| Samuel Webber - Mathematics - 1801
...denominator of the given aliquot part, aed the quotient will bi the logarithm of the root sought. 4. **Subtract the principal from the amount, and the remainder will be the interest.** ' EXAMPLES. i. What is the compound interest of 500!. for 4 years, at 5 per cent. per annum ? 1*05... | |
| Nathan Daboll - Arithmetic - 1813 - 240 pages
...956i 10s. 4,125d. in 8J years ? Ans. 51 per cent. CASE IV, The amount, principa , and rate per cent, **given, to find the time. RULE. — Subtract the principal from the amount;** divide the remainder by the product of the ratio and principal ; and the quotient will be the time.... | |
| Arithmetic - 1817 - 198 pages
...476 dois. CASE С. To find the rate per cent, when the amount, time and principal are given. ROLE. **Subtract the principal from the amount, and the remainder will be the interest** for the given time : then, As the principal, Is to one hundred pounds, or dollars, So is the interest... | |
| Arithmetic - 1818 - 251 pages
...^62.253750 in 3 years and 3 months ? Ans. 6 per cent. CASE IV. The AMOUNT, PRINCIPAL, and RATE per cent **given, to find the TIME. RULE Subtract the principal from the amount,** divide the remainder by the product of the rUio and principal ; and the quotient will be the time.... | |
| Nathan Daboll - Arithmetic - 1818 - 240 pages
...9561. 10s. 4,l25d. in 8| years ? Jlns. 5$ per cent. CASE IV. The amount, principal, and rate per cent, **given, to find the time. RULE. Subtract the principal from the amount** f divide the remainder by the product of the ratio and principal ; and the quotient will be the time.... | |
| Phinehas Merrill - Arithmetic - 1819 - 99 pages
...years. 3. Muii ¡ply this power by the principal, and the product •will be the amount required. 4. **Subtract the principal from the amount, and the remainder will be the interest.** EXAMPLES. 1. What is the Compound interest of £.500 for 4 years, at 5 per cent ? the amount is in-uutved... | |
| Arithmetic - 1824 - 198 pages
...Ans. 476 dollars. CASE 6. To find the rate per cent. when the amount, time and principal are given. **RULE. Subtract the principal from the amount, and the remainder will be the interest** for the given time: then, As the principal, Is to one hundred pounds or dollars, So rs the interest... | |
| Roswell Chamberlain Smith - Arithmetic - 1827 - 196 pages
...yrs. orno, and 15 days, at 6 per cent ? A. $600. CASE IV. — When the principal, rate, and amount are **given to find the time. RULE. — Subtract the principal from the amount, and** divide the remainder, by the interest of the given sum for 1 year. Exercises for the Slate. In what... | |
| Arithmetic - 1831 - 198 pages
...Ans. 476 dollars. CASE 6. To find the rate per cent, when the amount, time and principal are given. **RULE. Subtract the principal from the amount, and the remainder will be the interest** for the given time: then, As the principal, Is to one hundred pounds or dollars, So is the interest... | |
| Samuel Read Hall - Arithmetic - 1832 - 288 pages
...simple interest ; it will then show the amount of principal and interest for the given number of years. **Subtract the principal from the amount, and the remainder will be the** compound interest. When there are months and days, first find the compound interest for the given years... | |
| |