Εικόνες σελίδας
PDF
Ηλεκτρ. έκδοση

16. cts.
16. cis.

1b, cts.
2 at 12
1 at 12

3 at 12 1 at 11 2 at 11

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

3d Ans.s
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 resty 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.

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 S stands against the given quan

2 Mean rate, 88 487

[tity 10 30 S2

2 : 2 bushels of rye. As 8:10:: 10 : 123 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, asthe numbers which conpose the arsuier, zoll likewise satisfy the conditiosus of the question

2. How much water must be mixed with 100 gallons of rum, worth 78. 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 oats must be mixed with the 20 bushels of rye, that the provender may be worth 41 cents per

bushel ? Ans. 20 bushels of barley, and 6144 bushels of oats. 4. With 95 gallons of rum at 8s. per gallon, I mixed other rum at 68. 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 30 gals. water.

CASE III.

When the whole composition is linfited 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.

1. A grocer had four sorts of tea, at 1s. Ss. 6s. and 103. per lb. the worst would not sell, and the best were too dear; he therefore mixed 120 lb. and so much of each sort, as to sell it at us. per Ib.; how imuch of each sort did ne take?

[ocr errors]
[ocr errors]
[ocr errors]

16. ib.
1 As 12 : 12Q ::

s. 6 : 60 at 1 2 : 20 S 1 : 10 6 3 : 30 10

per

6 10

Sum, 12

1.20

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. 334 gals, water, and 662 gals, wine. S. 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?

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 vi 5 oz. fine, and has a mind to mis. a pound of it so that it shall be 8 oz fine; how much oi each sort must he take?

Ans. 4 of 5 oz. fine, and 7 of 10 oz. fine. 5. Brandy at 3s. Ed. and 5s. 9d. per gallon, is to be mixed, so that a hhd. of 65 gallons may be sold for 12. 128.; how many gallons must be taken of each ?

Ans. 14 gals. at 5s. Od. and 49 gals. at Ss. 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. So

$ 2, 4, 6, 8, &c. is an ascending 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 terris being given, to find the sum of all the terms.

* A 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 adraits of a varieiy of Problems; but most of themare best uiderstood by machraic process, and are here oritiede

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

EXAMPLES.

1. The first tern 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--9=143 the Ansieer. 2. How many strokes does the hammer of a cock strike, in twelve hours.

Ains, 3 S. A merchant sold on vards of cloth. viz. the first yard for 1 ct. the second 1072 cts. the third for S cts. &c. I demand what the cloth came to at that rate ?

Ans. $50k. 4. A man bought 19 yards of linen in arithmetical progression, for the first yard he gave is.and for the last yd. 1l. 17s. what did the wide come to? Ains. £18 1s.

5. A draper sold 100 yıls. of broadcloth, at 5 cts. for the first yarıl, 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. Imount, 90594, ihe average price is 82, 52cts. 5 mills per yard.

6. Suppose 144 oranges were laid 2 yards distant from each other, in a mght line, and a basketplace'i two yaris from the first orange, what lengths of ground will that boy travel over, who gathers then up singls, returning with them one by one to the basket?

-Ans. 25 miles, 5 furlong's, 180 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 the conmon difference.

EXAMPLES, 1. The extremes are 3 and 29, and the number of terms 14, what is the common dillerence:

3}Extremes

.

Ans. 4 years.

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 53; what was the common difference of their ages ?

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, increasing every day by an equal excess, so that the last day's journey may be 43 miles : Required the daily increase, 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 1001.: What is the common difference, and the sum of the whole debt?

Ans. 51. 145. 8d. common difference, and 9121. the whole debt.

PROBIERI HII. 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 tarms.

EXAMPLES.

1. If the extremes be 3 and 45, and the common difference 2; what is the number of terms ? Ans. 22.

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

« ΠροηγούμενηΣυνέχεια »