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2cwt. at $ 9.00 = $ 18.00
5 6 : $ 45.00 :: lcwt. : $ 9.00 Answer. 2. If 19 bushels of wheat at $ 1.00 per bushel should be mixed with 40 bushels of rye at $ 0.66 per bushel, and 11 bushels of barley at $0.50 per bushel, what would a bushel of the mixture be worth?
Ans. $ 0.72,77. 3. If 3 pounds of gold of 22 carats fine be mixed with 3 pounds of 20 carats fine, what is the fineness of the mixture ?
Ans. 21 carats. 4. If I mix 20 pounds of tea at 70 cents per pound with 15 pounds at 60 cents per pound, and 80 pounds at 40 cents per pound, what is the value of 1 pound of this mixture ?
Ans. $ 0.4715. Note. — If an ounce, or any other quantity, of pure gold be divided into 24 equal parts, these parts are called carats. But gold is often mixed with some baser metal, which is called the alloy; and the mixture is said to be so many carats fine, according to the proportion of pure gold contained in it; thus, if 22 carats of pure gold and 2 of alloy be mixed together, it is said to be 22 carats fine.
This rule teaches us how, from the prices of several articles given, to find how much of each must be mixed to bear a certain price.
CASE I. RULE. — Place the prices under each other, in the order of their value; connect the price of each ingredient, which is less in value than the intended compound, with one which is of greater value than the compound. Place the difference between the price and that of each simple, opposite to the price with which they are connected.
EXAMPLES 5. A merchant has spices, some at 18 cents a pound, some at 24 cents, some at 48 cents, and some at 60 cents. How much of each sort must he mix that he may sell the mixture at 40 cents a pound ?
cts. Ibs. cts.
20 at 18
24 8 • 24 Mean rate 40
16 6 48
> Answers. (60- 22 66 60 )
20+ 8 28 6 24 Mean rate 40
( 605 22 + 16 | 38 " 60 ) Explanation. — By connecting the less rate with the greater, and placing the differences between them and the mean rate alternately, 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 must be equal, and the compound will have the value of the proposed rate; 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 they may, and with how many soever every 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 others as we please by only multiplying or dividing each of the quantities found by 2, 3, or 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 treble, the half or third part, or any other equimultiples or parts of these quantities.
6. How much barley at 50 cents a bushel, and rye at 75 cents, and wheat at $ 1.00, must be mixed, that the composition may be worth 80 cents a bushel ?
Ans. 20 bushels of rye, 20 of barley, and 35 of wheat. 7. A goldsmith would mix gold of 19 carats fine with some of 15, 23, and 24 carats fine, that the compound may be 20 carats fine. What quantity of each must he take ?
Ans. 4oz. of 15 carats, 3oz. of 19, loz. of 23, and 5oz. of 24.
8. It is required to mix several sorts of wine at 60 cents, 80 cents, and $ 1.20, with water, that the mixture may be worth 75 cents per gallon; how much of each sort must be taken?
Ans. 45gals. of water, 5gals. of 60 cents, 15gals. of 80 cents, and 75gals, of $1.20.
CASE II. When one of the ingredients is limited to a certain quantity.
RULE. – Take the difference between each price and the mean rate, as before; then say, as the difference of that simple whose quantity is given is to the rest of the differences severally, so is the quantity given to the several quantities required.
EXAMPLES. 9. How much wine at 5s., at 5s. 6d., and at 6s. a gallon, must be mixed with 3 gallons at 4s. per gallon, so that the mixture may be worth 5s. 4d. per gallon? ( 48 7 8 +2= 10 Then 10:10 ::3:3 *2= 10
10:20::3:6 16 +4= 20. 10:20::3:6
16 * 4 = 20
Ans. 3gals. at 58., 6 at 5s. 6d., and 6 at 6s. 10. A grocer would mix teas at 12s., 10s., and 6s. per pound, with 20 pounds at 4s. per pound; how much of each sort must he take to make the composition worth 8s. per pound ?
Ans. 20lbs. at 4s., 10lbs. at 6s., 10lbs. at 10s., and 20lbs. at 12s.
