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4. Then, if only one difference stands against any rate, it will be the quantity belonging to that rate, but if there be several, their sum will be the quantity.

LXAMPLES.

Answon.

1. A merchant has spices, some at 9d. per Ib. some at 1s. some at 2s. and some at 24. 6d. per lb. bow much of each sort inust he mix, that he may sell the mixture at 18. 8d. per pound ? d. d. 16

d. lb. 9- 10 at 9

97 2. 12 4 12 | Gives the

d. / 12 10 22 20 24

8 24 Answer. Ol 20 24 SO '11 30

30 2. Agrocer would mix the following quantities of sugar; viz. at 10 cents, 13 cents, and 16 cts. per lb. ; what quantity of each sort must be taken to make a mixture worth 12 cents per pound? Ans. 5lb. at 10cts, 21b. at 13cts, and 216, at 16 cts.

per

1b. 3. A grucer has two sorts of tea, viz. at 9s. and at 158. per lb. how must he nux them so as to afford the composition for 12s. per Ib.?

Ans. He must mix an equal quantity of each sort. 4. A goldsmith would mix gold of 17 carats fine, with some of 19, 21, and 24 carats fine, so that the compound мау be 22 carats fine; wlat quantity of each must he take.

Ans. 2 of each of the first three sorts, and G of the last.

5. It is required to mix several sorts of ruin, viz. at šs. 78. and 9s. per gallon, with water at O per gallon together, so that the mixture may be worth 6s. per gallon; how much of each sort must the mixture consist of ? Ans. I gal. of Rum at 5s. I do. at 75. 6 du at 9s. and 3

gals. water. Or, 3 gals. rum at 5s. 6 do. at 78. 1 du. at 9.s. and 1 gal. water. 6. A grucer hath several sorts of sugar, viz. one sort at 12 cts. per Ib. another at 11 cts. a third at 9 cts. and a furth at 8 cts. per lb. ; I demnand how much of each sort must he mix together, that the whole quantity may be alturled at 10 conts per pound ?

lb.

cts.
lb. cts.

lb. cts.
(2 at 12
1 at 12

rs at 12 1 at 11 2 at 11

2 at 11 2d Ans. 1st. Ans.

3d Ans.
1 at 9
2 at 9

2 at 9
2 at 8
1 at 8

3 at 8 4th Ans. Slb. of each sort.*

CASE II. ALTERNATION PARTIAL. Or, when one of the ingredients is limited to a certain quantity, thence to find the several quantities of the rest, in proportion to the quantity given.

RULE. Take the difference between each price, and the mean rate, and place them alternately as in Case I. 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.

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1. A farmer would mix 10 bushels of wheat, at 70 cts. per bushel, with

at 48 cts. corn at 36 cts. and barley at 30 cts. per bushel, so that a bushel of the composition may be sold for 38 cents; what quantity of each must be taken.

70- 8 stands against the given quan

48 2 Mean rate, 38

[tity. 10 30 32

2 : 2 bushels of rye. As 8 : 10 : : 10 : 124 bushels of corn,

32 : 40 bushels of barley. * These

four answers arise from as many various ways of linking the rates of the ingredients together.

Questions in this rule admit of an infinite variety of answers : for after the quantities are found from different methods of linking; any other numbers in the same proportion between themselves, as the numbers which compose the answer, will likewise satisfy the conditions of the question 2. How much water must be mixed with 100 gallons of rum, worth 7s. 6d. per gallon, to reduce it to 6s. 3d. per gallon?

Ans. 20 gallons. 3. A farmer would mix 20 bushels of rye, at 65 cents per bushel, with barley, at 51 cts. and oats at 30 cts. per bushel ; how much barley and vats must be mixed with the 20 bushels of rye, that the provender may be worth

bushel ? Ans. 20 bushels of barley, and 61 11 bushels of oats. 4. With 95 gallons of rum at 8s. per gallon, I mixed other run at 6s. 8d. per gallon, and some water; then I found it stood me in 6s. 4d. per gallon; I demand how much rum and how much water I took ?

