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ALLIGATION.

ALLIGATION is the method of mixing two or more articles, of different qualities, so that the composition may be of a mean, or middle quality.

Alligation consists of two kinds, viz. Alligation Medial, and Alligation Alternate.

ALLIGATION MEDIAL,

Is when the quantities and prices of several articles are given, to find the mean price of the mixture, composed of those articles, or things given.

RULE. Find the value of all the articles, and then divide this value by the sum of the articles, which will give the price required.

EXAMPLES.

1. If 6 gallons of brandy, worth $1.25 a gallon, be mixed with 9 gallons worth 80 cents per gallon, and 5 gallons of whiskey, worth 40 cents a gallon, what is 1 gallon of this mixture worth? Ans. $.835, or 83 cents. 2. If a bushel of Indian corn, at 75 cents a bushel, be mixed with 5 bushels of rye, at 80 cents per bushel, and 15 bushels of oats, at 30 cents per bushel, what will be the value of a bushel of the mixture?

Ans. 44

cents.

3. A wine merchant mixed 12 gallons of wine, at 75 cents per gallon, with 24 gallons at 90 cents, and 16 gallons, at $1.10 cents; what is a gallon of this composition worth? Ans. $.92613. Questions. Alligation Medial is what? Recite the rule, &c.

ALLIGATION ALTERNATE,

Teaches the method of finding the quantity of each article given, at their respective rates, that must be taken to make a compound at any proposed rate.

RULE.

1. Write the given prices of the articles under each other, and the mean rate on the left hand of those.

2. Connect with a continued line each price that is less than the mean rate with one or more that is larger, and each price larger than the mean rate with one or more that is less.

3. Write the difference between the mean rate (or price), and the price of each article opposite to the price with which it is connected.

4. Then the sum of the differences, standing against any price, will express the relative quantity to be taken of that price.

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EXAMPLES.

cts.

9

1. A grocer would mix sugars, at 9 cents, 11 cents, and 13 cents per pound; what quantity of each kind must he take, that the mixture may be worth 12 cents a pound?

lb.

cts. lb. 1 at 9 9

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2. A merchant would mix wines, at $1.20, $1.50, $2, and $2.50 per gallon, so that the mixture should be worth $1.75 per gallon; what quantity of each kind must he take?

3. A grocer would mix rum at 80 cents, and at 70 cents a gallon, with water, that the mixture may be worth 75 cents a gallon; what quantity of each sort must he take?

Ans. 80 gal at 80 cts., 5 gal. at 70 cts. and 5 gal. of water.

CASE.

When the rates of all the articles, the quantity of one of them, and the mean rate of the whole mixture are given, 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 1. Then, as the difference standing against that article whose quantity is given, is to that quantity, so is each of the other differencos, severally, to the several quantities required.

EXAMPLES.

1. A merchant has 20 pounds of tea, at $1.04 per pound, which he would mix with some at 98 cents, some at 92 cents, and some at 80 cents per pound; how much of each_kind_must he take to mix with the 20 ̊pounds, that he may sell the mixture at 96 cents per pound?

104. 4 stands against the given quantity.

98. 16

lb.

lb. cts.

96

92- 8

16

80 at 98

80 2 As 4: 20 ::

8: 40 at 92 Ans. 2: 10 at 80

2. Bought a pipe of brandy, containing 120 gallons, at $1.30 a gallon; how much water must be mixed with it to reduce the first price to $1.10 a gallon?

Ans. 21, gals.

PERMUTATION.

PERMUTATION is the method of finding how many different changes can be made of any proposed number of things.

RULE.-Multiply all the terms of the natural series of numbers continually together, from one up to the given number, and the last product will be the changes required.

EXAMPLES.

1. Five gentlemen agreed to remain together as long as they could arrange themselves differently at dinner; how many days did they remain together? Ans. 120 days.

2. Christ church, in Boston, has 8 bells; how many changes may be rung on them? Ans. 40320 changes.

3. Seven gentlemen, who were travelling, met together by chance, at a certain inn upon the road, where they were so well pleased with their host and each other's company, that (in a frolic) they offered him 30 dollars, to stay at that place so long as they, together with him, could sit every day at dinner in a different order. The host, thinking that they could not sit in many different positions, imagined that he should make a good bargain, and readily entered into an agreement with them, and so made himself the eighth. I demand how many different positions they sat in; and how long they stayed at the inn?

Ans. They sat in 40320 positions; and the time was 110 years and 142.5days.

Questions-Permutation teaches what? What is the rule?

COMBINATION.

