EXAMPLES. 1. If £2 10s. be the wages of 15 men for 6 days, what will be the wages of 12 men for 18 11⁄2 men I days } days? 12 men : be :: 1××××55–55£=6£. 2s. 2d. 24qrs. Ans. NOTE. In this solution, having stated the question properly, inverted the two fractions in the first term and then arranged all the terms to be multiplied into each other, I take the shorter method of cancelling equal numerators and denominators, and abridging such as are divisible by the same number. The 12 cancels the 2 and the 6, and the 5 divides the 15, leaving a denominator 3 to be multiplied into the only remaining denominator, which is 3, and whose product is 9; then the only numerator above unity which is left, is 55 ; and, consequently, the whole is brought to the single improper fraction, which is the answer required. And thus let every similar question be managed. 2. In what time will $350 gain $261 at 5 per cent. per annum ? Ans. 18 months. NOTE.-For Double Rule of Three, Fellowship, &c. in decimals, let the scholar take, as exercises, any such questions in these rules, in whole numbers, as have fractions or compound numbers in them. SIMPLE INTEREST BY DECIMALS. A TABLE OF RATIOS. Ratio is the simple interest of £1 for 1 year, or, in Federal Money, of $1 for one year, at the rate per cent. agreed on. RULE.-Multiply the Principal, Ratio, and Time continually together, and the last product will be the interest required. EXAMPLES. 1. Required the interest of $425,65cts. for 6 years, at 6 per cent. per annum. $. cts. 425,65 Principal. 25,53 90 Interest for one year. 153,23 40 Ans. -$153,23cts. 4m. 2. What is the interest of £645 10s. for 3 years, at 6 per cent. per annum ? £645,5×,06×3=£116,190=£116 3s. 9d..2,4qrs. Ans. 3. Required the amount of $648,50cts. for 122 years, at 51 per cent. per annum. Ans. $1103,26cts.+ 4. What is the amount of $248,39cts. for 1 year at 6 per cent. per annum ? Ans. $270,7451. CASE 2.-The amount, time, and ratio given, to find the principal. RULE.-Multiply the ratio by the time; add unity to the product for a divisor, by which sum divide the amount, and the quotient will be the principal. EXAMPLES. 1. What principal will amount to $1235,97cts. 5m. in 5 years, at 6 per cent. per annum ? $ ,06×5+1=1,30)1235,975(950,75 Ans. 2. What principal will amount to £956 10s. 4,125d. in 8 years, at 5 per cent.? Ans. £645 15s. 3. What principal will amount to $1384,50cts. in 7 years, at 6 per cent. per annum? Ans. $975. CASE 3.-The amount, principal, and time given, to find the ratio. RULE.-Subtract the principal from the amount; divide the remainder by the product of the time and principal, and the quotient will be the ratio." EXAMPLES. 1. At what rate per cent. will $950,75 amount to $1235,975 in 5 years? From the amount=1235,975 Take the principal=950,75 950,75x5=4753,75)285,2250(,06-6 per cent. Ans. 2. At what rate per cent. will £543 amount to £705 18s. in 5 years? Ans. 6 per cent. 3. At what rate per cent. will $2124,25 amount to $3482,44234375 in 73 years? Ans. 81 per cent. CASE 4.-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 ratio and principal; and the quotient will be the time. 6 EXAMPLES. 1. In what time will $248,39 amount to $270,7451 at per cent. per annum? From the amount $270,7451 Take the principal 248,39 248,39,06 14,9034)22,3551(1,5-1 year. Ans. Ans. 5 years. 2. In what time will £543 amount to £705 18s. at 6 per cent. per aunum? 3. In what time will $2142,25 amount to $3482,44234375 at 81 per cent. per annum ? Ans. 7 years. TO CALCULATE INTEREST FOR DAYS. A TABLE OF RATIOS FOR DAYS. Rule. Multiply the principal by the given number of days and that product by the ratio for a year; divide the last product by 365, (the number of days in a year,) and it will give the interest required. Or, multiply the ratio for a day in the foregoing table by the principal, and that product by the given number of days; and the last product will be the interest required. EXAMPLES. 1. What is the interest of £360, 10s. for 146 days, at 6 per cent. ? 360,5x146x,06 £. £. s. d. qr. -8,652-8 13 0 1,92 Ans. 365 Or ,00016438356×360,5x146-£8,6519999+Ans. 2. What is the interest of 8780,40cts. for 100 days at 6 per cent. per annum? Ans. $12,82cts. 8m.+ 3. What is the interest of $481,75cts. for 25 days at 7 per cent. per annum ? Ans. $2,30cts. 9m.+ NOTE. The interest of any sum for 6 days, at 6 per cent. is just as many mills and decimals of a mill, as the principal contains dollars and decimals of a dollar. Therefore set down the principal, multiply it by the days, and divide the product by 6; the quotient will be the interest in mills and decimals of a mill. This is calling only 30 days a month. What is the interest of $231,84 for 100 days? 231,84×100 days? 6 3864m. Or $3,86 4; which is 5cts. 1m. too much; but when the time is less than 30 days, it gives the answer very exact, for ordinary sums. When interest is to be calculated on cash accounts, &c. where partial payments are made, it is the common practice to multiply the several balances into the days. they are at interest; then to multiply the sum of these products by the rate on the dollar, and divide the last product by 365; and thus cast the whole interest due on the account, &c. EXAMPLE. Lent John Joy, per bill on demand, dated 1st of June, 1821, $2000, of which I received back the 19th of August, $400; on the 15th of October, $600; on the 11th of December, $400; on the 17th of February, 1822, $200; and on the 1st of June, $400; how much interest is due on the bill, reckoning at 6 per cent. ? 365)23316,00(63,879+ Ans.=63,87 9 COMPOUND INTEREST BY DECIMALS. A table showing the amount of £1 or $1 at 5 and 6 per cent. per annum, compound interest, for 20 years. Yrs.5 per cent. 6 per ct. Yrs.5 per cent.6 per cent. 1 1,05000 1,06000 11 1,71033 1,89829 2,52695 3,02559 9 1,55132 1,68947 19 10 | 1,62889 | 1,79084 | 20 | 2,65329 | 3,20713 RULE.-Multiply the given principal continually by the amount of one pound, or one dollar, for one year, at the rate per cent. given, until the number of multiplications is equal to the given number of years, and the product will be the amount required. |