5. A refiner melts iclb. of gold of 20 carats fine with 10lb. of 18 carats fine ; how much alloy must he put to it to make it 22 carats fine ? Ans. It is not fine enough by 32 carats, so that no alloy must be put to it, but more gold. ALLIGATION ALTERNATE. Alligation alternate is the method of finding what quartity of any number of simples, 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. FULL I. 1 1. Write the rates of the simples in a column under each other. 2. Connect * DEMONSTRATION. By connecting the less rate to the greater, and placing the differences between them and the mean rate alternately, the quantities resulting are such, that there is précisely 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 : and the same will be true of any other two simples, managed according to the rule. In like manner, let the number of simples be what it may, and with how many soever each 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. Q. E. D. 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, or 4, &c. the ruason 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 or any 1 2. Connect or link with a continued line the rate of each simple, which is less than that of the compound, with one number of those, that are greater than the compound ; and each greater rate with one or any number of the less. 3. Write the difference between the mixture rate and that of each of the simples opposite the rates, with which they are respectively linked. 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 would mix wines at 145. 195. 15s. and 225. per gallon, so that the mixture may be worth 18s. the gallon : what quantity of each must be taken? 14 4 at 14.5. 1 at 155. 3 at 19s. 4 at 225. Or thus : 14 1+4. ! 5 at 145. .15 1 at 155. 22 1 18 19 4+3 7 at igs. 22 4 | 4 at 22s. 2. How much wine at 6s. per gallon and at 4ś. per gallon, must be mixed together, that the composition may be worth 55. per gallon ? Anş. 12 gallons, or equal quantities of each. 3. How also the double or treble, the į or } part, or any other ratio of these quantities, and so on, ad infinitum. Questions of this kind are called by algebraists indeterminate or unlimited problems, and, by an analytical process, theorems may be raised, that will give all the possible answers. 45. 8d. 3. How much corn at 2s. 6d. 35. 8d. per bushel, must be mixed together, that the compound may. be worth 39. rod. per bushel ? Ans. 12 at '2s. 6d. 12 at 38. 8d. 18 at 45. and 18 at 45. 8d. 4. A goldsmith has gold of 17, 18, 22 and 24 carats fine : how much must he take of each to make it 21 carats fine ? Ans. 3 of 1%, 1 of 18, 3 of 22 and 4 of 24. 5. It is required to mix brandy at 8s. wine at 7s. cider at is. and water at o per gallon together, so that the mixture may be worth 55. per gallon ? Ans. 9 of brandy, 9 of wine, 5 of cider, and 5 of water. RULE 2.* When the whole composition is limited to a certain quantity, find an answer as before by linking ; then say, as the sum of * A great number of questions might be here given relating to the specific gravity of metals, &c. but one of the most curious, with the operation at large, may serve as a sufficient specimen. Hiero, king of Syracuse, gave orders for a crown to be made entirely of pure gold ; but suspecting the workmen had debased it by mixing it with 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, the other of silver or copper, and each of the same weight with the former; and each being put separately into a vessel full of water, the quantity of water expell. ed by them determined their specific bulks : from which and their given weights, the exact quantities of gold and alloy in the crown may be determined. Suppose the weight of each crown to be rolb. and that the water expelled by the copper or silver was '921b. by the gold '52lb. and by the compound crown 641b. what will be the quantities of gold and alloy in the crown? The of the quantities, or differences thus determined, is to the given quantity, so is each ingredient, found by linking, to the required quantity of each. EXAMPLES 1. How many gallons of water at os. per gallon must be mixed with wine worth 39. per gallon, so as to fill a vessel of 100 gallons, and that a gallon may be afforded at 25. 6d. ? 6 30 { 30 36 I 2 24 Ans. 835; galions of wine, and 16 of water. 2. A grocer has currants at 4d. 6d. od. and 11d. per Ib. and he would make a mixture of 240lb. so that it may be afforded The rates of the simples are 92 and 52, and of the compound 64 ; therefore 12 of copper, 28 of gold. And the sum of these is 12+28=40, which should have been but 10 ; whence, by the rule, 40 : 10 :: 12 : 3lb. of copper, the answer. 40 : 10 :: 28 : 7lb. of gold, T affcnled at 8d. per pound ; how much of each sort niust he take ? Ans. 721b. at 4d. 24 at 6d. 48 at 9d. and 96 at ild. 3. How much gold of 15, of 17, of 18 and of 22 carats fine, must be mixed together to form a composition of 40 ounces of 20 carats fine ? Ans. 5 oz. of 15, of 17 and of 18, and 25 of 22. RULE 3." When one of the ingredients is limited to a certain quantity : take the difference between each price and the mean rate as before ; then, As the difference of that simple, whose qliantity is given, is to the rest of the differences severally, so is the quantity given to the several quantities required. EXAMPLES. 1. How much wine at 5s. at 5s. 6d. and ós. the gallon must be mixed with 3 gallons at 4s. per gallon, so that the mixture may be worth 58. 4d. per gallon ? 8+2=10 60 8.+2=10 48. 643 66- 15+4=20 IO IO : IO : 3.: 3. 3 6 6 Ans. 3 gallons at 5s. 6 at 5s. 6d. and 6 at 6s. : 2. A : 20 :: * In the very same manner questions may be wrought, when several of the ingredients are limited to certain quantities, by finding first for one limit and then for another. The two last rules can want no demonstration, as they evidently result from the first, the reason of which has been already explained. |