold) = $270,000, estimated inventory

at date of fire.

Use of gross profit test in verification of taxpayer's inventory.

Assessors make vise of the gross profit test to determine the approximate inventory and to check the item of inventory in the schedule filed by the taxpayer, since assessment dates seldom coincide with closing dates. The following forms have been given to the taxpayer to fill out, the date of assessment being May 1.

For Merchants

1. Book value of last inventory of stock of merchandise

2. Add purchases since last inventory to May 1

3. Add in-freight and cartage paid since last inventory to May 1

4. Total of above three items

Deduct from above total net result of following two items :

5. Amount of net sales from date of last inventory to May 1

6. Less: Gross profit on sales estimated at %

(Previous year % may be used where actual % is un- known.)

7. Net inventory of merchandise on May 1 (Item 4 less Item 6)

For Manufacturers

1. Book value of raw materials, finished goods, and work-in-process at

last inventory. Date

2. Add purchases of raw materials and finished goods since last inven-

tory to May 1

3. Add amount paid for in-freight and cartage from last inventory to

May 1

4. Add amount paid for labor and manufacturing expenses from last

inventory to May 1

5. Total of above four items

Deduct from above total the net result of the following two items :

6. Amount of net sales from date of last inventory to May 1

7. Less: Gross profit on sales estimated at %

(Previous year % may be used where actual % is not known.)

8. Net value of raw materials, goods-in-process, and finished goods on

May 1 (Item 5 - Item 7)

1 38 Gross Profit Computations

PROBLEMS

1. From the figures in the following tabulation, calculate the per cent of gross profit for each year, and by means of the average per cent of gross profit calculate the inventory at the end of the first half of the fifth year.

Opening Closing Per Cent of

Sales Purchases Inventory Inventory Gross Profit

First year $120,000 $90,000 $10,000

Second year 150 , 000 100 , 000 10 , 000 12 , 000 -~

Third year 165,000 110,000 12,000 10,000

Fourth year 180 , 000 122 , 000 10 , 000 11, 000

Fifth year (6 mo.) 95,000 62,000 11,000

2. From the following facts, find the inventory as of December 31 :

Inventory, January 15 following, $16,578.50.

Sales, December 31 to January 15, $2,890.00.

$765 of the above sales shipped and invoiced before December 31.

Purchases, December 31 to January 15, at cost, $1,256.50.

Average gross profit, 25% of cost.

3. The average gross profit of the X. Company for the past three years has been 45% of the sales. During the fourth year the sales amounted to $159,500. Goods were purchased to the amount of $105,000. Returned purchases totaled $5,000 for the period. Freight paid on purchases was $6,000. The inventory at the beginning of the period was $40,000. Current market prices are 10% above the purchase prices for the year. Find the cost of replacing the goods at the end of the year.

4. On April 30, the board of managers of the Ames Mercantile Company removed the superintendent on the general suspicion that his books misrepre- sented the true financial condition of the business. Prepare a statement showing the nature and the probable extent of the misrepresentations; also an approxi- mate statement of income and profit and loss for the four months ending April 30.

The following is a trial balance taken from the books, April 30:

Capital Stock $ 75,000

Furniture and Fixtures $ 10 , 000

Inventory, January 1 , 128, 600

Cash 15,450

Accounts Payable 39 , 000

Accounts Receivable 24,600

Loans Payable

Sales

Purchases

Salaries, Salesmen

Advertising

Salaries, Office

Rent

Interest

Insurance, January 1 to December 31

Stationery and Printing

Reserve for Depreciation of Furniture & Fixtures . Surplus, January 1

10,000

51,000

40,700

2,200

1,650

1,100

400

200

999

105

2,710

48,294

$226,004

Gross Profit Computations 139

An analysis of the Purchases, Sales, and Inventory accounts revealed the following :

Opening Closing Purchases Sales Inv't Inv't

Firstyear $122,000 $153,750 $101,000 $100,000

Second year 123 , 000 153 , 170 100 , 000 102 , 000

Third year 121,000 154,722 102,000 128,600

5.* The books of a concern recently burned out contained evidence of pur- chases, including inventory, to the amount of $200,000, and sales of $40,800, since the last closing. Upon investigation, however, the auditor ascertained that a sale of merchandise had been made just prior to the fire, and not recorded in the books, at an advance of two-fifths over cost less a 10% cash discount; the profit on the transaction was $31,928. The past history of the business indi- cated an average gross profit of 50% on cost of goods sold.

(a) What amount should be claimed as fire loss?

(b) What rate of gross profit do the transactions finally yield?

6.f The store and stock of the Diamond Jewelry Company was destroyed by fire on November 1. The safe was opened, and the books were recovered intact. The trial balance taken off was as follows:

Cash in Bank $ 1 ,000

Accounts Receivable 10 , 000

Accounts Payable $ 30,000

Merchandise Purchases 90 , 000

Furniture and Fixtures 7 , 500

Sales 110,000

General Expense 18,000

Insurance 1 , 500

Salaries 5 , 500

Real Estate— Store Lot 50,000

Store Building 35,000

Capital Stock 50,000

Surplus 28,500

$218,500 $218,500

The average gross profit as shown by the books and accepted by the insurance companies was 40% of sales. The insurance adjuster agreed to pay 75% of the book value of furniture and fixtures, 90% of the book value of the store building, and the entire loss on merchandise stock.

Draft journal entries to include the account against the insurance companies.

Installment sales of personal property. The large increase in sales of personal property on the installment plan, and the option that the government allows a taxpayer coming within the definition of an installment dealer to return his gross income from sales on the install- ment basis are indications of the growing importance of this subject.

The installment plan of selling was devised for the purpose of stim- ulating sales, whereas the installment basis of reporting income was devised for the purpose of deferring from year to year the income to be

* American Institute Examination. f C. P. A., Oklahoma.

140 Gross Profit Computations

realized from installment sales, with a view to the possible effect that this deferment might have upon the amount of federal income tax to be paid. It is, of course, essential that the latest Federal Income Tax Law be observed.

Computation of gross profit. The gross profit to be reported may be ascertained by taking that proportion of the total cash collections received in the taxable year from installment sales (such collections being allocated to the year against whose sales they apply) which the annual gross profit to be realized on the total installment sales made during each year bears to the gross contract price of all such sales made during that particular year.

Example. The books of the Model Credit Company, selling merchandise on the installment plan, show the following:

First Second Third Fourth

Year Year Year Year

Sales $ 80,000 $110,000 $130,000 $90,000

Cost of Sales

Inventory (old) $ 45,000 $ 40,000 $ 50,000 $48,000

Purchases 55,000 75,000 90,000 50,000

$100,000 $115,000 $140,000 $98,000

Less: Inventory (new) 40,000 50,000 48,000 40,000

Cost of sales $ 60,000 $ 65,000 $ 92,000 $58,000

Gross profit $ 20,000 $ 45,000 $ 38,000 $32,000

Collections were made in the fourth year on each year's contracts as follows:

First Second Third Fourth

Year Year Year Year

$1,600 $4,800 $25,000 $70,000

What was the gross profit to be reported for the fourth year?

Solution.

Per Cent of Gross Profit

First year, $20,000 (gross profit) -

- $ 80,000 (sal

es) = 25.00%

Second year, 45,000 " "

- 110,000 '

' =40.91%

Third year, 38,000 "

- 130,000 *

' =29.23%

Fourth year, 32,000 "

- 90,000 " =35.55%

Profit on Collections in Fourth Year

Collected on first-year contracts, $ 1,600 X 25.00% = $ 400.00

"second-year " 4,800X40.91%= 1,963.68

"third-year " 25,000X29.23%= 7,307.50

" fourth-year " 70,000 X 35.55% = 24,855.00

Gross profit realized in the fourth year $34,526.18

Reserve for unearned gross profit. The gross income to be realized on installment sales is credited to " Reserve for Unearned Gross Profit," and at this time this account is debited with the gross profit on collections. The balance of the account represents gross profit on install- ment sales contracts remaining unpaid at the date of closing.

Gross Profit Computations 141

Example. The books of the X.Y.Z. Company, selling merchandise on the installment plan, show the following:

First Second Third Fourth

Year Year Year Year

Sales $89,257.99 $111,825.86 $137,012.32 $97,912.26

Gross profits 29,962.89 48,068.37 38,128.63 39,168.71

Collections during the fourth

year on each year's accts.. 1,635.35 4,832.00 25,182.14 69,927.92

What amount should be credited to Reserve for Unearned Gross Profit to represent deferred income for the fourth year? What amount should be debited to Reserve for Unearned Gross Profit to represent income realized from the first, second, third, and fourth years' collections received in the fourth year?

Solution.

