CASE IV. 636. When the quantity of one of the simples is limited. 1. A miller has oats worth $.28, corn worth $.44, and barley worth $.90 per bushel. He wishes to form a mixture worth $.58 per bushel, and containing 100 bushels of corn. How many bushels of oats and barley may he take? 8 of barley. But as 100 bushels of corn, instead of 5, are required, we must take 100 = 20 times each of the other ingredients, in order that the gain and loss may be equal; and we shall therefore have 7 x 20 140 bushels of oats, and 8 × 20=160 bushels of barley. Hence the following = RULE. Find the proportional quantities by Case II or Case III. Divide the given quantity by the proportional quantity of this ingredient, and multiply each of the other proportional quantities by the quotient thus obtained. EXAMPLES FOR PRACTICE. 1. A dairyman bought 10 cows at $20 a head; how many must he buy at $16, $18, and $24 a head, so that the whole may cost him an average price of $22 a head? Ans. 10 at $16, 10 at $18, and 60 at $24. 2. Bought 12 yards of cloth for $15; how many yards must I buy at $11, and $ a yard, that the average price of the whole may be $11? Ans. 12 yards at $1 and 16 yards at $1. 3. How much water will dilute 9 gal. 2 per cent. strong to 84 per cent.? qt. 1 pt. of alcohol 96 Ans. 1 gal. 1 qt. 1 pt. 4. A grocer mixed teas worth $.30, $.55, and $.70 per pound respectively, forming a mixture worth $.45 per pound, having equal parts of the first two kinds, and 12 lbs. of the third kind; how many pounds of each of the first two kinds did he take? CASE V. 637. When the quantities of two or more of the ingredients are limited. 1. How many bushels of rye at $1.08, and of wheat at $1.44, must be mixed with 18 bushels of oats at $.48, 8 bushels of corn at $52, and 4 bushels of barley at $.85, that the mixture may be worth $.84 per bushel? ANALYSIS. Of the given quantities there are 18+ 8430 bushels, whose mean or average price we find by Case I to be $.54. We are therefore required to mix 30 bushels of grain worth $.54 per bushel, with rye at $1.08, and wheat at $1.44, to make a compound worth $.84 per bushel. Proceeding as in Case IV, we find there will be required 25 bushels of rye, and 5 bushels of wheat. Hence the following RULE. Consider those ingredients whose quantities and prices are given as forming a mixture, and find their mean price by Case I; then consider this mixture as a single ingredient whose quantity and price are known, and find the quantities of the other ingredients by Case IV. EXAMPLES FOR PRACTICE. 1. A gentleman bought 7 yards of cloth @ $2.20, and 7 yards @$2; how much must he buy @ $1.60, and @ $1.75 that the average price of the whole may be $1.80? 2. How much wine, at $1.75 a gallon, must be added to 60 gallons at $1.14, and 30 gallons at $1.26 a gallon, so that the mixture may be worth $1.57 a gallon? Ans. 195 gallons. 3. A farmer has 40 bushels of wheat worth $2 a bushel, and 70 bushels of corn worth $ a bushel. How many oats worth $1 a bushel must he mix with the wheat and corn, to make the mixture worth $1 a bushel? Ans. 63 bushels. CASE VI. 638. When the quantity of the whole compound is limited. 1. A tradesman has three kinds of tea rated at $.30, $.45, and $.60 per pound, respectively; what quantities of each should he take to form a mixture of 72 pounds, worth $.40 per pound? = 2224 112 672 ANALYSIS. By Case II, we find the proportional quantities to form the mixture to be 3 lb. at $.30, 2 lb. at $.45, and 1 lb. at $.60. Adding these proportional quantities, we find that they would form a mixture of 6 pounds. And since the required mixture is 12 times 6 pounds, we multiply each of the proportional terms by 12, and obtain for the required quantities, 36 lb. at $.30, 24 lb. at $.45, and 12 lb. at $.60. Hence the following RULE. Find the proportional numbers as in Case II or Case III. Divide the given quantity by the sum of the proportional quantities, and multiply each of the proportional quantities by the quotient thus obtained. EXAMPLES FOR PRACTICE. 1. A grocer has coffee worth 8 cts., 16 cts., and 24 cts. per pound respectively; how much of each kind must he use, to fill a cask holding 240 lb, that shall be worth 20 cts. a pound? Ans. 40 lb. at 8 cts., 40 lb. at 16 cts., and 160 lb. at 24 cts. 2. A man bought calves, sheep, and lambs, 154 in all, for $154. He paid $3 for each calf, $13 for each sheep, and $1 for each lamb; how many did he buy of each kind? Ans. 14 calves, 42 sheep, and 98 lambs. 3. A man paid $165 to 55 laborers, consisting of men, women, and boys; to the men he paid $5 a week, to the women $1 a week,and to the boys $1 a week; how many were there of each? INVOLUTION. 639. A Power is the product arising from multiplying a number by itself, or repeating it any number of times as a factor. 640. Involution of the process of raising a number to a given power. 641. The Square of a number is its second power. 642. The Cube of a number is its third power. 643. In the process of involution, we observe, I That the exponent of any power is equal to the number of times the root has been taken as a factor in continued multiplication. Hence: II. The product of any two or more powers of the same number is the power denoted by the sum of their exponents, and III. If any power of a number be raised to any given power, the result will be that power of the number denoted by the product of the exponents. 1. What is the 5th power of 6? tiplication; the final product, 7776, is the power required, (I). Or, we may first form the 2d and 3d powers; then the product of these two powers will be the 5th power required, (II). 2. What is the 6th power of 12? 122 OPERATION. = 144 144 2985984, Ans. ANALYSIS. We find the cube of the second power, which must be the 6th power, (III). 644. Hence for the involution of numbers we have the fol RULE. I Multiply the given number by itself in continued multiplication, till it has been takeu as many times as a factor as there are units in the exponent of the required power. Or, II. Multiply together two or more powers of the given number, the sum of whose exponents is equal to the exponent of the required power. Or, III Raise some power of the given number to such a power that the product of the two exponents shall be equal to the exponent of the required power. NOTES. 1. A fraction is involved to any power by involving each of its terms separately to the required power. 2. Mixed numbers should be reduced to improper fractions before involution. 3. When the number to be involved is a decimal, contracted multiplication may be applied with great advantage. EXAMPLES FOR PRACTICE. 1. What is the square of 79? Ans 6241. Ans. 16387.064. Ans. 796599 Ans. 1048576 6. Raise .4378565 to the 8th power, reserving 5 decimals. Ans. .00135 + 7. Raise 1.052578 to the 6th power, reserving 4 decimals 8. Involve .029 to the 5th power? Ans. 1.3600± Ans. .000000020511149. Find the value of each of the following expressions: |