66 ducts by the aggregate of the quantities, and the quotient 130 bottles which cost him 10 cents each. at 15 cents each. at 12 27 at 20 Now, 130 at 10 cts. = 1300 463)5737(12 39 75 66 15 66 1125 463 1107 926 1810 4210 2. A grocer has 4 lbs. of tea at 90 cents per lb., 8 lbs. at 75 cents, and 6 lbs. at 110 cents. to be mixed together, 'what will a pound of this mixture be worth? Ans. 90c. 3. A grocer has 2 cwt. of coffee at $25 per cwt.; 4 cwt. at $20.50 per cwt. and 7 cwt. at $18.624 per cwt. which he will mix together, what will 1 cwt. of this mixture be worth? Ans. $20.181 . CASE II. When the prices of all the simples, the quantity of one of them, and the mean price of the whole mixture are given to find the quantities of all the rest. Rule 1.- Place the mean rate and the several prices, link them and take their differences, as in the preceding case. 2. As the difference of the same name with the quantity given is to the differences respectively, so is the given quantity to the several required quantities. 1. What quantity of coffee at 20 cents, and at 16 cts. per lb. must be mixed with 35 lb. at 14 cents to make a mixture worth 18 cents per lb.? 145 2 Mean rate 18 16 2 20 4 + 2 = 6 Then, as 2 : 35 : : 2 = 35 at 16. 2 : 35 :: 6 105 at 20. 2. How much tea at 86 cents, at 94 cents, and at 105 cents per lb. ought to be mixed with 6 lbs. at 75 ets. per lb. for a mixture, to sell at 92 cts. per lb.? Ans. 18 lbs. at $1.05, 51 lbs. at 94c., 39 lbs. at 86c. CASE III. When the prices of the several simples, the quantity to be compounded, and the mean price are given to find the quantity of each simple. RULE 1.—Link the several prices and take their differences as before. 2d. As the sum of the differences is to the difference opposite each price, so is the quantity to be compounded to the quantity required. 1. A grocer has three sorts of sugar, viz: 10, 11, and 8 cents per 1b. how much of each sort must he take? 8 1 + 2 = 3 Mean rate 9 10 1 1 11 =1=1 : 40 : Sum of differences, 5 8 at 10 5 1 : : 40 cm 8 at 11 2. A vintner has wine at 130 cts. at 160 cts, and at 180 cts. per gallon, and he would have 32 gallons worth 145 cents per gallon, I demand how much of each sort he must have? Ans. 20 gals. at $1.30, 6 gals, at $1.60 and 6 gals. at $1.80. ARITHMETICAL PROGRESSION. When a series of numbers or quantities increase or decrease by a constant difference, it is called Arithmetical progression; as, 1, 2, 3, 4, 5 6; 1, 3, 5, 7, 9, 11; 6,5, 4, 3, 2, 1; 11, 9, 7,5, 3, 1. There are five things to be particularly attended to in Arithmetical Progression; the first term, the last term, the number of terms, the common difference, and the sum of all the terms. CASE I. The first term, common difference, and number of terms being given to find the last term, and sum of all the terms. RULE 1.-Multiply the number of terms, less one, by the common difference, and to that product, add the first term, the sum is the last term. 2. Add the first and last terms together, and multiply the sum by the number of terms, and half the product will be the sum of all the terms. 1. A person sold 40 yards of muslin at 2 cents for the first yard, 4 cents for the second, increasing 2 cents every yard, what did they amount to? Ans. $16.40. OPERATION. Nos. of terms 40 – 1 - 39 X 2 = 78 + 2 = 80 last term. 1st term or extreme 2 Last do. or second extreme 80 82 x 40 = 380 = $16.40 2. A butcher bought 75 sheep, and gave 6 cents for the first, 8 for the second, &c., what did he give for the last, and what did the whole number cost him? Ans. For the last $1.54, the whole $60. 3. A travels uniformly at the rate of 6 miles an hour, and sets off upon his journey 3 hours and 20 minutes before B; B follows him at the rate of 5 miles the first hour, 6 the second, 7 the third, and so on. In how many hours will B overtake A? Ans. 8 hours. CASE II. When the first and last terms (or two ertremes) are given to find the common difference. Rule.- Divide the difference of the extremes by the number of terms less 1; the quotient will be the common difference. 1. If the ages of 12 persons are equally different, the youngest is 18 years and the eldest 40, what is the common difference of their age? Ans. 2 common difference. Illustration of the above question. 40 12–1 = 11)22(2 common difference. 22 2. The extremes are 3 and 45, and the number of terms is 22, what is the common difference? Ans. 2. 3. A man received “charity” from 10 different persons, the first 4 cents, the last 49 cents, what was the common difference, and what did the man receive? Ans. he received $2.65; com. dif. 5 cts. 4. The extremes are 3 and 39, and the sum of the series 399; what is the common difference? Ans. 2. GEOMETRICAL PROGRESSION. Geometrical Progression is the increase of a series of numbers by a common multiplier, or decrease by a common divisor; as 2, 4, 8, 16, 32; 32, 16, 8, 4, 2; the ratio is the number by which the series increases or decreases. CASE I. To find the last term and sum of the series. RULE.—Raise the ratio to the power whose index is 1 Less than the number of terms given. 2. Multiply the product by the first term, and the result will be the last term. 3. Multiply the last term by the ratio; from the product subtract the first term, and divide the remainder by the ratio less 1, for the sum of the series. 1. If I buy 16 cords of wood, and agree to 2 cents for the first, 4 for the second, 8 for the third, &c., doubling the price to the last, what will it cost me? 1st. 2nd. 3rd. 4th. power 1, 2, 3, 4, ratio 2, 4, 8, 16 fourth power 16 $1310.70 Answer. 2. A person at the birth of his son, deposited in bank 1 cent, towards his fortune, promising to double it at the return of every birthday, until he was 21 years of age, what was his portion? Ans. $20,971.51. |