(a) Per Cent of Gross Profit

First year $29,962.89 -h $ 89,257.99 = 33.57%

Second year 48,068.37

Third year 38,128.63

Fourth year 39,168.71

111,825.86 = 42.98% 137,012.32 = 27.83%

97,912.26 = 40.00%

Profit on Collections

First-year accounts $ 1,635.35 X 33.57% = $ 548.99

Second-year accounts 4,832.00 X 42.98% = 2,076.79

Third-year accounts 25,182.14 X 27.83 % = 7,008.19

Fourth-year accounts 69,927.92 X 40.00% = 27,971 . 17

Gross profit realized in 4th yr $37, 605 . 14

Journal entries

Installment Sales Contracts $ 97,912.26

Cost of Sales $ 58,743.55

Reserve for Unearned Gross Profit 39, 168.71

Cash $101,577.41

Installment Sales Contracts $101,577.41

Reserve for Unearned Gross Profit $ 37,605. 14

Realized Gross Profit on Installment Sales $ 37 , 605 . 14

Bad debts. The bad debts written off during the year should be allocated by years, and a charge should be made to Reserve for Unearned Gross Profit for the percentage of gross profit in each year's write-off, and to Profit and Loss (Bad Debts) for the remainder, the entire credit being made to Installment Sales Contracts.

Example. During the fourth year, bad accounts were written off as follows:

First-Yr. Accts. Second-Yr. Accts. Third-Yr. Accts. Fourth-Yr. Accts.

$67.65 $141.05 $65.62 $126.25

What amount should be charged to these accounts: Profit and Loss (Bad Debts), and Reserve for Unearned Gross Profit?

142 Gross Profit Computations

Solution.

Unrealized

Profit Remainder

$ 67.65 X 33.57% $ 22.71 $ 44.94

141.05X42.98% 60.62 80.43

65.62X27.83% 18.26 47.36

126.25 X 40.00% 50.50 75.75

$152.09 $248.48

Reserve for Unearned Gross Profit $152.09

Profit and Loss (Bad Debts) 248.48

Installment Sales Contracts $400 . 57

PROBLEMS

1. The X.Y.Z. Company's books for the 5th year showed :

Sales $128,642.60

Gross profit 42,975. 12

Collections were made in the fifth year on each year's contracts as follows:

1st Yr. 2nd Yr. 3rd Yr. 4th Yr. 5th Yr.

$230.60 $1,590.31 $9,326.80 $21,256.30 $82,327.58

Calculate: (a) the per cent of gross profit for the fifth year; (6) the amount to be credited to Reserve for Unearned Gross Profit; (c) the amount to be debited to Reserve for Unearned Gross Profit. Use the rates given in the solution on page 141 for the first four years.

2. The analysis of bad debts written off during the 5th year was:

lst-Yr. Acct. 2nd-Yr. Acct. Srd-Yr. Acct. 4th-Yr. Acct. 5th-Yr. AccL $8.35 $209.75 $910.40 $150.80 $470.62

What amounts should be charged to Reserve for Unearned Gross Profit and to Profit and Loss, respectively?

3. Results for the 6th year:

Sales $140,695.39

Gross profit 54,541 .07

Collections were made in the sixth year on each year's contracts as follows:

1st Yr. 2nd Yr. 3rd Yr. 4th Yr. 5th Yr. 6th Yr.

$62.70 $492.54 $2,798.30 $4,689.30 $2,657.80 $90,275.89

(a) Calculate the per cent of gross profit for the 6th year.

(6) Calculate for the 6th year the gross profit on collections made.

4. Accounts receivable were written off as follows:

1st Yr. 2nd Yr. 3rd Yr. 4th Yr. 5th Yr. 6th Yr.

$52.83 $31.50 $51.10 $150.00 $163.82 $108.28

Compute the charges to be made to Profit and Loss (Bad Debts) and to Reserve for Unearned Gross Profit.

Gross Profit Computations 143

5.* The "A & B" Company is engaged in the business of retailing musical merchandise. The majority of the sales consist of installment sales of pianos and hi-fi sets, on which the initial payment is less than 25% of the sales price and the balance is payable in monthly installments over a period of three to five years. The company was incorporated and began business on January 1. The following schedules are submitted on the various classes of merchandise:

Sales

Piano Install- Hi-fi Install- Other

ment Sales ment Sales Mdse Sales

First year $148,650.00 $92,475.00 $38,337.60

Second year 163,520.00 88,535.00 39,543.50

Third year 180,400.00 94,256.00 40,731.15

Purchases

First year 106,322.37 67,432.18 27,108.88

Second year 120,987.41 55, 116.92 27,224.35

Third year 140,125.25 60,013.22 27,469.33

Inventories

First year 20,103. 14 10,248.31 8,323.64

Second year 32,105.86 15,012.83 15,299.41

Third year 39,294.44 18, 144.77 13,521.31

Attention is called to the fact that " Other Merchandise Sales" are sales for cash, or credit sales other than installment sales.

No adjustments to Deferred Income account are made until the end of the year. Additions to this income are made at the end of the year on the basis of the balance due on the current year's installment sales, and deductions are made on the basis of cash received during the current year on installment sales of previous years. On December 31 of the third year, the unpaid balances on third-year piano installment sales amount to $110,425.50, and on third-year hi-fi installment sales to $60,475.00 — exclusive of accrued interest. The follow- ing amounts were received during the third year on installment sales of previous years :

On first-year piano installment sales $30 , 285 . 00

On second-year piano intallment sales 42 , 413 . 00

On first-year hi-fi installment sales 25 , 386 . 00

On second-year hi-fi installment sales 26 , 285 . 00

The above amounts are exclusive of interest, which is credited direct to interest revenue.

Fractional percentages may be disregarded in the computation of ratios — over J of 1% should be added, and less than i of 1% should be dropped.

On first-year hi-fi installment sales, uncollectible balances amounting to $399.00 were charged on the books of the company to expense and credited to installment sales contracts.

Federal income taxes paid in the third year were charged to surplus.

C. P. A., Michigan.

144 Gross Profit Computations

Depreciation is calculated at the following rates:

Buildings 2 % Furniture and Fixtures 10 %

Auto Trucks 25%

The following is a copy of the trial balance as of December 31, end of third year, before closing and before apportionment of deferred income on installment

sales :

Cash $ 15,327.48

Notes Receivable 2,000.00

Accounts Receivable 20, 842 . 11

Installment Sales Contracts 265,418.50

Inventories 62,418.10

Government Bonds 5,000.00

Real Estate 10,000.00

Buildings 40,000 .00

Furniture and Fixtures 4 , 500 . 00

Auto Trucks 3,000.00

Notes Payable $ 50,000.00

Accounts Payable 13,458.25

Deferred Income on Installment Sales 83 , 245 . 70

Reserve for Depreciation, Buildings. 1,600.00

Reserve for Depreciation, Fur. & Fix 900.00

Reserve for Depreciation, Trucks 1,500.00

Capital Stock 150,000.00

Surplus 75,556.21

Sales 315,387.15

Piano Rentals 1 ,785 .00

Interest on Installment Sales 2 , 035 . 23

Interest on Government Bonds 237 . 50

Cash Discounts on Purchases 2 , 452 . 07

Purchases 227,607.80

Salaries, Officers 14,000.00

Salaries, Store 8, 101 . 46

Light and Heat 717.68

Advertising 4 , 015 . 71

Truck Expense 508 . 53

Sundry Store Expense 2 ,239 . 17

Salaries, Office 2,020.00

Traveling Expense 648 . 50

Postage 472 . 30

Telephone and Telegraph 441 . 40

Insurance 1 , 309 . 06

Real Estate and Personal Property Taxes . . 2 , 029 . 69

Bad Debts, Accounts Receivable 709 . 66

Bad Debts (first-year hi-fi installment sales) 399.00

Repairs, Sundry 365 . 68

Donations . 200 . 00

Cash Discounts on Sales 444 . 48

State Franchise Tax 187 . 80

Capital Stock Tax 233 . 00

Interest Paid 3 , 000 . 00

,157.11 $698,157.11

Gross Profit Computations 145

You are asked to give: (a) the net taxable income (for federal tax purposes) for the third year; (b) a balance sheet of the "A & B" Company as of January 1, beginning of fourth year.

Deferring income; its effect on tax. The statement was made in the second paragraph of this subject that the installment basis of account- ing defers income with a view to the possible effect that deferment may have on the amount of federal income tax to be paid. Since the income is deferred, the tax is deferred (not saved).

The amount of profit realized and to be realized from the sales of a particular year, if not taxed in that particular year, will be taxed even- tually, and the saving of tax results from a possible reduction in the rate of tax or from the spreading of taxable income over several years. If it is anticipated that the rate of tax will be increased, it may not be wise to defer the income.

Second, a change from the accrual to the installment basis results in double taxation, for Section 44 (c) of the Internal Revenue Code pro- vides as follows: "If a taxpayer entitled to the benefits of subsection (a) elects for any taxable year to report his net income on the installment basis, then in computing his income for the year of change or any sub- sequent year, amounts actually received during any such year on account of sales or other dispositions of property made in any prior year shall not be excluded.' '

The amount of gross income which may be deferred on installment sales is governed by:

(1) The terms of sale;

(2) Annual increase, if any, in sales;

(3) Per cent of year's sales collected in the current year; and

(4) Fluctuation of gross profits.