11. How much port wine at $ 1.75 per gallon, and temperance wine at $ 1.25 per gallon, must be mixed with 20 gallons of water, that the whole may be sold at $ 1.00 per gallon?
Ans. 20gals. port wine, and 20gals. temperance wine. 12. How much gold of 15, 17, and 22 carats fine must be mixed with 5 ounces of 18 carats fine, so that the composition may be 20 carats fine ?
Ans. 5oz. of 15 carats, 5oz. of 17, and 25oz. of 22.
CASE III.* When the sum and quality of the ingredients are given. Rule. — Find an answer as before, by linking; then say, as the sum 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.
* To this case belongs the curious fact of King Hiero's crown.
Hiero, King of Syracuse, gave orders for a crown to be made of pure gold; but suspecting that 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, the one of pure gold, the other of copper, and each of the same weight of the former; and by putting each separately into a vessel full of water, the quantity of water expelled by them determined their specific gravi
EXAMPLES. 13. How many gallons of water must be mixed with wine at $ 1.50 per gallon, so as to fill a vessel containing 100 gallons, that it may be sold at $ 1.20 per gallon?
gals. gals. gals. gals.
30 150: 100 :: 30: 20 water 120 150 120 150 : 100 :: 120 : 80 wine )
14. A merchant has sugar at 8 cents, 10 cents, 12 cents, and 20 cents per pound ; with these he would fill a hogshead that would contain 200 pounds. How much of each kind must he take, that he may sell the mixture at 15 cents per pound?
Ans. 33flbs. of 8, 10, and 12 cts., and 100lbs. of 20 cts.
PERMUTATIONS AND COMBINATIONS.
THE permutation of quantities is the showing how many different ways the order or position of any given number of things may be changed.
The combination of quantities is the showing how often a less number of things can be taken out of a greater, and combined together, without considering their places or the order they stand in.
CASE I. To find the number of permutations or changes that can be made of any given number of things, all different from each other.
Rule. -- Multiply all the terms of the natural series of numbers, from
ties ; 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 10 pounds, and that the water expelled by the copper was 92 pounds, by the gold 52 pounds, and by the compound crown 64 pounds : what will be the quantities of gold and alloy in the crown? The rates of the simples are 92 and 52, and of the compound 64; there1 up to the given number, continually together, and the last product will be the answer required.
592- 12 of copper, 40 : 10 :: 12 : 31bs. of copper) A 04 622 28 of gold, 40 : 10::28 : 7lbs. of gold S
This rule may be illustrated by inquiring how many different numbers may be formed from the figures of the following number, 789, making use of the three figures in each number.
789, 798, 879, 897, 978, 987. It will be perceived, that six are all the permutations the above number will admit of. By adopting the rule, we find the same answer.
1 X2 X3 = 6 Ans. 1. How many changes may be rung on 6 bells?
OPERATION. 1 X 2 X 3 X 4 X 5 X 6=720 changes. Ans. 2. For how many days can 10 persons be placed in a differ. ent position at dinner?
the Ans. 3628800. 3. How many changes may be rung on 12 bells, and how long would they be in ringing, supposing 10 changes to be rung in one minute, and the year to consist of 365 days, 5 hours, and 49 minutes ? Ans. 479001600, and 9ly. 26d. 22h. 41m.
4. How many changes or variations will the letters of the alphabet admit of? Ans. 403291461126605635584000000.
CASE II. Any number of different things being given, to find how many changes can be made out of them, by taking any given number of quantities at a time.
RULE. – Take a series of numbers, beginning at the number of things given, and decreasing by 1, to the number of quantities to be taken at a time; the product of all the terms will be the answer required.
Illustration. — The above rule may be illustrated by inquiring how many different numbers can be made by a selection of any three figures from the following number, 1234.
OPERATION. 123, 124, 132, 134, 142, 143, 213, 214, 231, 234, 241, 243, 312, 314, 321, 324, 341, 342, 412, 413, 421, 423, 431, 432.
By examining the above, it will be perceived that there are 24 different numbers or permutations; and this number may be obtained by multiplying the number of things given, 4, by the next lower number, and that product by the next lower, and