1. Divide 54 by 15. EXAMPLES. •54=§4=,& and •15=!4=; &÷3=&×¥=37-344-3-506493 the quotient. IS the method of mixing two or more simples of different qualities, so that the composition may be of a mean or middle quality; It consists of two kinds, viz. Alligation Medial, and Alligation Al ternate. ALLIGATION MEDIÁL Is, when the quantities and prices of several things are given, to find the mean price of the mixture compounded of those things. RULE. As the sum of the quantities, or the whole composition, is to their total value; so is any part of the composition to its mean price or value. EXAMPLES. 1. A Tobacconist would mix 60% of tobacco, at 6d. per ib with 50 at 1s. 40 at 1s. 6d. and 30 at 2s. per : What is 11 of this mixture worth? d. 1 0 2 10 50 1 2. A farmer would mix 20 bushels of wheat at $1 per bushel, 16 bushels of rye at 75c. per bushel, 12 bushels of barley at 50c. * per bushel, and 8 bushels of oats at 40c. per bushel: What is the value of one bushel of this mixture? Ans. 73c. 5m. 3. A wine merchant mixes 12 gallons of wine, at 75c. per gallon, with 24 gallons at 90c. and 16 gallons at $1 10c.: What is a gallon of this composition worth? Ans. 92c. 6m. 4. A goldsmith melted together 8oz. of gold of 22 carats fine, 1 8oz. of 21 carats fine, and 10oz. of 18 carats fine: Pray what is the quality, or fineness of the composition? 8×22+20×21+10×18. 8+20+10 5. A refiner melts 51b of gold of 20 carats fine with 3 of 18 carats fine: How much alloy must be put to it, to make it 22 carats fine? 22-5×20+8x18÷÷5+8=3,3 Answer. It is not fine enough by 3, carats, so that no alloy must be added, but more gold. ALLIGATION ALTERNATE*. Is the method of finding what quantity of each of the ingredients, whose rates are given, will compose a mixture of a given rate: So that it is the reverse of Alligation Medial, and may be proved by it. CASE I. The whole work of this case consists in linking the extremes truly together and taking the differences between them and the mean price, which differences are the quantities sought. RULE. 1. Place the several prices of the simples, being reduced to one denomination, in a column under each other, the least uppermost, and so gradually downward, as they increase with a line of Demon. By connecting the less rate with the greater, and placing the dif ference between them and the mean rate alternately, or one after the other in turn, the quantities resulting are such, that there is precisely as much gained by one quantity as is lost by the other, and therefore the gain and loss, upon the whole, are equal, and are exactly the proposed rate. In like manner, let the number of simples be what it may, and with how many soever, each one is linked, since it is always a less with a greater than the mean price, there will be an equal balance of loss and gain between every two, and consequently an equal balance on the whole. It is obvious from the rule, that questions of this sort admit of a great variety of answers; for having found one answer, we may find as many more as we please, by only multiplying or dividing each of the quantities found, by 2, 3, 4, &c. the reason of which is evident; for if two quantities of two simples make a balance of loss and gain with respect to the mean price, so must also the double or triple, the half or third part, or any other ratio of these quantities, and so on ad infinitum. If any one of the simples be of little or no value with respect to the rest, its rate is supposed to be nothing, as water mixed with wine, and alloy with gold and silver. connection at the left hand, and the mean price at the left hand of all. 2. Connect, with a continued line, the price of each simple, or ingredient, which is less than that of the compound, with one or any number of those which are greater than the compound, and each greater rate or price with one or any number of the less. 3. Place the difference, between the mean price (or mixture rate) and that of each of the simples, opposite to the rates with which they are connected. 4. Then, if only one difference stand against any rate, it will be the quantity belonging to that rate; but if there be several, their sum will be the quantity. EXAMPLES. 1. A merchant has spices, some at 1s. 6d. per , some at 2s. some at 4s. and some at 5s. per : How much of each sort must he mix that he may sell the mixture at 3s. 4d. per ? s. d. per i d. I s. d. Note. These seven answers arise from as many various ways of linking the rates of the ingredients together. 2. *A merchant has Canary wine, at 3s. per gallon, Sherry, at 2s. 1d. and Claret at 1s. 5d. per gallon: How much of each soit must he take, to sell it at 2s. 4d. per gallon? 