Example. Assume the terms of sale to be 10% down, and 10% a month; the annual increase in sales to be $10,000; the per cent of year's sales collected, and the sales throughout the year, to be uniform, and the per cent of gross profit to be fixed.

Gross Profit

Year Sales on Sales

First $50,000 30%

Second 60,000 30%

Third 70,000 30%

Fourth 80,000 30%

Fifth 90,000 30%

Since it has been assumed that the sales are uniform throughout the }-ear and that collections are met promptly, the second year's business may be analyzed as follows:

146

Gross Profit Computations

Down Payments

Jan 10% of $ 5,000 = $ 500

Feb 10% of 5,000 =

Mar 10% of 5,000 =

Apr 10% of 5,000 =

May 10% of 5,000 =

June 10% of 5,000 =

July 10% of 5,000 =

Aug 10% of 5,000 =

Sept 10% of 5,000 =

Oct 10% of 5,000 =

Nov 10% of 5,000 =

Dec 10% of 5,000 =

Year's sales $60,000

Down payments $ 6 , 000

Install, payments 31,500

Total payments $37 , 500

Ratio of payments to sales: $37,500 ■*■ $60,000 = 62

Installment Paymenti

500

10% of $ 5,000 = $

500

10% of 10,000 =

1,000

500

10% of 15,000 =

1,500

500

10% of 20,000 =

2,000

500

10% of 25,000 =

2,500

500

10% of 30,000 =

3,000

500

10% of 35,000 =

3,500

500

10% of 40,000 =

4,000

500

10% of 45,000 =

4,500

500

10% of 45,000 =

4,500

500

10% of 45,000 =

4,500

$31,500

5%.

Fourth

Fifth

Year

$80,000

$90,000

30%

24,000

27,000

A comparison of the income to be reported on the accrual basis and on the installment basis may be made as follows:

Accrual Basis

Second Third

Year Year

Sales $60,000 $70,000

Gross profit (%) 30% 30%

Gross profit ($) 18,000 21,000

Installment Basis

Collections:

lst-year accounts $18,750

2nd-year accounts 37 , 500 $22 , 500

3rd-year accounts 43 , 750

4th-year accounts

5th-year accounts

Gross income to be reported:

30% of lst-year coll $ 5,625

30% of 2nd-year coll 11,250 $6,750

30% of 3rd-year coll 13, 125

30% of 4th-year coll

30% of 5th-year coll

Total income reported $16,875 $19,875

Income deferred $ 1,125 $ 1,125

$26,250 50,000

$ 7,875 15,000

$30,000 56,250

$ 9,000 16,875

$22,875 $25,875

$1,125 $ 1,125

It may be observed from the foregoing analysis that with an annual increase of $10,000 in sales, and with a constant gross profit ratio of 30% the amount of income deferred from year to year is $1,125.

With an annual increase of $20,000 in sales, and other conditions the same, the amount of income deferred would be $2,250 (2 X $1,125).

Gross Profit Computations 147

PROBLEMS

1. Assume the terms of sale to be 10% down and 5% a month, the annual increase in sales $10,000, the per cent of year's sales collected and the sales throughout the year uniform, and the per cent of gross profit fixed1.

Gross Profit

Year Sales on Sales

First $50,000 30%

Second 60,000 30%

Third 70,000 30%

Fourth 80,000 30%

Fifth 90,000 30%

Show the amount of income deferred when the installment basis is used.

2. If the terms of payment were 5% down and 5% a month, and other con- ditions were the same as in Problem 1, what would be the amount of income deferred each year?

Analysis of Statements

Financial and operating ratios. An analysis of the financial and the operating ratios of a business means a study of the relationships that are expressed in the statistics presented. Well-known and com- monly used ratios are those of expenses and earnings to sales, and of earnings on capital employed. Other ratios, relationships, and turnovers that are indicators of the condition of a business should also be considered.

A summary of financial and operating ratios, relationships, and turn- overs would include the following:

(1) Ratio of costs and expenses to net sales.

(2) Ratio of gross profit to net sales.

(3) Ratio of operating profit to net sales.

(4) Ratio of net profit to net sales.

(5) Ratio of operating profit to total capital employed.

(6) Ratio of net profit to net worth.

(7) Earnings on common stockholders' investments.

(8) Working capital ratio.

(9) Sources of capital.

(10) Manner in which capital is invested.

(11) Turnover of total capital employed.

(12) Turnover of inventories.

(13) Turnover of accounts receivable.

(14) Turnover of fixed property investment.

There are many other ratios which are important measures of effi- ciency, but of which only brief mention can be made in this chapter. Depending on the type of business being analyzed, these other ratios

148

Analysis of Statements 149

might include the labor turnover, the unit of output per operative, the average wage per man, the average wage per hour, and other statistics.

Costs, expenses, and profits. Costs, expenses, and profits should be expressed as per cents of money values and, where possible, should be expressed in terms of dollars per production unit, such as the ton, pound, yard, or gallon. The per cents, compared with those of previous years, show whether sales prices have been adjusted proportionately to costs of production and distribution. The unit prices supplement the per cents and afford a direct comparison.

Ratio of gross profit to net sales. The ratio of gross profit to net sales is an indication of the spread between the cost of production and the selling price. The gross profit must be as large as possible, for out of it must come the expenses of selling, administration, finance, and other charges, before a net return is realized on capital.

Ratio of operating profit to net sales. The ratio of operating profit to net sales expresses the basic relationship between profits and sales. Operating profits represent the gain before the deduction of federal taxes, interest on borrowed money, and extraordinary losses but do not include miscellaneous income not attributable to ordinary oper- ations.

Ratio of net profit to net sales. The ratio of net profit to net sales indicates the margin of profit on the selling price. The rapidity of stock turnover, and the capital invested in accounts receivable, in inven- tory, and in plant should be considered with this ratio.

Ratio of operating profit to total capital employed. The ratio of operating profit to total capital employed forms a ready basis for a comparison of the operating results of a business or of several plants under a single control. Capital employed includes plant, inventories, accounts receivable, cash balances, and so forth, regardless of the source of such capital, and is readily determined by referring to the asset side of the balance sheet.

Ratio of net profit to net worth. The ratio of net profit to net worth expresses the measure of earnings available to the stockholders or proprietors, and is the final indicator of the success or failure of any business.

Earnings on common stockholders' investments. The earn- ings on common stockholders' investments are based on the stockholders' share of the net profit, in relation to their interest in the net worth of the business. There are two ways in which these earnings may be stated: (a) as a per cent of the amount of such investments, and (b) in dollars earned per share outstanding.

Example. The following profit and loss statement, together with certain other facts, is presented to illustrate items 1-7 in the summary on page 148. The numbers in parentheses refer to the numbered ratios in the summary.

150 Analysis of Statements

Blank Mercantile Company

Profit and Loss Statement For the Twelve Months' Period Ended December 31, 19 — .

Sales :

Gross Sales $693,004 . 10

Less: Sales Rebates and

Allowances $ 870.64

Prepaid Freight 200. 2b

1,070.89

Net Sales $691,933.21 100.00%

Cost of Sales :

Inventory, beginning of

year $107,278.46

Purchases $624,225.28

Freight 16,271.98

$640,497.26 Less : Purchase Rebates and

Allowances 630.81

639,866.45

$747,144.91

Inventory, end of year. . . . 124,814.04

Cost of Sales 622,330.87 89.94 (1)

Gross Profit $69,602.34 10.06% (2)

Delivery Expenses:

Salaries of Drivers $ 3 , 414 . 34

Dep'n on Equipment 2 , 839 . 57

Auto Repairs 1 , 562 . 53

Gasoline and Oil 1,479.27

Drivers' Expenses 119.40

Drayage 66.84

Total $ 9,481.95 1.37 (1)

Selling Expenses:

Salesmen's Salaries $ 11,812.50

Salesmen's Expenses 1 , 942 . 06

Advertising 844 . 32

Telephone and Telegraph . . 642 . 57

Total " $ 15,241.45 2.20 (1)

General Expenses:

Salaries $ 8,722.33

Expenses 613.36

Executive Salaries 3 , 600 . 00

Taxes (other than federal) . 1 , 906 . 23

Insurance 1 , 723 . 46

Depreciation 1 , 259 . 54

Light, Heat, and Water ... 829 . 49

Printing and Stationery . . . 444 . 50

Postage 408.52

Collections 219.76

Repairs 115.91

Storage 22 . 69

Miscellaneous 380.50

Total 20,246.29 2.93 (1)

Total Expense 44,969.69 6.50% (1)

Net Operating Profit $ 24,632.65 3.56% (S)

Analysis of Statements

151

Additions to Income :

Discount on Purchases

9,565.86

Interest Earned

563.32

Bad Debts Recovered

102.53

Total

10,231.71 $ 34,864.36

1.48 5.04%

Deductions from Income :

Discount on Sales

4,771.92

Interest Paid for Money

Borrowed

4,373.16

Interest Paid on Building

Contract

3,010.00

Bad Debts Reserve

1,283.91

Donations

162.00

Total

13,600.99

1.97

Net Profit

$ 21,263.37

3.07% (4)

Supplemental

Total Capital Used (see Balance Sheet, below) $276,317.34

Ratio of Profit to Capital

Net Worth (beginning of year) 124,252 . 36

Ratio of Profit to Net Worth

Common Stock Outstanding 113,400.00

Number of Shares ($50.00 par value) 2,268

Per cent earned

Dollars earned per share

7.69% (5)

17.11

18.75 9.38

(6)

(7)

Working capital ratio. This ratio is probably the best-known measure applied to financial statements, because more than any other it has been stressed by bankers and businessmen. It is computed by dividing the amount of the current assets by the amount of the current liabilities. If the quotient is 2, the current assets are said to be in a "2 to 1 " ratio; that is, in a ratio of $2 of current assets to each $1 of current liabilities.