COMBINATION teaches how many different ways a less number of things may be combined out of a larger; thus, out of the letters a, b, c, d, are 6 different combinations of 2, viz. ab, ac, ad, dc, db, bc. Thus, 4×3=12

1x2= 2

6 Ans.

RULE.-1. Take a series of as many terms decreasing by 1, from the number, out of which the election is to be made; and find the product of all its terms for a dividend.

2. Take a series beginning with 1, and increasing by 1, up to the number to be combined, and find the product of all its terms for a divisor.

3. Divide the dividend by the divisor, and the quotient will be the number sought.

EXAMPLES.

1. How many combinations may be made of 5 letters in 10?

Questions.-Combination teaches what? Recite the rule.

Ans. 252.

ARITHMETICAL PROGRESSION.

ANY rank or series of numbers, more than two, increasing or decreasing by uniform difference, are said to be in Arithmetical Progression.

When the numbers are formed by a continual addition of the common difference, they form an ascending series; but when they are formed by a continual subtraction of the common difference, they form a descending series.

3, 5, 7, 9, 11, 13, 15, &c. is an ascending series. Thus, 15, 13, 11, 9, 7, 5, 3, &c. is a descending series.

The numbers which form the series are called the terms of the series.

The first and last terms are called the extremes, and the other terms the means.

There are five denominations in arithmetical progression, any three of which being given, the other two may be found, viz. 1st. The first term.

2d. The last term.

3d. The number of terms.

4th. The common difference, or ratio.

5th. The sum of all the terms.

When the first term, the number of terms, and the common difference, are given, to find the last term,

Multiply the number of terms, less 1, by the common difference. and add the first term to the product for the last term.

1. A man bought 100 yards of cloth, giving 4 cents for the first yard, 7 for the second, 10 for the third, and so on, with a common difference of 3 cents; what was the cost of the last yard? Ans. 301 cents=$3.01.

2. There are, in a certain triangular field, 41 rows of corn; the first row, in one corner, is a single hill, the second contains 3 hills, and so on, with a common difference of 2; what is the number of hills in the last row? Ans. 81 hills.

3. A man on journey travelled 20 miles the first day, 24 the second, and so on, increasing 4 miles every day-how far did he travel the twelfth day? Ans. 64 miles.

4. There is an arithmetical series consisting of 18 terms; the first term is 4, the common difference 12; what is the largest term?

Ans. 208. 5. If the first term be 3, the common difference 2, and the number of terms 19; what is the last term? Ans. 39.

Questions.-What is Arithmetical Progression? The numbers that form the series, what are they called? What are the respective names of the five denominations in arithmetical progression; When the first term, common difference, and number of terms given, how is the last term found? When is the series ascending? When descending?

When the extremes and number of terms are given, to find the common difference.

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

1. If the extremes be 5 and 605, and the number of terms 151; what is the common difference?

Ans. 4. 2. A man had 8 sons, whose ages differed alike; the youngest was 10 years old, and the eldest 45; what was the common difference of their ages? Ans. 5 years.

Question.-How is the common difference found when the extremes and terms are given?

When the extremes and number of terms are given, to find the sum of all the terms. Multiply half the sum of the extreme, by the number of terms, and the quotient will be the answer, (or &c.) 1. If the extremes be 5 and 605, and the number of terms 151, what is the sum of the series?

Ans. 46055.

2. How many times does a common clock strike in 12 hours?

Ans. 78. 3. How many strokes do the clocks of Venice strike, which go to 24 o'clock, in the compass of one day? Ans. 300.

Questions. Having the first and last terms, and the number of terms given, how is the sum of all the terms found? How can the rule be inverted so as to admit of three different methods of solutions, to each question?

The last term, the number of terms, and the common difference being given, to find the first term. Multiply the number of terms, less 1, by the common difference, which product subtract from the last term, and the remainder is the first term.

1. If the common difference be 2, the number of terms 19, and the last term 39, what is the first term? Ans. 3. 2. A man takes out of his pocket, at 8 several times, so many several numbers of dollars, every one exceeding the former by 6; the last was 54, what was the first? Ans. $12.

Question. Having the last term, the number of terms, and the common difference given, how is the first term found?

Given the extremes (the first and last terms,) and the common difference, to find the number of terms. Divide the difference of the extremes, by the common difference, and the quotient increased by I will be the number of terms required.

1. The extremes are 2 and 53, and the common difference 3; what is the number of terms? Ans. 18 terms.

2. A man going a journey, travelled the first day 7 miles, the last day 51 miles, and each day increased his journey by 4 miles; how many days did he travel, and how far?

Ans. 12 days, and 348 miles. Questions-Having the extremes and common difference given, how is the number of terms found? Having 7 and 51 extremes given, and having obtained (12) the extremes, how is the sum of the terms 348 found?

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