Interest on the same, up to Oct. 6, 1838, is $620.000 40.386 660.386 Interest from Oct. 6, 1838, to March 4, 1839, is 17.247 Interest from March 4, 1839, to Dec. 11, 1839, is 28-414 The third indorsement is 555.947 107.770 448.177 Interest from Dec. 11, 1839, to July 20, 1840, is 19.085 The fourth indorsement is 467.262 200.500 266.762 4.409 Ans $271.171 Interest from July 20, 1840, to Oct. 15, 1840, is UTICA, May 1, 1836. 3. For value received, I promise to pay ISAAC CLARK, or order, three hundred and forty-nine dollars, ninety nine cents, and eight mills, with interest at 6 per cent. The amount of the note, or principal, is. $349.998 } $4.998 is less than the interest then due, Indorsement August 22, 1838, . 15.000 4. For value received, I promise to pay PETER SMITH, or order, one hundred and eight dollars and forty-three cents, on demand, with interest at 7 per cent. JOHN SAVEALL. 5. For value received, I promise to pay F. Gould, or bearer, one hundred and forty-three dollars and fifty cents, on demand, with interest How much remains due Dec. 28, 1840, the interest being 7 per cent.? Ans. $60.866. 6. A note of $486 is dated Sept. 7, 1831, on which, March 22, 1832, there was paid $125 What was the balance due April 19, 1834, the interest being 7 per cent.? Ans. $144.404. 65. The principal, the rate per cent., the time, and the interest, are so related to each other, that any three of them being given, the remaining one can be found. Problem I. Given the principal, the rate per cent., and the time, to find the interest. The rule for this problem has already been given under Case III., Art. 63; it is equivalent to the following RULE. Multiply the interest of $1 for the given time and given rate per cent., by the number of dollars in the principal. Problem II. Given the time, the rate per cent., and the interest, to find the principal. By the reverse of the last problem, we obtain this RULE. Divide the given interest by the interest of $1 for the given time and given rate per cent.; and the quotient will be the number of dollars in the principal. EXAMPLES. 1. The interest on a certain principal for 9 months and 10 days, at 4 per cent., is $101605. What was the principal? In this example, we find the interest of $1 for 9 months and 10 days, at 4 per cent., to be $0.035; .. $1·01605, divided by $0.035, gives 29.03 for the number of dollars in the principal required. 2. What principal will, in 1 year, 7 months, and 15 days, at 6 per cent., give $9.75 interest? Ans. $100. 3. What principal will, in 7 years and 9 days, at 6 per cent., give $16.86 interest? Ans. $40. 4. What principal will, in 3 years and 6 months, at 5 per cent., give $92.75 interest? Ans. $530 5. What principal will, in 3 months and 9 days, at 8 per cent., give $90, interest? Ans. $4090-909. Problem III. Given the principal, the time, and the interest, to find the rate per cent. RULE. Divide the given interest by the interest of the given principal, for the given time, at one per cent. EXAMPLES. 1. The interest of $100 for 9 months and 10 days, is $3.50. What is the rate per cent.? In this example, we find the interest of $100 for 9 months and 10 days, at 6 per cent., to be $4.663. The interest at 1 per cent. is of $4·663=$0·777; therefore, dividing $3.50 by $0.77%, we obtain 41 for the rate per cent. required. 2. At what rate per cent. will $530, in 3 years and 6 months, give $92.75, interest? Ans. 5 per cent. $1941, in 1 year, 7 3. At what rate per cent. will months, and 13 days, give $2·200339}, interest? Ans. 7 per cent. 4. At what rate per cent. will $5.37, in 4 years and 12 days, give $1.73272, interest? Ans. 8 per cent. |