What the working capital ratio should be depends upon differences in types of business, location, and other factors the effect of which is to vary somewhat the proportions involved. While some lines of trade may be expected to maintain a 2-to-l ratio, others may necessitate a proportion as high as 10 to 1.

The rapidity with which receivables and inventory are turned is a factor bearing on the adequacy of the working capital ratio. With respect to accounts receivable, there is a range of turnover from 3 days in some of the retail chain stores to 80 or 90 days in coal and heavy manufacturing industries. The turnover of inventories is most rapid in such industries as slaughtering and meat packing, retail chain stores, chemical products, and iron and steel, while the turnover of inventories is found to be slow in such industries as tobacco products, machinery manufacturing, leather products, and rubber goods.

Example. The following balance sheet is presented to illustrate the working capital ratio. It will also be referred to in later paragraphs, where the computation of other ratios is discussed.

1 52 Analysis of Statements

Blank Mercantile Company

Balance Sheet December 31, 19 — .

Assets Current :

Cash in Banks $ 13,598.85

Cash on Hand 4,113.24 $ 17,712.09

Accounts Receivable — Customers $ 64 , 832 . 57

Accounts Receivable — Others 647 . 92

Notes Receivable — Customers 5,329.91

Notes Receivable — Others 227 . 31

Securities 1,274.34

Accrued Interest 32 . 98

Railroad Claims 93 . 76

$ 72,438.79

Less: Reserve for Bad Debts 1,890.06 70,548.73

Merchandise Inventory 124,814.04

Total $213,074.86

Fixed :

Land $ 3,450.00

Warehouse Building $ 50,373.48

Warehouse Equipment 545 . 77

Delivery Equipment 14,090.39

Furniture and Fixtures 2 , 488 . 85

$ 67,498.49 Less: Accumulated Depreciation. ..... 9,152.48 58,346.01

Total $ 61,796.01

Deferred Charges:

Prepaid Insurance $ 1,298. 13

Prepaid Interest 148.34

Total 1,446.47

$276,317.34 Liabilities Current :

Payroll $ 1,131.77

Accounts Payable 16 , 177 . 08

Notes Payable— Banks 50,000.00

Notes Payable— Others 17,600.00

Notes Payable— Stockholders 11 , 700 . 00

Accrued Taxes 1 , 575 . 17

Accrued Interest— Notes 1 , 393 . 92

Accrued Interest — Contracts 3,010.00

Total $102,587.94

Fixed:

Warehouse Contract for Deed 43,000.00

Net Worth:

Capital Stock— Common $113,400.00

Surplus 17,329.40 130,729.40

$276,317.34

In the foregoing balance sheet, the current assets are stated at $213,074.86. and the current liabilities are stated at $102,587.94.

213,074.86 ^- 102,587.94 = 2.077.

The ratio of working capital is, therefore, 2.077.

Analysis of Statements 153

Sources of capital. The sources of capital may be stated in a gen- eral way under four headings, as follows :

(1) Short-term borrowings and credits.

(2) Long-term borrowings and credits.

(3) Stockholders' investments.

(4) Surplus (earnings left in the business).

Summarizing the liability section of the foregoing blance sheet and dividing each section total by the total of all sections, the ratio of capital supplied by each source is as shown in the right-hand column of the fol- lowing tabulation :

Amount Per Cent

Current Liabilities $102,587.94 37.13

Fixed Liabilities 43,000.00 15.56

Capital Stock— Common 113,400.00 41.04

Surplus 17,329.40 6.27

$276,317.34 100.00

Manner in which capital is invested. The manner in which the capital is employed in the business is shown by a summary of the asset sections.

Amount Per Cent

Current Assets $213,074.86 77.12

Fixed Assets 61,796.01 22.36

Deferred Charges 1 , 446 . 47 .52

$276,317.34 100 . 00

Turnover of total capital employed. This item expresses the relation of the net sales to the total capital employed. The average capital employed throughout the year should be used, but, in the absence of monthly statements, the capital at the beginning of the year and the capital at the end of the year should be added and divided by 2 to give an estimate of the average capital employed. In arriving at this average, investments not employed in operations should be eliminated from the total assets, for, as a rule, they represent a surplus not required in the conduct of the business. Income from such investments should be eliminated from the statement of earnings before the ratio is computed.

Total assets at beginning of year $246,351 .89

Total assets at end of year 276 , 317 . 34

2)$522,669.23 Average capital employed (securities not eliminated, as

the amount was negligible) $261 , 334 . 61

The turnover of total capital employed is therefore:

$691,933.21 (net sales) ^- $261,334.61 (average capital) = 2.64.

Turnover of inventories. The subject of inventory turnover was presented in Chapter 5.

154 Analysis of Statements

The rate of turnover is computed as follows:

$622,330.87 (cost of sales) -J- $112,131.69 (average inventory) = 5.55.

Turnover of accounts receivable. The normal credit period, whether it be 30, 60, or 90 days, is compared with the average number of days' sales uncollected obtained from the following formula, as a means of judging the efficiency of the collection department:

Accounts receivable at end of fiscal period . ...

-— — - — - — = r-^ X Days in fiscal period

Sales for fiscal period

= Average number of days' sales uncollected.

The Accounts Receivable account showed $64,832.57 of outstanding accounts at the close of the fiscal period. The sales for the fiscal period of 12 months amounted to $691,933.21, and the average term of credit granted at time of sale was 30 days. The average number of days' sales represented in standing accounts is computed as follows:

64,832.57

691,933.21

X 365 = 34.

If the average number of days' sales uncollected is greater than the average term of credit, the presence of overdue accounts is indicated. This is true of the example just given.

Turnover of fixed property investment. This turnover expresses the relationship between the volume of business done and the capital invested in plant and equipment. Large investments in property and equipment increase the expense burden through charges for depreciation, insurance, taxes, and so forth, and may make a favorable or an unfavor- able operating statement, depending on the volume of business handled.

The number of dollars of sales for each dollar of fixed property invest- ment is calculated as follows:

$691,933.21 (net sales) + $58,346.01 (net fixed property investment) = 11.86.

PROBLEMS

1. From the balance sheets and supplemental information, determine the ratios named following the balance sheets.

Assets This Year Last Year

Current Assets $215,003.48 $213,074.86

Fixed Assets— Net 57,535.04 61,796.01

Deferred Charges 1,193.59 1,446.47

Total $273,732.11 $276,317.34

Liabilities

Current Liabilities $ 86,229.30 $102,587.94

Fixed Liabilities 38,000.00 43,000.00

Total Liabilities $124,229.30 $145,587.94

Analysis of Statements

155

Net Worth

Capital Stock $124,300.00

Surplus 25,202.81

Total Net Worth $149,502.81

$114,300.00 16,429.40

$130,729.40

Total $273,732.11 $276,317.34

Annual Sales $688,167.98 $691,933.21

Annual Expense 47,340.74 44,969.69

Ratios

Current Ratio - —

Worth to Debt -

Worth to Fixed Assets

Sales to Fixed Assets

Sales to Current Debt -

Sales to Worth

Expense to Sales (%) - -

2. The United Manufacturing Company's card in the credit file of the Second National Bank contained the data for the year ended January 31, 1960, and from their balance sheet and profit and loss statement you have entered the comparative figures for the year ended January 31, 1961. Compute the com- parative ratios for 1961. (See Fig. 8-1.)

COMPARATIVE RATIOS

COMPARATIVE FINANCIAL STATEMENTS

1/31 1/31 1960 1961 19 19 19

ASSETS i^go jgs1! 19 19 19

FIXED ASSETS TO TANGIBLE NET WORTH

24. 8

CASH

3,206 1 1,862

ACCOUNTS RECEIVABLE

45,199

42,267

CURRENT DEBT TO TANGIBLE NET WORTH

48.0

NOTES. TRADE ACCEPT RECV.

INVENTORIES

89,342

83,218

NET WORKING CAPITAL REP BY FUNDED DEBTS

NET SALES TO

.INVENTORY

4.3

NET WORKING CAPITAL REP BY INVENTORY

107.7

TOTAL CURRENT .