Ans. 95 gals. rum at 6s. 8d. and so gals. water.

41 cents per

CASE III.

When the whole composition is limited to a given quantity.

RULE.

Place the difference between the mean rate, and the several prices alternately, as in Case I.; then, As the sum of the quantities, or difference 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.

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1. A grocer had four sorts of tea, at 1s. 3s. 6s. and 10s. per lb. the worst would not sell, and the best were too

he therefore mixed 120 15. and so much of each sort, as to sell it at 45. per Ib. ; kow much of each sort did he take ? S. 16.

1b. 1- 6

C6 : 60 at 17 2 lb. lb. 2 : 20 3

Ibe 1 As 12 : 120 : : 1: 10

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6 10

S : 30 10

S.

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Sum, 12

12C

2. How much water at 0 per gallon, must be mixed with wine at 90 cents per gallon, so as to fill a vessel of 100 gallons, which may be afforded at 60 cents per gallon :

Ans. 394 gals. water, and 664 gals. wine. 3. A grocer having sugars at 8 cts. 16 cts. and 24 cts. per pound, would make a composition of 240 lb. worth. 20 cts. per Ib. without gain or less ; what quantity of each must be taken ?

Ans. 40 lb, at 8 cts. 40 at 16 cts. and 160 at 24 cts. 4. A goldsmith had two sorts of silver bullion, one of 10 oz. and the other of 5 07. fine, and has a mind to mix a pound of it so that it shall be 8 oz fine; how much of each sort must he take ?

Ans. 4 of 5 oz. fine, and 7 of 10 oz. fine. 5. Brandy at 3s. 6d. and 55. 9d. per gallon, is to be mixed, so that a hhd. of 63 gallons may be sold for 12. 12s.; how many gallons must be taken of each ?

Ans. 14 gals. at 5s. 9d. and 49 gals. at 3s. 6d.

ARITHMETICAL PROGRESSION. ANY rank of numbers more than two, increasing by common excess, or decreasing by common difference, is said to be in Arithmetical Progression.

$ 2, 4, 6, 8, &c. is an ascending arithmetical series

28, 6, 4, 2, &c. is a descending arithmetical series : The numbers which form the series, are called the terms of the progression; the first and last terms of which are called the extremes.

PROBLEM I. The first term, the last term, and the number of terms being given, to find the sum of all the terms.

*A series in progression includes five parts, viz. the first torm, last term, number of terms, cominon difference, and sum of the series.

By having any three of these parts givwn, the other time may be found, which admits of a variety of Problems ; hest most of them are best understood by an algebraic process and are here omitted.

RULE. Multiply the sum of the extremes by the number of terms, and half the product will be the answer.

EXAMPLES.

1. The first terin of an arithmetical series is 8, the last term y, and the number of terms 11; required the sum of the series.

23+26 sum of the extremos.

Then 26x11228143 the Answer. 2. How many strokes does the hammer of a clock trike, in twelve hours ?

Ans. 78. 3. A merchant sold 100 yards of cloth, viz. the first yard for 1 ct. the second for 2 cts. the third for 3 cts. &c. I demand what the cloth came to at that rate ?

Ans. $501. 4. A man bought 19 yards of linen in arithmetical progression, for the first yard he gave 1s. and for the last yd. H. 178. what did the whole come to ? Ans. £18 is.

5. A draper sold 100 yards of broadcloth, at 5 cts. for the first yard, 10 cts. for the second, 15 for the third, &c. increasing 5 cents for every yard ; what did the whole amount tv, and what did it average per yard ?

Ans. Amount 8252), and the arerage price is 82, 52cts. 5 mills per yard.

6. Suppose 144 oranges were laid 2 yards distant from cach other, in a right line, and a basket placed two yards from the first orange, what length of greund will that boy travel over, who gathers them up singly, returning with them one by one to the basket ?

Ans. 23 miles, 5 furlongs, 180 yds.

PROBLEM II. The first term, the last term, and the number of terms given, in find the coinmon difference.

RULE. Divide the difference of the extremes by the number of terms less 1, and the quotient will be the common dif. ference.

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