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II. To find the amount of a note at compound interest, when there have been partial payments.

RULE.

Find the amount of the principal, and from it subtract the amount of the indorsements.

7. $144. .

Haverhill, Sept. 25, 1839. For value received, I promise to pay Charles Northend, or order, on demand, one hundred forty-four dollars, with interest.

John Small, Jr. Attest, Q. Jones.

On this note are the following indorsements.
Jan. 1, 1840. Received thirty dollars.
June 30, 1841. Received eighty dollars.
Feb. 7, 1842. Received ten dollars.

What is due on the above note at compound interest, Oct. 4, 1842 ?

Ans. $ 40.02.

OPERATION BY COMPOUND INTEREST.

Principal

$ 144.00 Interest from Sept. 25, 1839, to Oct. 4, 1842 27.76

Amount 171.76 First payment

$ 30.00 Interest from Jan. 1, 1840, to Oct. 4, 1842 5.23 Second payment

80.00 Interest from June 30, 1841, to Oct. 4, 1842 6.12 Third payment

10.00 Interest from Feb. 7, 1842, to Oct. 4, 1842 39

Amount $ 131.74 Remains due, Oct. 4, 1842

$ 40.02 Section 41.

EQUATION OF PAYMENTS.

OPERATION.

When several sums of money, to be paid at different times, are reduced to a mean time for the payment of the whole, without gain or loss to the debtor or creditor, it is called Equation of Payments. 1. John Jones owes Samuel Gray $ 100 ; $ 20 of which is to be paid in 2 months ; $ 40 in 6 months ; $ 30 in 8 months; and $ 10 in 12 months ; what is the equated time for the payment of the whole sum ?

Ans. 6mo. 12da.

By analysis, $20 for 2 $20 x 2 40

months is the same, as $ 40 x 6 =240

$ 40 for 1 month ; and $30 X8 - 240

* 40 for 6 months is the $10 x12 120

same,

$1 for 240 $ 100 100)6 40 (6 mo.

months ; and $ 30 for 8 600

months is the same, as

$1 for 240 months ; and 40

$ 10 for 12 months is the 30

same,

$1 for 120 100)1200 (1 2 da. months; therefore, $1 1200

for 40 +240 +240 +

120 = 640 months is the same, as $ 20 for 2 months, $ 40 for 6 months, $ 30 for 8 months, and $ 10 for 12 months ; but $ 20 + $ 40+ $30 + $ 10 are $ 100 ; therefore, $1 for 640 months is the same, as $ 100 for ido of 640 months, which is 6 months and 12 days, as before. Hence the following

as

as

RULE.

Multiply each payment by the time at which it is due, then divide the sum of the products by the sum of the payments, and the quotient will be the true time required.

2. John Smith owes a merchant, in Boston, $ 1000, $250 of which is to be paid in 4 months, $ 350 in 8 months, and the remainder in 12 months ; what is the equated time for the payment of the whole sum ?

Ans. 8mo. 18da. Note. The following, example will illustrate the method, the merchants practise to find the medium time of payment of goods sold on credit. 3. Purchased of James Brown, at sundry times, and on

various terms of credit, as by the statement annexed. When is the medium time of payment ? Jan. 1, a bill amounting to $ 360, on 3 months' credit. Jan. 15do. do.

186, on 4 months' credit. March 1, do. do.

450, on 4 months' credit. May 15, do. do.

300, on 3 months' credit. do. do.

500, on 5 months' credit.

June 20,

FORM OF STATEMENT.

91=

Due April 1, $360

May 15, $186 x 45= 8370
July 1, $450 X

40950
Aug. 15, $300 x 136 = 40800
Noy. 20, $ 500 X 233=116 500
1796

) 2066 20 (11548 days. 1796

2 7 0 2
1796

9060
8980

80 The medium time of payment will be 116 days from April 1, which will be July 25. 4. Sold S. Dana several parcels of goods, at sundry times, and on various terms of credit, as by the following statement. Jan: 7, 1841, a bill amounting to $ 375.60, on 4 months. Apr. 18, 1841, do. do.

687.25, on 4 months. June 7, 1841, do. do.

568.50, on 6 months. Sept. 25, 1841, do.

do.

300.00, on 6 months. Nov. 5, 1941, do. do.

675.75, on 9 months. Dec. 1, 1841, do.

do.

100.00, on 3 months. What is the equated time for payment of all the bills ?

Ans. Dec. 24.

Section 42.

PROPORTION.

PROPORTION is the likeness or equalities of ratios. Thus, because 4 has the same ratio to 8, that 6 has to 12, we say such numbers are proportionals.

If, therefore, any four numbers whatever be taken, the first is said to have the same ratio or relation to the second, that the third has to the fourth, when the first number, or term, contains the second, as many times, as the third contains the fourth ; or, when the second contains the first, as many times, as the fourth does the third. Thus, 9 has the same ratio to 3, that 12 has to 4, because 9 contains 3, as many times, as 12 does 4. And 10 has the same ratio to 5, that 12 has to 6, because 10 contains 5, as many times, as 12 does 6. Ratios are represented by colons ; and equalities of ratios by double colons.

The first and third terms are called antecedents, and the second and fourth are called consequents; also, the first and fourth terms are called extremes, and the second and third are called means.

Whatever four numbers are proportionals, if their antecedents and consequents be multiplied or divided by the same numbers, they are still proportionals; and, if the terms of one proportion be multiplied or divided by the corresponding terms of another proportion, their products and quotients are still proportionals.

If the product of the extremes be equal to the product of the means, it is evident, that if any three of the four proportionals be given, the other may be obtained ; for, if the product of the means be divided by one of the extremes, the quotient will be the other extreme ; and, if the product of the extremes be divided by one of the means, the quotient will be the other mean. Hence the following

RULE.

State the question by making that number, which is of the same name or quality as the answer required, the third term ;

* L

then, if the answer required is to be greater than the third term, make the second term greater than the first ; but if the answer is to be less than the third term, make the second less than the first.

Reduce the first and second terms to the lowest denomination mentioned in either, and the third term to the louest denomination mentioned in it.

Multiply the second and third terms together, and divide their product by the first, and the quotient is the answer in the same denomination to which the third is reduced.

If any thing remains, after division, reduce it to the next lower denomination, and divide as before.

If either of the terms consists of fractions, state the ques. tion as in whole numbers, and reduce the mixed numbers to improper fractions, compound fractions to simple ones, and invert the first term, and then multiply the three terms continually together, and the product is the answer to the ques. tion. Or, the fractions may be reduced to a common denominator ; and their numerators may be used as whole numbers. For when fractions are reduced to a common denominator, their value is as their numerators.

Note 1. It may be observed in Proportion, that the third term is the quantity, whose price or value is wanted, and that the second term is the value of the first; when, therefore, the second term is multiplied by the third, the product is as much more than the answer, as the first term is greater than unity; therefore, by dividing the product by the first term, we have the value of the quantity required.

lbs.

lbs.

cts.

Note 2. The papil should perform every question by analysis, previous to his performing it by Proportion. 1. If 7lbs. of sugar cost 56 cents, what cost 36lbs. ?

In stating this question, we make 7 : 36 :: 56

56 cents the third term, because the 5 6

answer will be in cents. And, as 216

we perceive from the nature of the 180

question, that the answer or fourth

term will be more than 56 cents, we 7) 2016

know, that of the other two terms, $ 2.8 8 Ans.

the second must be larger than the

first, we therefore make 36lbs. the second term, and 7lbs. the first term.

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