168,709

127,349

DUE FROM AFFILIATE OR SUBS7

INVENTORY COVERED BY CURRENT DEBT

61.3

LAND. BUILDINGS

MACHINERY-FIXTURES

28,244

49,248

AVERAGE COLLECTION PERIOD

42. 5

NOTES. ACC7S (OFf ICERS. PMTNEK)

2,716

.2,156

Organization Expense

999

TURNOVER OF TANGIBLE NET WORTH

3.4

TURNOVER OF

NET WORKING CAPITAL

4.7

TOTAL ASSETS

168,709

179,754

WET PROFITS ON NET SALES

.52

LIABILITIES

NET PROFITS ON TANGIBLE NET WORTH

1.8

ACCOUNTS PAYABLE-TRADE

34,037

32,671

ACCEPT., NOTES PAYABLE

NET PROFITS ON NET WORKING CAPITAL

2.4

BANKS PAYABLE

10,000

13,040

PAYABLE AFFILIATE OR SUBS'Y

CURRENT ASSETS

TO CURRENT DEBT

2.5

ACCRUALS

2,557

.2,793

Due Officers

8,130

5,321

TOTAL DEBT INCLUDING N. W. TANGIBLE NET WORTH

148.0

TOTAL CURRENT

54,725

55,825

SALES

388,553

394,774

MORTGAGES

EXPENSES

CHATTEL MORTGAGES

Deferred Bank Loan

8,280

NET PROFIT

2,048

1,664

WORKING CAPITAL

83,022

71,523

TANGIBLE NET WORTH

113,983

114,649

FIXED ASSETS

28,244

49,244

TOTAL LIABILITIES

54,725

64,105

CAPITAL STOCK

103,100

FUNDED DEBT

8,280

SURPLUS

10,883

12.548

TOTAL LIABILITIES & NET WORTH

168,709

179,754

Fig. 8-1.

3. From the data given in the following balance sheet and profit and loss statement, together with the supplemental data, compute the 14 financial and operating ratios relationships, and turnovers outlined in the preceding sections of this chapter.

156 Analysis of Statements

Blank Mercantile Company

Balance Sheet December 31, 19 —

Assets Current :

Cash in Banks $ 13,771.58

Cash on Hand 3,616.34 $ 17,387.92

Accounts Receivable — Customers $ 59,424.48

Accounts Receivable — Others 704 . 30

Notes Receivable — Customers 3 , 746 . 76

Notes Receivable— Others 272 . 19

Securities 994.64

Accrued Interest 52 . 30

Railroad Claims 50 . 95

$ 65,245.62

Less: Reserve for Bad Debts 3,852.57 61,393.05

Merchandise Inventory 136,222.51

Total $215,003.48

Fixed:

Land $ 3,450.00

Warehouse Building $ 50, 180.55

Warehouse Equipment 545 . 77 ■'

Delivery Equipment 14,090.39

Furniture and Fixtures 2 , 503 . 85

$ 67,320.56 Less: Accumulated Depreciation 13,235.52 54,085.04 57,535.04

Deferred Charges:

Prepaid Insurance 1, 193.59

Total $273,732.11

Liabilities Current :

Payroll $ 1,116.17

Accounts Payable 13,325.73

Notes Payable— Banks 24,000.00

Notes Payable— Others 14,500.00

Notes Payable— Stockholders 26,700.00

Accrued Taxes 1,641.97

Accrued Interest — Notes 2,285.33

Accrued Interest — Contracts 2 , 660 . 00

Total $ 86,229.20

Fixed:

Warehouse Contract for Deed 38,000.00

Net Worth:

Capital Stock— Common $123,400.00

Surplus 26,102.91 149,502.91

Total. - $273,732.11

Analysis of Statements 157

Blank Mercantile Company

Profit and Loss Statement For the Year Ended December 31, 19 —

Sales :

Gross Sales $689,361 .43

Less: Sales Rebates and Allow- ances $ 1,059.89

Prepaid Freight 133.56

1,193.45

Net Sales $688,167.98 100.00%

Cost of Sales:

Inventory, beginning of year $124,814.04

Purchases $611,332.45

Freight 15,184.68

$626,517.13 Less: Pur. Rebates and Allow- ances 1,392.74

625,124.39

$749,938.43

Inventory, end of year 136,222.51

Cost of Sales 613,715.92 %

Gross Profit $ 74,452.06 _ %

Delivery Expenses: ■

Salaries of Drivers $ 3 , 874 . 27

Dep'n on Equipment 2 , 818 . 08

Auto Repairs 1 , 430 . 61

Gasoline and Oil 1 ,231 .29

Drivers' Expenses 125 . 35

Drayage 52 . 91

Total $ 9,532.51 %

Selling Expenses:

Salesmen's Salaries $ 12,300.00

Salesmen's Expenses 2,015.78

Advertising 1,357.83

Telephone and Telegraph 536.21

Total 16,209.82 %

General Expenses:

Salaries $ 8, 797 .50

Expenses 265.43

Executive Salaries 4, 175 .00

Taxes (other than federal) .... 2,069. 17

Insurance 1,937.82

Depreciation 1,264.96

Light, Heat, and Water 826.33

Printing and Stationery 516.70

Postage 486.85

Collections 238.65

Repairs 106.47

Storage 18.29

Miscellaneous 895 . 24

Total $ 21,598.41 %

Total Expense 47,340.74 %

Net Operating Profit $ 27,111.32 %

158 Analysis of Statements

Additions to Income:

Discount on Purchases

Interest Earned

.... $

9,759.20

1,348.60

10.65

Bad Debts Recovered

Total

11,118.45

— %

Deductions from Income:

Discount on Sales

Interest Paid for Money Borrowed

.... $

$ 38,229.77

4,523.98 4,443.87 2,660.00 3,446.80 269 . 20

Interest Paid on Building Contract

Bad Debts Reserve

Donations

Total

15,343.85

Net Profit

$ 22,885.92

Supplemental

Total Capital Employed (see Balance Sheet) $

Ratio of Profit to Capital %

Net Worth (beginning of year) 130,729.40

Ratio of Profit to Net Worth %

Common Stock Outstanding (see Balance Sheet)

Number of Shares ($50.00 per value)

Per Cent Earned %

Dollars Earned Per Share

Goodwill

Definition. Goodwill is an intangible asset, and may be denned in general terms as the value of any benefits or advantages which may accrue to a business from its being soundly established, bearing a good reputation, having a favorable location, and so forth. It results in the earning of a higher rate of net income than that of less fortunate concerns in the same line of business.

Basis of valuation. When two or more businesses are consolidated or merged, the payment made for each business depends upon:

(1) The value of the net assets of each business.

(2) The earning power of each business.

A committee should be formed, consisting of members from each of the businesses being consolidated or merged (proprietorship, firm, or corporation) ; this committee should have the assistance of an appraiser and an accountant in the preparation of a report dealing with the net assets and the earning power.

The report should contain a balance sheet of each business, stating the values at which it is proposed to take over the assets, and stating the liabilities to be assumed.

The value of the fixed assets and of the inventory should be determined by the appraiser. The accountant, after making an audit, should submit the other balance sheet items.

Earning power determined from profit and loss statements. The following points should receive consideration when earning power is being determined from profit and loss statements:

(1) Number of years included. The value of goodwill depends to some extent on whether profits have been uniform year after year, oi- ls?

160 Goodwill

have steadily increased or decreased, or have fluctuated from year to year. Therefore, in order to show the trend of profits, it is necessary to have profit and loss statements for several years. A statement of average profits is insufficient, as it does not show the trend.

(2) Adjustments to correct profits. Adjustments may be necessary to correct errors, such as:

(a) Wrong classification of capital and revenue expenditures.

(6) Omission of provision for depreciation, bad debts, and so forth.

(d) Anticipation of profits on consignments and sales for future delivery.

(3) Uniformity of methods.

(a) If the methods of computing the manufacturing costs are not uniform, the cost statements should be revised and put on a uniform basis.

(6) The depreciation charges should be analyzed as to method and rate. If different methods and rates have been used, adjust- ments should be made so that the charges will have been cal- culated on a uniform basis.

(c) There may be a wide difference in the management salaries paid by the consolidating companies for the same services. The salaries should be adjusted. In a single proprietorship or partnership, salaries may not have been paid or credited; in that case they should be included at an arbitrary figure.

(d) If, in a partnership, interest on capital has been charged as an expense, the entries should be reversed and the item of interest on capital thus eliminated.

(4) Eliminations. Eliminations may have to be made for extra- ordinary and non-operating profits or losses.

Methods of valuing goodwill. Goodwill may be valued on the basis of:

(1) An appraisal of goodwill.

(2) A number of years' purchase price of the net profits.

(3) A number of years' purchase price of excess profits over interest on net assets.

Capitalization of profits in excess of interest on net assets is usually calculated as follows :

Net assets $100,000.00

Profits 10,000.00

Interest on net assets @ 6% 6,000.00

Excess of profits over interest 4,000.00

Excess capitalized at 20% (4,000 -*- .20) 20,000.00

Goodwill 161.

Case illustrations. The following four cases of goodwill valuation, taken from reports of consolidations, show how goodwill has been valued in practice.