36 3--1114 at 30 *Note, the 2d and 34 questions admit but of one way of linking, and so but of one answer; yet all numbers in the same proportion between themselves, as the numbers which compose the answer, will likewise satisfy the condition of the question. 3. How much barley at 40c. rye at 60c. and wheat at 30c. per bushel, must be mixed together, that the compound may be worth 621c. per bushel? Ans. 17 bushels of barley, 17 of rye, and 25 of wheat. 4. A goldsmith would mix gold of 19 carats fine, with some of 16, 18, 23 and 24 carats fiue, so that the compound may be 21 carats fine: What quantity of each must he take? Ans. 5oz. of 16 carats fine, 5oz. of 18, 5oz. of 19, 10oz. of 23, and 10oz. of 24 carats fine. 5. It is required to mix several sorts of wine, at 60c. 90c. and $1 15c. per gallon, with water, that the mixture may be worth 75c. per gallon Of how much of each sort must the composition consist? : Ans. 40galls. of water, 15galls. of wine, at 60c. 15galls. do. at 90c. and 75galls. do. at $1 15c. CASE II. When the rates of all the ingredients, the quantity of but one of them, and the mean rate of the whole mixture are given, to find the several quantities of the rest, in proportion to the quantity given. RULE. Take the differences between each price, and the mean rate, and place them alternately, as in Case 1. Then, as the difference standing against that simple, whose quantity is given, is to that quantity, so is each of the other differences, severally, to the several quantities required. EXAMPLES. 1. A merchant has 4015 of tea, at Cs. per b, which he would mix with some at 5s. Ed. some at 5s. 2d. and some at 4s. 6d. How much of each sort must he take, to mix with the 4015, that he may sell the mixture at 5s. 5d. per th? 2. A farmer being determined to mix 20 bushels of oats, at 60c. per bushel, with barley, at 75c. rye, at $1, and wheat, at $1 25c. per bushel; I demand the quantity of each, which must be mixed with the 20 bushels of oats, that the whole quantity may be worth 90c. per bushel? Ans. 70 of barley, 60 of rye, and 30 of wheat, (or 20 of each.) 3. How much gold of 16, 20 and 24 carats fine, and how much alloy, must be mixed with 10oz. of 18 carats fine, that the composition may be 22 carats fine. Ans. 10oz. of 16 carats fine, 10 of 20, 170 of 21, and 10 of alloy. When the rates of the several ingredients, the quantity to be compounded, and the mean rate of the whole mixture are given, to find how much of each sort will make up the quantity. RULE. Place the differences between the mean rate, and the several prices alternately, as in Case 1; then, as the sum of the quantities, or differences thus determined, is to the given quantity, or whole composition; so is the difference of each rate, to the required quantity of each rate. EXAMPLES. 1. Suppose I have 4 sorts of currants, at 8d. 12d. 18d. and 22d. per ; the worst will not sell, and the best are too dear; I therefore conclude to mix 120 and so much of each sort as to sell them at 16d. per ; how much of each sort must I take? d. (6:36 at 8d. 2. A goldsmith has several sorts of gold; viz. of 15, 17, 20 and 22 carats fine, and would melt together, of all these sorts, so much as may make a mass of 40oz. 18 carats fine; how much of each sort is required? Ans. 16oz. 15 carats fine, 8oz. 17, 4oz. 20, and 12oz. of 22 carats fine. To this Case belongs that curious question concerning king Uiero's crown. Tiero, king of Syracuse, gave orders for a crown to be made entirely of pure gold; but suspecting the workmen had debased it, by mixing with it silver or copper, he recommended the discovery of the fraud to the famous Archimedes, and desired to know the exact quantity of alloy in the crown. Archimedes, in order to detect the imposition, procured two other masses, one of pure gold, and the other of silver, or copper, and each of the same weight with the former; and by putting each separately into a vessel full of water, the quantity of water expelled by them, determined their specific bulks; from which, and their given weights, it is easier to determine the quantities of gold and alloy in the crown by this case of Alligation, than by an Algebraic process. Suppose the weight of each mass to have been 51b, the weight of the water expelled by the alloy, 23oz. by the gold, 130z, and by the crown 160z. that is, that their specifick bulks were as 23, 13, and 16; then, what were the quantities of gold and alloy respectively in the crown? Here, the rates of the simples are 23 and 13, and of the compound 16, whence, 17 of gold? And the sum of these is 74-3-10, which should have 233 of alloy been but 5, whenee, by the rule, 137 16 { 23. 10:5 :: $7:31lb. of sold the Answer, 3:1b. of alloy |