CASE II.

438. To find what quantities of different kinds must be taken to form a mixture of a given value.

1. Let it be required to find what quantities of coffee, at 14 cents, 16 cents, 19 cents, and 22 cents a pound, must be taken to form a mixture worth 18 cents a pound.

If we take 1 lb. at 14 cents to form a mixture worth 18 cents, we gain 4 cents, and to gain 1 cent we must take of a lb.

If we take 1 lb. at 22 cents, we shall lose 4 cents, and to lose 1 cent, so as to balance exactly what we have just gained, we must take † of a lb.

In like manner, we find we must take of a lb. of the 16 cent kind to gain 1 cent, and to lose 1 cent, to exactly balance that gain, we must take of the 19 cent kind 1 lb.

Therefore, we may take lb. of the coffee at 14 cents, lb. at 16 cents, 1 lb. at 19 cents, and 4 lb. at 22 cents, or by multiplying these proportionals by 4, the least common multiple of the denominators of the fractions, which will make all the quantities integral, we may take 1 lb. at 14 cents, 2 lb. at 16 cents, 4 lb. at 19 cents, and 1 lb. at 22 cents, to form a mixture worth 18 cents a pound.

Also, since proportional quantities will remain proportional, when multiplied or divided by any quantity, an indefinite number of results, all answering the conditions of the question, may be obtained.

RULE. Find how much must be taken of a kind whose value is less than the given average, to gain 1 of that average; also, how much must be taken of a kind whose value is greater than the given average, to lose 1.

In like manner, compare the value of each of the other kinds with the average value.

If there are fractions in the results, multiply each of the

Explain the operation. What is the Rule?

numbers by the least common multiple of their denominators, and the several products will be the required parts, in integers.

We may, also, clear of fractions simply by multiplying each pair of results corresponding to couplets of quantities compared, by the least common multiple of the denominators of their own fractions.

Examples.

2. How much sugar, at 10, 14, 17, and 18 cents a pound, may be mixed together, so that the mixture shall be worth 16 cents a pound? Ans. 1, 3, 6, and 3. 3. How much rice, at 4, 6, and 11 cents a pound, may be mixed, so that the compound shall be worth 7 cents a pound?

4. A grocer wishes to mix brandy at 3, 5, and 7 dollars a gallon, with water, so that the mixture may be worth $4 a gallon; how many gallons of each kind may be taken?

5. A farmer bought pigs at $6 each, sheep at $9 each, and colts at $10 each; how many may he have bought if he paid for all on an average of $8 apiece ? Ans. 2, 2, and 1.

439. To find the quantities of other kinds, when the quantity of one kind in the mixture is limited.

1. Let it be required to find how much coffee, at 16, 20, and 24 cents a pound, must be mixed with 10 lb., at 18 cents, so that the mixture may be worth 22 cents.

By comparing the price of the articles with the average price (Art. 438), we find, by taking & lb. of the 16 cent article, lb. of the 18 cent, and lb. of the 20 cent, there is a total gain of 3 cents, which may

be

Explain the operation.

balanced by taking }+1+1=11⁄2 lb. of the 24 cent, at a corresponding loss of 3 cents.

But, of the 18 cent kind it is required to take 10 lb., or 40 times lb. ; and taking the other proportional parts also 40 times as large, we have, as the required mixture, 63 lb. of the 16 cent article, 10 lb. of the 18 cent, 20 lb. of the 20 cent, and 60 lb. of the 24 cent.

RULE. Find the proportional quantities as in the preceding case, and take these proportional quantities as many times as large as the proportional quantity found of the proposed rate is contained times in the given quantity.

Examples,

2. A merchant has sugar at 9, 10, 13, and 14 cents a pound; how much may be mixed with 30 pounds of the 14 cent kind, to make a mixture worth 12 cents per pound?

3. How much water of no value may be mixed with 50 liters of wine, at $.90 per liter, to reduce the price to $.75 per liter? Ans. 10 liters.

4. Sold some cows at 80, 60, and 40 dollars each; also, 30 at 100 dollars each, and found that the whole averaged 70 dollars each; how many may there have been sold of the first three kinds? Ans. 1, 1, and 30.

CASE IV.

440. To find the quantity of each kind, when the entire quantity is limited.

1. How much sugar, of kinds at 10, 12, 18, and 20 cents a pound, must be taken to fill a box with a mixture of 200 pounds at 15 cents a pound?

We find the proportional quantities (Art. 438) to be cent kind, lb. of the 12 cent, lb. of the 18 cent, and 20 cent.

lb. of the 10 lb. of the

The sum of these quantities is only 18 lb.; but the entire quantity of the proposed mixture must be 200 lb.; hence, each of the proportional parts must be taken as many times as large, as 18 lb. is contained times in 200 lb., or $75 times.

Takinglb., lb., lb., and lb., respectively, 15 times, we have 371⁄2 lb. of the 10 cent kind, 62 lb. of the 12 cent, 62 lb. of the 18 cent, and 371⁄2 lb. of the 20 cent.

RULE. Find the proportional quantities as in preceding cases, and take each of these quantities as many times as large as their sum is contained times in the given entire quantity.

Examples.

2. What quantities of tea, worth $.96, $.90, and $.78 per pound, respectively, may be taken to form a mixture of 112 pounds, at $.88 per pound? Ans. 1633, 6738, 2733

3. A farmer has oats worth 40, 60, and 70 cents a bushel; what quantity of each kind may he take to make 40 bushels worth 50 cents a bushel?

4. A jeweler has gold 18, 19, and 24 carats fine; what quantity of each may he take to make 1 pound 20 carats fine?

Ans. 1, 1, 4.

5. A grocer has sugar worth 7 cents a pound, which he would mix with some at 8 cents a pound, some at 10 cents, and some at 11 cents a pound. How much of each kind may he take to make a mixture of 90 pounds worth 9 cents a pound?

6. How much wine at $2.40, $2.60, $2.80, and $2.90 per gallon, may be taken to make a hogshead worth $2.70 per gallon? Ans. 18, 9, 9, 27.

What is the rule? What is medial proportion? (436) What is it sometimes called? (436) What is the rule in Case I.? (437) In Case II.? (438) In Case III.? (439)

SERIES.*

441. A Series is a succession of numbers that depend on one another according to some fixed law.

The TERMS of a Series are the numbers composing it;

The EXTREMES are the first and last terms; and

The MEANS are all the terms between the first and last.

442. A series is increasing when the terms increase from left to right, and decreasing when they decrease from left to right.

ARITHMETICAL SERIES.

443. An Arithmetical Series, or Progression, is a series in which the terms vary by a common difference.

2, 4, 6, 8, 10, 12,

Thus,

is an increasing arithmetical series, in which 2 is the common difference.

444. The first term, the last term, the number of terms, the common difference, and the sum of the terms, in an arithmetical series, are so related that, any three of them being given, the other two may be found.

CASE I.

445. To find any term in an arithmetical series.

Let 3 the first term, 2 the common difference, and 5 number of terms. Then,

the

2d term = 32; 3d term = 3 + 2 × 2; 4th term =3+2×3; 5th term = 3+2 X 4.

That is, when the first term, the common difference, and the number of terms are given, to find any term of the series,

To the first term add the product of the common difference by the number of terms which precede the required term, if the

What is a Series? The Terms of a series? The Extremes? The Means? An Arithmetical Series?

* Optional.