Case 1. The goodwill of the consolidating units was fixed at the sum of the profits for the two preceding years, plus an additional 10%.

Case 2. The goodwill was based on the total profits for the five years preceding, less five years' interest on the net worth.

Case 3. The goodwill was the average annual earnings for the four years preceding consolidation, less the following deductions:

(a) Profits on favorable contracts about to expire.

(6) $100,000 for the estimated value of services rendered by the

retiring president, (c) 6% interest on actual capital invested.

The remainder was capitalized on a 10% basis.

Case 4. From the net profits of each company the following items were deducted:

(a) 7% on capital actually employed.

(b) 14% on sales.

(d) 4% depreciation on frame buildings.

(e) 8% depreciation on machinery.

The remainder was capitalized at 20%, or 5 times the amount of such earnings in excess of 7% on capital and other deductions agreed upon.

Valuation by appraisal. There is no particular problem in the calculation of the value of goodwill by appraisal. It may be appraised by a disinterested party; or, more often, it is the amount on which the vendor and the vendee agree. They usually appraise the net assets, and agree that the purchase price shall be a certain amount in excess of the value of the net assets. This excess is the payment for goodwill.

Valuation by number of years' purchase price of net profits. The goodwill may be estimated at so many years' purchase price of the net or gross profits of any one year, or at so many years' purchase price of the average profits of a number of years.

Example. The consideration of the sale of a business, as agreed to between the parties, is four years' purchase price of the average profits for the preceding three years, plus the net value of the assets.

Net value of assets $100 ,000

Profits of preceding three years:

1st year $20,000

2nd year 15,000

3rd year 28,000

What is the selling price of the business?

162 Goodwill

Solution.

Net value of assets $100 , 000

Profits of preceding three years:

1st year $20,000

2nd year 15,000

3rd year 28,000

$63,000

$63,000 -5- 3 = $21,000, average profits for three ~ years.

$21,000 X 4 (goodwill) 84,000

Selling price $184,000

Valuation on basis of excess of profits over interest on net assets. The value of goodwill is calculated under this method by, first, deducting from the average profits a fair return of interest on the capital invested, and, second, by multiplying the remainder of the profits, or the excess, by an agreed number of years' purchase price.

Example. A agrees to buy a certain business, and to pay for it in cash. He agrees to give dollar for dollar of the value of the net assets, plus a six years' purchase price of the excess of the profits over the interest on capital at 6%. Net assets are valued at $100,000, and average profits are $18,000. What is the purchase price of the business, including goodwill?

Solution.

Net assets $100,000

Profits, average $18,000

$100,000 X .06 6,000

Excess profits $12,000

$12,000 X 6 (goodwill) 72,000

Purchase price $172,000

It would be more favorable to the seller to determine the value by using a higher rate of interest and capitalizing the excess profits at this rate ; thus :

Net assets $100,000

Profits, average $18,000

$100,000 X .08 8,000

Excess profits $10,000

$10,000 -J- .08 (goodwill) 125,000

Purchase price $225,000

$225,000 - $172,000 = $53,000, advantage to the seller.

In the foregoing example, the goodwill represents the capitalization of that portion of the profits which is not attributable to the net tangible assets. The rate to be used depends largely on the kind of business under consideration. In some lines of business the per cent may be as low as 6% or 8%; in others it may be 10%; and in still others, 15% or even 20%.

Basis of stock allotment. Since most phases of the calculation of the value of goodwill are found in consolidations, an example of consolida-

Goodwill 163

tion is given. The matter of stock allotment is included, because when an agreement has been reached as to the valuation of the assets and as to the earning power of each of the businesses, the next question to decide is the method of making payment.

The following three typical methods will be presented :

(1) Payment entirely in common stock.

(2) Payment in preferred stock for the net assets; payment in common stock for the goodwill.

(3) Payment in bonds for the fixed assets, or for an agreed percentage thereof; payment in preferred stock for the balance of the net assets; payment in common stock for the goodwill.

In the allotment of securities, the fundamental rule is to distribute them in such a manner that, if the income of the consolidation is the same as the combined income of the several businesses, each of the old busi- nesses, or the former owners or stockholders thereof, will receive the same net income as before the consolidation.

To illustrate how this principle would operate under each of the three methods outlined, assume that three companies are to be consolidated on the basis of the following statements:

Net assets $40,000

Average earnings 4 , 000

Rate of income on net assets . 10 %

Common stock only. When only common stock is to be issued, it must be issued in the ratio of the net earnings if the income of the con- solidation is to be distributed in the ratio in which the companies con- tributed earnings. To determine the amount of stock which is to be issued, capitalize the earnings by dividing the income of each company by a rate of income agreed upon. Thus, if it is agreed that the rate be 10%, the distribution of common stock is made as follows:

ABC Total

Stock to be issued:

A: $4,000 + .10 $40,000

B: $12,000 + .10 $120,000

C: $20,000 -5- .10 ... $200,000

Total $360,000

Less net assets transferred 40,000 60,000 120,000 220,000

Goodwill 0 $ 60,000 $ 80,000 $140,000

Ten per cent was chosen as the basic rate, because it was the lowest rate earned by any one of the three companies.

If the profits of the consolidation amount to $36,000, it will be possible to pay a 10% dividend, which would be distributed as follows:

Total

$60,000

$120,000

$220,000

12,000

20,000

36,000

20%

16f%

U4 Goodwill

A: 10% of $40,000 $ 4,000

B: 10% of $120,000 12,000

C: 10% of $200,000 20,000

$36,000

This is an equitable division, so far as profits are concerned. How- ever, it is objectionable because it gives each old company an interest in the assets which is proportionate to the profits earned before the consolida- tion, instead of an interest proportionate to the assets contributed. This might work a hardship in case of liquidation.

Net Assets Goodwill Total Fraction

A $ 40,000 0 $ 40,000 ^

B 60,000 $60,000 120,000 £•§■§■

C 120,000 80,000 200,000 £££

$220,000 $140,000 $360,000

Assume that after a number of years it is decided to liquidate the consolidated company, and that in the meantime, all of the profits have been paid out as dividends. The goodwill has no realizable value, so there is $220,000 to be distributed as follows :

Former stockholders of A : -ffo of $220,000 $ 24 , 444 . 45

Former stockholders of B : £§£ of 220,000 73 , 333 . 33

Former stockholders of C: -§££ of 220,000 122 , 222 . 22

$220,000.00

The former stockholders of A would lose, and the former stockholders of B and C would profit.

Former

Stockholders Assets Liquidating

of Company Contributed Dividend Gain Loss

A $40,000 $24,444.45 $15,555.55

B 60,000 73,333 33 $13,333.33

C 120,000 122,222.22 2,222.22

$220,000 $220,000.00 $15,555.55 $15,555.55

Preferred stock for net assets. In order to avoid giving an advan- tage to one or more companies at the expense of the others, it is advisable to issue preferred stock for the net assets, and common stock for the good- will. The goodwill should be allotted to the several companies in the ratio of the excess of the profits contributed over the dividends on the preferred stock.

Assume that in the above illustration 6% stock, preferred as to assets, is to be issued for the net assets, and that common stock is to be issued for the goodwill.

Goodwill 165

ABC Total

Earnings $4,000 $12,000 $20,000 $36,000

Less dividends on preferred

A: 6% of $ 40,000 2,400

£:6%of 60,000 3,600

C:6%of 120,000 7,200

Excess earnings $1,600 $ 8,400 $12,800

Common stock should be issued in the ratio of the excess earnings. If five years' purchase of the excess profits were agreed upon, the distri- bution of stock would be :

ABC Total

Preferred stock $40,000 $60,000 $120,000 $220,000

Common stock 8,000 42,000 64,000 114,000

Assuming profits of $36,000 as before, the distribution of dividends would be :

Profits $36,000

Preferred dividends: 6% of $220,000 13,200

Balance available for common stock dividends $22 , 800

Then, $22,800 -i- $114,000 = 20%, the rate per cent which could be paid on the common stock.

ABC

Preferred dividends:

6% of $ 40,000 $2,400

6% of 60,000 $ 3,600

6% of 120,000 $ 7,200

Common dividends :

20% of $ 8,000 1 ,600

20% of 42,000 8,400

20% of 64,000 12,800

Total dividends $4,000 $12,000 $20,000

These dividends are in each case equal to the profits contributed to the consolidation by the several companies.

It is important to note that goodwill should be based on the profits contributed minus the profits to be returned as preferred dividends, and not on the total profits.

Bonds, preferred stock, and common stock. If bonds are issued for a percentage of the net assets, preferred stock for the remaining net assets, and common stock for the goodwill, the goodwill should be based on the profits turned in minus the bond interest and the preferred dividends.

Assume that 5% bonds are to be issued for 80% of the net assets, 6% preferred stock for the remaining net assets, and common stock for the goodwill, which is to be computed by capitalizing at 20% the earnings

166 Goodwill

of each company in excess of bond interest and preferred dividends to be paid to former stockholders. The issues of the three classes of securities would be computed as follows:

ABC Total

Bonds :

A: 80% of $ 40,000.. . .

