OR 2. Multiply the number found on the right of the decimal. point in the table, by the principal, and the product is the interest required. EXAMPLES. 1. What is the interest of $400 forborn 3 years, at 5 per cent per annum? Tabular number for 3 years = 1.20 Do. for 6 months. .02499 2. What will be the interest of $210 35 cents, for 9 years, at 6 per cent per annum? Ans. $123 05c. 4.75 m. 3. What will be the interest on £45 10s. for 1 year and 7 months, at 6 per cent? Ans. £4 6s. 5d. 1 qr. 4. Required the interest of £896 15s. for 7 months, at 5 per cent per annum. Ans. 27 2s. 11d. 3 qrs. CASE III. To find the Rebate or Present worth of any given sum for years and months. RULE. Multiply the Tabular number under the given rate and opposite the time by the principal, and the product will be the present worth. EXAMPLES. 1. What is the rebate, or present worth of $ 100 due 1 year hence, discounting at 6 per cent per annum ? Tabular number = .943396 100 $94.33.9 Ans. 2. What is the present worth of $180 50 cents, due 5 years hence, at 5 per cent per annum ? Ans. $138 84c. 6m. £95 6s. 11d. 3. How much ready money will pay a debt of £112 10s. due 3 years hence, discounting at 6 per cent? 4. How much ready money is equal in value to £315 8s. due 7 months hence, allowing 5 per cent discount? 1 Ans. £306 15s. 4d. Note. When the discount is required, subtract the present worth from the principal, and the remainder is the discount. 5. What is the discount on £500, due 7 years hence, at 6 per cent per annum? Ans. 147 17s. 9d. COMPOUND INTEREST BY DECIMALS. RULE. 1. Find the amount of $1, or £1, for one year at the given rate per cent. 2. Involve the amount, thus found, to such a power, as is denoted by the number of years, and multiply this power by the principal, or given suni, and the product will be the amount required. 3. 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 per annum? The amount of 11. for 1 year = 1.05 and 1.054 X 500 = 607.753125 = the amount. 500 107.753125 = £107 15s. 03d. interest required. 2. What is the amount of £760 10s. for 4 years, at 4 per ct.? Ans. 8891. 13s. 6d. A Table of the amount of $1. or 11. at 6 per cent per annum, for When the given time consists of years and months, seek the amount of $1 &c. in the table for years, and the amount of $1 &c. in the foregoing table for the months, and the continual product of these tabular numbers into the principal, will give the amount required. Note. Subtract the principal from the amount, and the remainder is the compound interest. EXAMPLES 1. Required the amonnt of £480 for 5 years and 6 months, at 6 per cent per annum compound interest. Tabular number of 17. for 5 years = 1 338225 Do. 1.77782931 480 = prin❜l. Ans. £661.2341 &c. 2. What will $100 amount to, forborn 7 years and 10 months, at 6 per cent per annum ? Ans. $157 82c. 3m. 3. What is the compound interest of £210 50 cents, for 3 Ans. $29 48c. 7m. years, at 6 per cent? 4. What is the compound interest of 801. 4s. for 9 years and 4 months, at 6 per cent per annum? Ans. 471. 19s. Cad. Another method of computing Compound Interest for years, months, and days. RULE. To the Logarithm of the principal, add the several logarithms answering to the number of years, months, and days, found in the following tables, and their sum will be the logarithm of the amount required, Logarithmick Tables, at 6 per cent per annum, for years, months and days. Y. | dec. pts. | Y. | dec. pts. | Y. | dec. pts. | Y. | dec.pis. ,278366 ,025306 21 ,531426 31 ,784586 ,050612 12 ,303672 22 ,556732 3,075918 13 32 ,809792 5 7 ,328978 23 ,582038 ,101224 14 ,354284 24 ,607344 34 ,860404 ,126530 15 ,379690 25 ,632650 35 ,885710 151836 16 ,404896 26 ,657956 $6 177142 17 ,430202 27 ,683262 37 33 ,835098 ,911016 ,936322 8 ,202448 18,455058 28 M. dec. pts. M. | dec. pts. | M. | dec. pts. | M. | dec. pts. I ,002160 ,004321 4. ,008600 7 ,014940 10 ,021189 1. Required the amount of $436 50 cents, for 3 years, 8 months and 12 days. Log. of principal Ans. $541 75 = 2. 6399842 0. 0170330 2. 7337922 Amount requir'd 2. What will $175 amount to, in lo years and 5 months, at 6 per cent per annum ? CASE II. Ans. $321 28 cts. The amount, rate, and time given, to find the principal. RULE. 1. Divide the amount of the given sum by the amount of $I, or £1, for the given time, and the quotient will be the principal: OR 2. Multiply the PRESENT WORTH of $1, or £1, for the given number of years, at the given rate, by the amount; the product will be the principal. EXAMPLES. 1. What principal at 6 per cent per annum, will amount to $757.4856, in 4 years? 2. What principal at 6 per cent. per annum, will amount to £757 9s. 84. in 4 years? Ans, 6ccl. ARITHMETICAL PROGRESSION. Any rank of numbers, increasing by a common excess, and decreasing by a common difference, is said to be in ARITHMETICAL PROGRESSION. Thus S1.2.3.4.5.6.7 2 Ascending series. S14.12.10.8.6 &c. Desending series. The numbers which form the series, are called the TERMS of the progression; the first and last terms of which are called the EXTREMES. Any three of the five following terms being given, the other two may be readily found. 1. The first term. 2. The last term. 3. The number of terms. The first term, the last term, and the number of terms being given, to find the sum of all the terms. RULE. Multiply the sum of the extremes by the number of terms and half the product will be the answer. EXAMPLES. 1. The first term of an arithmetical progression is 5, the last term, 60, and the number of terms 12, required the sum of the series. 560 X 12 2 380 Ans. 2. The first term of an arithmetical progression is 3, the last term 112, and the number of terms 18; required the sum of the series. Ans. 1035 3. How many strokes do the clocks of Venice, (which go to 24 o'clock,) strike in the compass of a day? Ans. 300 4. Suppose a man lay up 1 cent the first day of the year, 2 cents the second, and 3 the third day, and so on in arithmetical progression, every day increasing i cent; how much will he have saved at the year's end? Ans. $667,95 cents. 5. A merchant bought 100 yards of cloth in arithmetical progression; he gave 5 cents for the first yard, and 1 dollar for the last, what did the cloth amount to? Ans. $52,50 cts. 6. If 100 stones be placed in a right line, exactly a yard asunder, and the first a yard from a basket, what length of ground will that man go, who gathers them up singly, returning with them one by one to the basket? Ans. 5 miles, 233 rods, 2 yards. |