CASE II. When one of the ingredients is limited to a certain quanti. ty, we have this RULE. Find the proportionate quantities of each ingredient, by Case 1., in the same manner as though there was no limitation ; then as the difference against the simple whose quantity is given, is to each of the other differences, so is the given quantity of that simple to the quantity required of each of the other simples. Examples. . 1. A person wishes to mix 10 bushels of wheat, worth $1 per bushel, with rye, worth 70 cents per bushel, and oats, worth 30 cents per bushel, so that the mixture may be worth 60 cents per bushel. How many bushels of rye and oats must he use ? Proceeding according to Case I., we find the proportionate numbers to be 30, 30, and 50. Hence, 30 : 30 : : 10 : 10 30 : 50 :: 10 : 161 So that he must make use of 10 bushels of rye, and 163 bushels of oats. 2. A grocer has 90 pounds of tea, worth 90 cents per pound, which he wishes to mix with three other qualities, valued at 80 cents, 70 cents, and 60 cents per pound. How much must he take of these three kinds so as to be able to sell the mix. ture at 85 cents per pound? Ans. 10 pounds of each. 3. A wine-merchant wishes to mix 100 gallons of wine, worth $1 per gallon, with wines worth $1.10, $1.20, and $1.30 per gallon, so that the mixture shall be worth $1.15 per gal. lon. How many gallons must he use ? Note.--This question, like question 2, under Case I., will admit of seven answers, arising from the seven ways in which the prices of the ingredients may be connected. 4. How much gold at 16, 20, and 24 carats fine, and how much alloy, must be mixed with 10 ounces of 18 carats fine, that the composition may be 22 carats fine ? 5. A merchant has 90 pounds of spice worth 86 cents per pound, which he wishes to mix with three other sorts which are worth 30, 40, and 50 cents per pound, respectively. How many pounds must be used so that the compound may be worth 55 cents per pound ? CASE III. When two or more of the ingredients are limited in quantity, we have the following Find, as in Alligation Medial, what will be the rate of a mixture made of the given quantities of the limited ingredients only; then consider this as the rate of a limited ingredient, whose quantity is the sum of the quantities of the limited ingre. dients, from which, and the rates of the unlimited ingredients, proceed to calculate the several quantilies required, as in Case II. Examples. 1. I have 36 gallons of wine, at 24 cents a gallon, 8 gallons at 52 cents, and 4 gallons at 88 cents, and would mix the whole with three other kinds of wine, one at $1.25, one at 86 cents, and the other at 90 cents per gallon. How many gal. Jons of the last sorts must I use, so that the mixture may be worth $1 per gallon? By Alligation Medial, we find as follows: 36 gallons at 24 cents=$8.64 6 52 " = 4.16 48 gallons come to. $16.32 Therefore, one gallon of this mixture is worth 34 cents. Now by Case II., of Alligation Alternate, we have 0.02 Therefore, I must take 48 gallons of the two sorts, which are worth 86 and 90 cents per gallon, and 1724 gallons of the sort which is worth $1.25 per gallon. 2. With 63 gallons of wine, worth 8 shillings per gallon, I mixed other wine at 7 shillings, and some water : I then found that it was worth 6 shillings per gallon. How much wine and water did I mix with the 63 gallons ? 3. A person wishes to mix 10 bushels of wheat, at 70 cents per bushel, with rye at 48 cents, corn at 36 cents, and barley at 30 cents per bushel, so that a bushel of the mixture may be worth 38 cents. What quantity of each must be taken ? 4. With 50 pounds of tea, worth 75 cents per pound, I wish to mix other teas worth 90 and 95 cents per pound, so as to be able to afford the mixture at 80 cents per pound. How many pounds of the 90 and 95 cent teas must I use? CASE IV. When the whole compound is limited to a certain quantity, we have this RULE. Find the proportional parts, as in Case I. ; then as the sum of the proportional parts thus obtained, is to the given quantity, so is the proportionate quantity of each ingredient, to its required quantity. Examples. 1. Having three sorts of raisins at 9, 12, and 18 cents per pound, what quantity of each sort must I take to fill a cask of 210 pounds, so that its contents may be worth 14 cents per pound? SOLUTION. 15 : 210 :: 4 : 56 210 Therefore, I must take 56 pounds each, at 9 and 12 cents per pound, and 98 pounds at 18 cents per pound. 2. A goldsmith has gold of 14, 18, and 20 carats fine; and would mix of all these sorts so much as to make a mass of 50 ounces, which shall be 16 carats fine. How much of each sort is required ? 5 30 oz. at 14 carats fine; and 10 " oz. each, at 18 and 20 carats fine. 3. How much water must be mixed with brandy, worth $1.60 per gallon, to reduce the price to $1.20 per gallon, provided I fill a cask of 120 gallons ? 4. A grocer has currants at 6, 8, 10, and 12 cents per pound ; and he would make a mixture of 840 pounds, so that a pound of it may be worth 9 cents. How many pounds of each sort may be taken ? In this case, the seven sets of answers arise from the seven ways in which the ratios are capable of being alligated. We may add, that when there are two rates less than the mean, and two greater than the mean, there must be seven ways in which they may be alligated, and consequently they will give rise to seven distinct sets of answers. Questions in Alligation belong properly to that branch of algebra called Indeterminate Analysis, and will, in many ca. |