$32,000

£:80%of 60,000....

$48,000

C:80%of 120,000....

$96,000

Total bonds

$176,000

Preferred Stock:

A: 20% of $ 40,000.. . .

8,000

B: 20% of 60,000....

12,000

C:20%of 120,000....

24,000

Total preferred stock . . .

44,000

Common Stock:

A : Earnings

$ 4,000

Bond interest

.. $1,600

Pfd. dividend

480

2,080

Excess

$ 1,920

$1,920 -*- .20

$ 9,600

B : Earnings

$12,000

Bond interest

. . $2,400

Pfd. dividend

720

3,120

Excess

$ 8,880

$8,880 -s- .20

$44,400

C: Earnings

$20,000

Bond interest

. . $4,800

Pfd. dividend

. . 1 , 440

6,240

Excess

$13,760

$13,760 -f- .20

$68,800

Total common stock ....

$122,800

With profits of $36,000 before allowance for bond interest and pre- ferred dividends, the former stockholders would receive interest and divi- dends as follows:

ABC Total

Bond interest:

5% of $32,000 $1,600

5% of 48,000 $2,400

5% of 96,000 $ 4,800

Total $ 8,800

Preferred dividends:

6% of $ 8,000 480

6% of 12,000 720

6% of 24,000 1,440

Total 2,640

Common dividends:

20% of $ 9,600 1,920

20% of 44,400 8,880

20% of 68,800 13,760

Total 24,560

Total distribution $4,000 $12,000 $20,000 $36,000

Goodwill 167

Conclusion. The illustrations given are merely indicative of the principles to be borne in mind in the distribution of stock and other secur- ities; they cannot be accepted as procedures to be invariably followed, for several reasons.

First, in the illustrations, the profits of the consolidation are assumed to be the same as the combined profits of the separate companies before they were consolidated. However, consolidations are usually made with the object of increasing profits; hence the question is raised as to how the additional profits should be divided. Should the preferred stock be participating or non-participating?

Second, the question of control involves the matter of the voting rights of the several classes of stock.

These and other considerations would tend to cause modifications in the methods described, but the illustrations serve to indicate the basic principles which must be followed in security allotment in order that the stockholders of the several consolidating companies may preserve their interests in the assets and earnings of the consolidation.

PROBLEMS

1. A, B, and C are about to consolidate. The following data are presented:

ABC Total

Net Assets $250 , 000 $150 , 000 $600 ,000 $1 , 000 , 000

Average Profits 50,000 15,000 150,000 215,000

Interest Rate 10%

Profit Rate 20% 10% 25%

Prepare tabulations showing the stock distribution:

(a) Preferred stock for the net assets, and common stock for the goodwill.

(6) Show the possible disadvantage of issuing only common stock.

2. Using the data in Problem 1, show the security allotment if 5% bonds are issued for 80% of the net assets, 6% preferred stock for the remainder, and common stock for the goodwill, which is to be based on excess earnings capitalized at 15%.

3. A, B, and C call upon you to draw up plans for their consolidation. They submit the following information:

Assets ABC

Plants $350,000 $200,000 $180,000

Materials 100,000 20,000 20,000

Accounts Receivable 80,000 60,000 40,000

Cash 20,000 10,000 10,000

Liabilities

Accounts Payable 70 , 000 30 , 000 20 , 000

Capital 350,000 200,000 100,000

Surplus 130,000 60,000 130,000

Average income 30,000 35,000 40,000

Upon your recommendation, the consolidation will issue: (a) 6% bonds for the fixed assets ; (b) 7 % preferred stock for the remaining net assets ; (c) common

168 Goodwill

stock for the goodwill, which is to be based on excess profits capitalized at 10%. Assuming that the consolidation will have net profits amounting to $105,000, prepare statements showing the allotment of securities and the distribution of profits.

4.* The net worth and profits of three companies are as follows:

Capital $100,000

Profits 50,000

Capital 200,000

Profits 50,000

Capital 250,000

Profits 50,000

(a) Give your theory of how a consolidation should be made. (6) Show the respective interests of X, of Y, and of Z in the consolidated company, using a factor of 6% to represent the normal value of money.

5.f A has agreed to sell to B the goodwill of the X. Y. Company on the basis of three years' profits of the business, which are to be determined by you, on sound principles of accounting and as accurately as possible, from the following statement handed you by A. You are required to compute the value of the goodwill, but are not expected to take into account any considerations except those presented by the statement.

Credits 1st Year 2nd Year 3rd Year Sales (selling prices substantially uniform through- out period) $638,400 $602,500 $ 564,000

Estimated value of construction work performed

and charged to property 110,000 77,600 154,000

Appreciation of real estate upon revaluation by

experts 80,000

Profit on sale of Bethlehem Steel Co. stock 85,000

Inventory at end of period :

Production material at cost 72 , 000

Finished goods at selling prices 76 , 500

$896,900 Debits

Production materials purchased $233,000

Production labor 50 , 850

Production expense (including depreciation) 66 , 750

Selling expenses 52 , 500

Interest 96,000

Cost of construction work 74 , 600

Inventory at beginning of period :

Production material at cost 51 , 400

Finished goods at selling prices 54,900

$680,000 Balance, being profit claimed by A $216 , 900 $246 , 950 $ 243 , 700

103,100

106,600

114,000

150,000

$977,200

,059,600

$252,400

220 , 300

61,400

60,900

69,300

70,300

55,650

62 , 800

94,000

98,500

49,000

86,000

72,000

103,100

76,500

114,000

$730,250

815,900

* C. P. A., Michigan.

t American Institute Examination.

Goodwill 169

6.* In the preceding problem, does the basis used for arriving at the value of the goodwill — three years' profits — appear to you to be reasonable in view of the facts disclosed to you? If not, what advice would you offer upon the question if A or B were your client?

7.* A and B are partners in business and have the following statement:

Store $15,000 Accounts Payable $10,000

Accounts Receivable 12,000 Bills Payable 5,000

Cash 9,000 A's Capital 30,000

Furniture and Fixtures 2,800 £'s Capital 35,000

Merchandise 37,000

Miscellaneous Equipment 4 , 200

$80,000 $80,000

C is admitted as a special partner, under the following arrangement: C is to contribute $30,000, and is to be entitled to one-third of the profits for 1 year. Before the contribution is made, the following changes are to be made in the books: store to be marked down 5%; allowance for doubtful accounts to be created, amounting to 2%; merchandise to be revalued at $35,000; furniture and fixtures to be revalued at $2,500. At the end of the year, the goodwill is to be fixed at 3 times the net profits for the year in excess of $20,000, this good- will to be set up on the books and the corresponding credit to be to A and B equally. A, B, and C are each to draw $3,000 in cash, and the remaining profits are to be carried to their capital accounts.

During the year, the following transactions took place:

Merchandise bought on credit $240 , 000

Cash purchases 25 , 000

Cash sales 125,000

Sales on credit 175 ,000

Accounts payable paid (face, $245,000; discount, 2%) 240, 100

Accounts receivable collected (face, $170,000; all net except

$50,000, on which 2% was allowed) 169,000

Buying expenses, paid cash 1 , 500

Selling expenses, paid cash 21 ,000

Delivery expenses, paid cash 9 , 000

Management expenses, paid cash 4 , 500

Miscellaneous expenses, paid cash 3 , 000

Interest on notes payable, paid cash 250

The partners each withdrew $3,000 cash, as agreed.

When the books were closed for the purpose of determining the profits and goodwill, the following were agreed upon:

Value of merchandise on hand $60 , 000

Depreciation on store 285

Additional allowance for doubtful debts 165

Furniture and fixtures written down 200

The goodwill having been estimated and duly entered, C then contributes enough cash to make his capital account equal one-third of the total capital.

Prepare statements showing how the accounts are to be adjusted, and pre- pare the balance sheet after the final adjustment.

* American Institute Examination.

Business Finance

Stock rights. Corporations, in undertaking to secure additional capital, not infrequently offer additional stock to their stockholders at a price below the prevailing market quotation of the outstanding shares.

This privilege of subscribing has value as long as the market price of the old stock remains higher than the offering price of the new stock; and if the stockholders prefer not to exercise the rights, they may sell them in the market for whatever they will bring.

Example. A corporation has a capital stock of $100,000, divided into 1,000 common shares of $100 par value. The entire amount is outstanding, and the market quotation is $150. Finding that $50,000 additional capital is needed, the directors decide to offer to the stockholders 500 shares of new common stock at $125. They accordingly announce on August 1 that stock- holders of record as of September 1 will have the privilege of subscribing for the new issue in the proportion of one share of the new stock for every two shares of the old stock held on the latter date. The subscriptions are pay- able on or before October 1 following, and transferable warrants for the rights are to be issued as soon as practicable after September 1. What is the value of a right?

