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in. cts.

lb. cts.

1b. cts. 2 at 12 1 at 12

r3 at 12 1 at 11 2 at 11

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

3d Ans.
1 at 9
2 at 9

2 at 9 2 at 8

1 at 8

3 at 8 4th Ans. 3lb. 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 dif: ferences, severally, to the several quantities required.

EXAMPLES.

.

1. A farmer would mix 10 bushels of wheat, at 70 cts. per bushel, with rye 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 quan489 2

[tity. Mean rate,

38

36 10
30 32

2: 2} bushels of rye. As 8:10 : : 10 : 121 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 an. swers : for after the quantities are found from different methods of linking ; any other numbers in the same propor.

fion between themselves, as the numbers which oompose the * answer, will likewise satisfy the conditions of the question, per gallon?

2. How much water must be mixed with 100 gallons of rum, worth 78. 6d. per gallon, to reduce it to os. 3d.

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 inay be worth 41 cents per bushel?

Ans. 20 bushels of barley, and 61 bushels of oats. 4. With 95 gallons of rum at 8s. per gallon, I mixed other rum 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 50 gals. water.

CASE III.

When the whole composition is limited to a given quantity.

RULE.

Place the difference between the mean rate, and the geveral 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.

1. A grocer had four sorts of tea, at 19. 3s. 6s. and 10s. per Ib. the worst would not sell, and the best were too dear; he therefore mixed 120 Ib. and so much of each sort, as to sell it at 4s. per Ib. ; how much of each sort did he take? 1b.

lb. 1. 6

C6 : 60 at 17

7 16. lb. 2 : 20 3 6

1 As 12 : 120 : : 1: 10 10

3 : SO

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3

6 per. Ib.

21318

10

Sum, 12

120

a

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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 gailon?

Ans. 33. gals, water, and 66 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 lb. without gain or loss ; what quantity of each must be taken ?

ns. 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 oz. 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 5s. 9d. per gallon, is to be mixed, so that a hhd. of 63 gallons may be sold for 12l. 125.; how many gallons must be taken of each ?

Ans. 14 gals. at 58. 9d. and 49 gals. at Ss. 6d.

a

a

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.

52, 4, 6, 8, &c. is an ascending arithmetical series : So

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

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

By having any three of these parts given, the other two may be found, which admits of a variety of Problems; but most of themare 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 term of an arithmetical series is 3, the last term 23, and the number of terms 11; required the sum of the series.

23+3=26 sum of the extremes.

Then 26x11:2*143 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 loct. the second for 2 cts. the third for 3 cts. &c. I demand what the cloth came to at that rate ?

Ans. $50. 4. A man bought 19 yards of linen in arithmetical progression, for the first yard he gave 1s. and for the last yil. il. 17s. what did the whole come to ? Ans. 618 ls.

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 to, and what did it average per yard? Ans. Jmount 8252, and the average price is S2, 52cts.

. 5 mills per yard.

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

ins. 23 miles, 5 furlongs, 130 yds.

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

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

EXAMPLES

292 -35

1. The extremes are 3 and 29, and the number of terms 14, what is the common difference?

Extremes. Number of terms less 1=13)26(2 Ans. 2. A man had 9 sons, whose several

ages

differed alike, the youngest was 3 years old, and the oldest 35; what was the common difference of their

ages

?
Ans. 4

years 3. A man is to travel from New-London to a certain place in 9 days, and to go but 3 miles the first day, in. creasing every day by an equal excess, so that the last day's journey may be 43 miles : Required the daily ine crease, and the length of the whole journey?

Ans. The daily increase is 5, and the whole journey 207 miles.

4. A debt is to be discharged at 16 different payments (in arithmetical progression, the first payment is to be 141. the last 100l.: What is the common difference, and the sum of the whole debt?'

Ans. 5l. 14s. 8d. common difference, and 9121. the whole

debt.

PROBLEM III. Given the first term, last term, and common difference, to

find the number of terms.

RULE. Divide the difference of the extremes by the common difference, and the quotient increased by i is the number of terms.

EXAMPLES.

1. If the extremes be 3 and 45, an: the common dif. ference %; what is the number of terms ? Ans. 22.

2. A man going a journey, travelled the first day five miles, the last day 45 inilos, and each day increased his journey by 4 miles; how many days did he travel, and how far i Ans. 11 days, and the wkole distance travelled 275 miles,

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