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amount for the second period of time, and so continue for the whole time. The last amount is the amount required.

2. The last amount minus the given principal is the compound interest.

592. Examples for the Slate. At compound interest, what is the amount 128. Of $ 200 for 3 years at 6%? 129. Of $ 350.50 for 4 years at 5%? 130. Of $ 2000 for 3 years 11 months at 6%?

131. Of $ 2000 for 1 y. 6 mo. at 7%, interest compounded semi-annually ? NOTE. Take interest at 34 % for three intervals of time.

132. What is the compound interest of $40 for 1 y. 2 mo, at 6%, interest compounded semi-annually ?

133. What is the compound interest of $ 900 for 1 y. 1 mo, at 6%, interest compounded quarterly ?

593. The work of computing compound interest may be shortened by the use of the following

TABLE, Showing the amount of $1 at compound interest from 1 year to 10 years, at 594. ILLUSTRATIVE EXAMPLE. What is the compound interest of $ 1000 for 2 y. 4 mo. at 7%?

3, 4, 41, 5, 6, and 7 per cent.

Years. 3 per cent. 4 per cent. 41 per cent. 5 per cent. 6 per cent. 7 per cent.

1. 2 3. 4. 5.

1.030000 1.040000 | 1.045000 1.050000 1.060000 1.070000 1.060900 1.081600 1.092025 1.102500 1.123600 1.144900 1.092727 1.124864 1.141166 | 1.157625 | 1.191016 1.225043 1.125509 1.169859 | 1.192519 1.215506 | 1.262477 | 1.310796 1.159274 | 1.216653 | 1.246182 1.276282 | 1.338226 1.402552

6. 7. 8. 9. 10.

1.194052 1.265319 | 1.302260 1.340096 1.418519 1.500730 1.229874 1.315932 1.360862 | 1.407100 1.503630 1.605781 1.266770 1.368569 1.422101 1.477455 1.593848 1.718186 1.304773 1.423312 1.486095 1.551328 1.689479 | 1.838459 1.343916 1.480244 | 1.552969 1.628895 1.790848 1.967151

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NOTE. In the above operation, the amount of $ 1000 for 2 years is first found, and the amount for the months is then obtained by multiplying by 1.023. It would be equally well to find the amount of $1 for the entire time, and then multiply that amount by 1000.

595. Examples for the Slate.

Using the preceding table, find the amount at compound interest

134. Of $ 200 for 2 y. 4 mo. at 7 %. 135. Of $ 580 for 7 y. 10 mo. at 6%.

136. What is the compound interest of $300 for 3 y. 2 mo. 6 d. at 8%, interest payable semi-annually ?

137. What is the compound interest of $ 380 for 1 y. 10 mo. 22 d. at 6%, interest payable semi-annually ?

138. If at the age of 25 years, a person puts $ 1000 on interest, compounding annually at 6%, what will be the amount due him when he is 40 years old ?

NOTE. First find by the table the amount for 10 years, then find the amount of that amount for 5 years more.

For additional examples in compound interest, see page 253.

AVERAGE OR EQUATION OF PAYMENTS.

596. ILLUSTRATIVE EXAMPLE. A debtor owes to one person the following sums at the dates specified : Oct. 1, $262; Oct. 10, $ 220; Nov. 6, $ 250. At what date may he pay the total of these items without loss of interest to either

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WRITTEN WORK.

Due.

Items.

66 10,

36

Explanation. — To do this
Days. Interest.

example, we may suppose all Oct. 1, $262 0

the items to be paid at the

earliest date at which any 220 9 $ 0.66

item becomes due, viz. Oct. 1. Nov. 6, 250

3.00

This would involve a loss 1 day's int. of 732 0.244) 3.66 (15 to the debtor of interest on

$ 220 from Oct. 1 to Oct. 10 Oct. 1 + 15 d. = Oct. 16. Ans.

(9 days), and on $ 250 from

Oct. 1 to Nov. 6 (36 days).
The interest of $ 220 for days at 12%* is $0.66
250

12% is 3.00
Total interest

$ 3.66 That no loss may result, the total of the items, $732, should be paid as many days after Oct. 1 as will be required for $ 732 at 12% to gain $3.66 of interest. To find this time, we divide $3.66 by the interest of $ 732 for 1 day at 12% (Art. 562, note), and have for a quotient 15.

15 days after Oct. 1 is Oct. 16. Ans. Oct. 16.

66 36

66

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597. The process of finding the time when the payment of several items, due at different times, may be made at once, without loss of interest to either party, is average, or equation of payments.

598. The date at which several sums due at different times

may be paid at once is the average date or equated time of payment.

1

* Any per cent may be baken, but 12 per cent (1% a month) is taken for convenience, the interest then being for every month 0.01 of the principal, and for every 3 days 0.001 of the principal.

AVERAGE OR EQUATION OF PAYMENTS.

235

599. From the foregoing operation may be derived

Rule I.

To find the average time for the payment of several sums due at different times :

1. Select some convenient date; for example, the earliest date at which any item matures.

2. Compute the interest on each item from the selected date to the date of its maturity.

3. Add the interests thus found; divide their sum by the interest of the sum of the items for one day; the quotient will express the number of days from the selected date to the average date of payment.

4. Add this number to the selected date ; the result will be the average date required.

600. The foregoing illustrative example performed by

The Product Method.

WRITTEN WORK.

Explanation. — To do this example Days. Products.

by the product method, we select some 0 x 262= 00

date, for example the earliest date at 9 x 220= 1980

which any item becomes due, and sup36 x 250= 9000

pose all the items to be paid at this 732)10980 (15

date. This would involve a loss to the

debtor of interest on $ 220 for 9 days, Oct. 1 +15 d. = Oct. 16. Ans. and on $ 250 for 36 days.

The interest on $ 220 for 9 days equals the interest on $1 for 1980 days ; the interest on $ 250 for 36 days equals the interest on $1 for 9000 days, which together equals the interest on $ 1 for 10980 days, but $ 732 is the sum to be paid, and the time required for the interest on this sum to equal the interest on $1 for 10980 days will be 732 of 10980 days, which is 15 days.

15 days after Oct. 1 is Oct. 16. Ans. Oct. 16.

601. From the preceding operation may be derived

Rule II. To find the average date for the payment of several sums due at different dates :

1. Select some convenient date; for example, the earliest date at which any item matures.

2. Multiply the time each item has to run by the number of dollars in the item.

3. Divide the sum of the products thus obtained by the number of dollars in the sum of the items; the quotient will express the time from the selected date to the average date of payment.

4. Add this time to the selected date ; the result will be the average date required.

602. Proof.

Find the sum of the interests on all items due BEFORE the average date, from the date at which they are severally due to the average date; also find the sum of the interests on all items due AFTER the average date from that date to the dates at which they are severally due. If these sums are equal, or differ by less than half a day's interest on the sum of all the items, the result is correct.

NOTE I. The examples in this book are performed by the interest method, which has the advantage of brevity when the accountant uses interest tables. The pupil will perform the work by either or by both methods, as directed by the teacher.

Note II. Any date may be selected from which to average an account. The last day of the month previous to the earliest day at which any item becomes due is a convenient date.

Note III. When any item contains cents, if less than 50, disregard them, if 50 or more, increase the units of dollars by $ 1.

Note IV. When a quotient contains a fraction of a day, if less than , disregard it; if for more, call it 1 day.

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