2. What principal put at interest will amount to $5000 in 13 years at 5 per cent. compound interest ? 3. At what rate per cent. must $378.57 be put at compound interest, that it may amount to $500 in 5 years f Solving the equation 4 = p (1 + r.)" making r the unknown quantity, it becomes 4. In what time will $284.37 amount to 750 at 7 per cent.” Making n the unknown quantity, the equation d = p (1 + r.)" becomes log. 4. . . . . . . 0421177 log. 0.029384 - - - - 8.4681 11 5. What will be the compound interest of $947 for 4 years and 3 months at 5% per cent. * 6. What will $157.63 amount to in 17 years at 43 per cent.” 7. A note was given the 15th of March 1804, for $58,46, at , the rate of 6 per cent. compound interest; and it was paid the 19th of Oct. 1823. To how much had it amounted P 8. A note was given the 13th of Nov. 1807, for $456.33, and was paid the 23d of Sept. 1819. The sum paid was $894.40. What per cent. was allowed at compound interest ? 9. In what time will the principal p be doubled, or become 2p, at 6 per cent. compound interest ? In what time will it be tripled P JNote. In order to solve the above question, put 2 p in the place of .4 for the first, 3 p for the second, and find the value of n. i The principles of compound interest will apply to the following questions concerning the increase of population. 10. The number of the inhabitants of the United States in A. D. 1790 was 3,929,000, and in 1800, 5,306,000. What rate per cent. for the whole time wis the increase ? What per cent. per year : 11. Suppose the rate of increase to remain the same for the next 10 years, what would be the number of inhabitants in 1810 P 12. At the same rate, in what time would the number of inhabitants be doubled after 1800? 13. The number of inhabitants in 1810 by the census was 7,240,000. What was the annual rate of increase ? 14. At the above rate, what would be the number in 1820? 15. At the above rate, in what time would the number in 1810 be doubled P 16. The number of inhabitants by the census of 1820, was 9,638,000. What was the annual rate of increase from 1810 to 1820? 17. At the same rate, what is the number in 1825 ° 18. At the same rate, what will be the number in 1830? 19. At the same rate, in what time will the number in 1820 be doubled P 20. In what time will the number in 1820 be tripled: 21. When will the number of inhabitants, by the rate of the last census, be 50,000,000 ° LIII. 1. Suppose a man puts $10 a year into the savings bank for 15 years, and that the rate of interest which the bank is able to divide annually is 5 per cent. How much money will he have in the bank at the end of the 15th year? Suppose a = the sum put in annually, r = the rate of interest, t = the time, .4 = the amount. ~ According to the above rule of compound interest, the sum a at first deposited will amount to a (r. -- 1)"; that deposited the second year will amount to a (r. -- 1)"; that deposited the third year will amount to a (r. -- 1)*; that deposited the last year will amount to a (r + 1)". Hence we have whose largest term is (r. -- 1)", the smallest r + 1, and the ratio r + 1. The sum of this progression, Art. XLVII. is The same result may be obtained by another course of reasoning. The amount of the sum a for one year is a -- a r. Adding a to this, it becomes 2 a + a r. 3 a + 3 a r + a r". The amount of this for 1 year is 3 a + 3 a r + a r" + 3 a r + 3 a r" + a r" w = 3 a + 6 a r + 4 a r" + a r", = a (3 + 6 r + 4 r" + r"), This is the amount at the end of the third year before the addition is made to the capital. The law is now sufficiently manifest. With a little alteration, the quantity 3 + 6 r + 4* + r" may be rendered the 4th power of 1 + r. The three last coefficients are already right. If we add 1 to the quantity it becomes This is now the 4th power of 1 + r, and it may be written (1 + r.)". Subtract the 1 which was added last, and it becomes (1 + r.)‘– 1. Divide this by r, because it was multiplied by r, and it beCOines (1 + r.)" — 1 r - Subtract 1 again, because 1 was added previous to multiplying by r ; and it becomes The particular question given above may now be solved by logarithms, using this formula. log: (1 + r) = 1.05 - e - 0.021 189 Multiply by t- 15 - - 15 o 105945 log. (1 + r.)” = 2.079 - - .317835 Subtract 1 I log. 1,079 . . . . 0.033021 log. (1 + r.) - - - e 0.021 189 log. a = 10 . . . 1.000000 Arith. Com. log. r = los e e 1.301030 .Ans. $226.59 - . . . 2.355.240 |