Explanation. According to the conditions of this offer, every holder of two of the old shares at the close of business on September 1 will be entitled to subscribe to one of the new shares. He will therefore come into possession of two " rights," as that term is used on the New York Stock Exchange, or one right for every old share held. (On some stock exchanges the term right indicates the privilege of subscribing to one share of the new issue.)

Trading in the rights will begin following the declaration of the directors on August 1, and will continue until October 1. Until the warrants are in the hands of the stockholders — during the period from August 1 to September 1 — the trading will be on a " when issued" basis; that is, delivery and payment for the rights will be made when the warrants are available. During this time

170

Business Finance 171

the stock will sell "rights-on"; that is, the market value of the shares will include the value of the rights.

With the delivery of the warrants on September 1, the stock will sell " ex-rights," its price no longer including the value of the rights. With the issuance of the warrants, and until October 1, trading in the rights will be for immediate delivery and payment ; that is, delivery and payment the day after the sale is made.

Should a holder of two shares exercise his privilege of subscription, he would own:

2 shares @ $150 $300

1 share @ 125 J25

3 shares @ 141.67 $425

It will be noticed that the difference between the market price of the old stock and the average price of the three shares is $8.33, the value of a right; also that the difference between the market price of the old stock and the offering price of the new stock is $25.00, or three times the value of a right.

Therefore, the following formula may be used:

Formula

Market price — Offering price __. , . .

Number of rights to purchase 1 share + 1

Substitution

During the second period — that is, while the stock is quoted ex-rights — the value of the rights may be ascertained as follows:

Formula

Market price — Offering price __ . , . . ,

r= r , • rr , r r~T — = Value of a nSht-

JN umber ol rights to purchase 1 share

It will be found that the market price of the rights during the second period will tend to coincide with this value. Any appreciable difference opens an opportunity for a profit.

The foregoing discussion and example apply only to values on the market. The profit or loss resulting from the sale of rights, and the profit or loss from the sale of stock acquired by the exercise of rights, are gov- erned by Section 29.22(a)-8, Regulations 111.

Sale of stock and rights, federal income tax. Ordinarily, a stockholder derives no taxable income from the receipt of rights to sub- scribe for stock, nor from the exercise of such rights, but if he sells the rights instead of exercising them, he may derive taxable income, or sustain a loss.

The following rule is stated in Sec. 29.22(a)-8, Regulations 111.

"(1) If the shareholder does not exercise, but sells, his rights to sub- scribe, the cost or other basis, properly adjusted, of the stock in respect

172 Business Finance

of which the rights are acquired shall be apportioned between the rights and the stock in proportion to the respective values thereof at the time the rights are issued, and the basis for determining gain or loss from the sale of a right on one hand or a share of stock on the other will be the quotient of the cost or other basis, properly adjusted, assigned to the rights or the stock, divided, as the case may be, by the number of rights acquired or by the number of shares held."

Example. A purchased 100 shares of stock at $125.00 a share, and in the follow- ing year the corporation increased its capital by 20% A, therefore, received 100 rights, entitling him to subscribe to 20 additional shares of stock; the subscription price was $100.00 a share. Assume that at the time that the rights were issued the stock had a fair market value of $120.00 a share, and that the rights had a fair market value of $3.00 each. If, instead of sub- scribing for the additional shares, A sold the rights at $4.00 each, his taxable gain would be computed as follows:

100 shares @ $125.00. . . . $12,500.00, cost of stock in respect of

which rights were issued. 100 shares @ $120.00. . . . $12,000.00, market value of old stock. 100 rights @ $3.00 $300.00, market value of rights.

12,000 of 12,500 $12,195.12, cost of old stock apportioned

1^)300 ^0 sucn stock after issuance of rights. 300

of 12,500 $304.88, cost of old stock apportioned

12'300 to rights.

100 rights @ $4.00 $400.00, sales price of rights.

$400.00 - $304.88 $95.12, profit on sale of rights.

For the purpose of determining the gain or loss from the subsequent sale of the stock in respect of which the rights were issued, the adjusted cost of each share is $121.95— that is, $12,195.12 4- 100.

Rule 2 of Sec. 29.22(a)-8, Regulations 111, states:

"(2) If the shareholder exercises his rights to subscribe, the basis for determining gain or loss from a subsequent sale of a share of the stock in respect of which the rights were acquired shall be determined as in paragraph (1). The basis for determining gain or loss from a subsequent sale of a share of the stock obtained through exercising the rights shall be determined by dividing the part of the cost or other basis, properly adjusted, of the old shares assigned to the rights, plus the subscription price of the new shares, by the number of new shares acquired."

Example. A purchased 100 shares of stock at $125.00 a share, and in the follow- ing year the corporation increased its capital by 20%. A, therefore, received 100 rights entitling him to subscribe to 20 additional shares of stock; the sub- scription price was $100.00 a share. Assume that at the time that the rights were issued the stock had a fair market value of $120.00 a share, and that the rights had a fair market value of $3.00 each. A exercised his rights to sub- scribe, and later sold for $140.00 a share 10 of the 20 shares thus acquired. The profit is computed as follows:

Business Finance 173

Cost of old stock apportioned to rights in accordance

with the computation in the example under Rule 1 ... $ 304 . 88

Subscription price of 20 shares at $100.00 a share 2,000.00

Basis for determining gain or loss from sale of shares

acquired by exercise of rights $2,304.88

$2,304.88 4- 20 = $115.24, basis for determining gain or loss from sale of each share of stock acquired by exercise of rights.

Proceeds of sale :

10 shares @ $140.00 $1,400.00

Cost of stock sold :

10 shares @ $115.24 1,152.40

Profit $ 247.60

The basis for determining the gain or loss from the subsequent sale of the remaining 10 shares of stock acquired on subscription is $115.24 a share; and the basis for determining the gain or loss on the stock in respect of which the rights were issued is $121.95 a share — that is, $12,195.12 -5- 100, as in the example under Rule 1.

PROBLEMS

1. A company has a capital stock of $1,000,000, divided into 10,000 common shares of $100 par value. The entire amount is outstanding, and the market quotation is $150 a share. Finding that $500,000 of additional capital is needed, the directors decide to offer to the stockholders 5,000 shares of new common stock at par. What is the approximate market value of a right if each stock- holder may subscribe for one share of new stock for every two shares of the old stock held?

2. A corporation offered, at $100 a share, one share of its new stock for each six shares held. The stock was selling at $185 a share when the offer was announced. What was the approximate market value of a right?

3. W owned 100 shares of Purity Baking Common that cost him $13,400. Later, he received rights to subscribe to additional stock, but since he did not care to increase his investment, he sold the rights at 3f less commission, receiving therefore $357.30. At the date when the stock was quoted ex-rights, the average market values were:

Stock 124£

Rights 3f

(a) What was W's loss on the sale of the rights?

(b) What was the carrying value of the stock?

4. Smith owned 100 shares of common stock in the W. Corporation, which offered rights to subscribe to new common stock at $100 a share, the basis of the offering being one share for each five shares held. The average market values on the date when the stock sold ex-rights were:

Stock 150.50

Rights 11.8125

Smith later sold his rights at $14.50.

174

Business Finance

(a) If Smith paid $120 a share for the original 100 shares, what is his profit on the sale of the rights?

(b) What is the carrying value of the 100 shares?

Working capital. One of the most difficult problems for anyone entering a new business is to know how much money will be required to finance the enterprise until the receipts will equal or exceed the disburse- ments. While this is strictly a question of finance, the accountant is often called upon to deal with it.

Example. A manufacturer gives you the following data, and requests that you estimate the amount of working capital required to finance the making and selling of an article :

Selling price, each $100

Cost to make, each 60

Selling expenses, each 20

Overhead, each 10

Net profit, each 10

Sales, first month 50 articles

" second month 100 "

" third month 150 "

" fourth month 200

" each month thereafter 200

All the sales are installment sales, the payments being $10 per month. Assume arbitrarily that the complete cost of $90 on each article is incurred at the time that the sale is made.

What will be the largest amount of capital required, and in which month will it be required?

Solution.

Working

Total Costs Total Receipts Deficiency Capital

Months Each Month Each Month Each Month Required

First $4,500 $ 500 $4,000 $4,000

Second 9,000 1,500 7,500 11,500

Third 13,500 3,000 10,500 22,000

Fourth 18,000 5,000 13,000 35,000

Fifth 18,000 7,000 11,000 46,000

Sixth 18,000 9,000 9,000 55,000

Seventh 18,000 11,000 7,000 62,000

Eighth 18,000 13,000 5,000 67,000

Ninth 18,000 15,000 3,000 70,000

Tenth 18,000 17,000 1,000 71,000

Eleventh 18,000 18,500 500f 70,500

Twelfth 18,000 19,500 l,500f 69,000

f Receipts from collections are more than the costs for the month.

The above table shows in the last column the amount of working capital required to finance the business by months. The greatest amount required is found to be $71,000 in the tenth month. Thereafter, the collections are greater than the costs.

PROBLEMS

1. A company is about to be formed for the purpose of manufacturing a specialty. After careful investigation, the following estimates have been made:

Business Finance 175

Selling

Record · ID 646038
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