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II. TO DETERMINE THE QUALITY OF AN ARTICLE, OF WHICH A GIVEN

QUANTITY IS EXCHANGED FOR A GIVEN QUANTITY OF ANOTHER.

Rule. Find the value of the goods received, and divide it by the number of units in the quantity exchanged; the result will be the price of an unit of the quantity.

EXAMPLES. 1. Exchanged 84 gallons of brandy, at 24s. per gallon, for 504 yards of silk; what was the price of the silk per yard ? Value of brandy = 84 X 24 s. = 2016 s.

2016 .. Price per yard of silk =

S. = 4s. 504

2. Received in barter a cow valued at 6 guineas, 20 sheep valued at 2 guineas, £100 in cash, and 150 quarters of barley, for 110 quarters of wheat at 69s. per quarter; what was the price of the barley per quarter ?

Value of wheat = 110 X 698. = £379: 10:0

Value of cow
Value of sheep
Cash

11 1111

6: 6:0 42 : 0:0 100: 0:0

148 : 6:0

Value of barley =

£231 : 4:0

231} 1156 2312 ... Price per quarters -£= -£= s.

150 750 75

23
=30s. 9 d.

25

$

III. THE NETT AND BARTERING PRICES OF ONE ARTICLE BEING GIVEN,
TO DETERMINE THE BARTERING PRICE OF ANOTHER, THE NETT PRICE
BEING KNOWN, OR vice versa.

Rule. State: as the nett price of one is to its bartering price, so is the nett price of the other to its bartering price. Or as the bartering price of one is to its nett price, so is the bartering price of the other to its nett price.

EXAMPLE The nett price of an article is 5s. but in barter is raised to 5s. 4d. What should be the bartering price of an article, valued at 3s. Od. ready money ? 5's.

34 s.

Ans.
16 15
Ans. =- - X - 58. = 48.

3

As

:

5} s.

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Def. Alligation is the Rule by which is determined the rate of value of a mixture, from the rates of the ingredients of which it is composed.

Rule. Multiply the quantities of each ingredient, expressed in terms of the same unit, by the rates also expressed in the same unit; divide the sum of the products by the sum of the quantities; the quotient will be the price of an unit of quantity in terms of an unit of price.

EXAMPLE. A wine merchant mixes 30 gallons of wine at 158.a gallon with 40 gallons at 16s.; and with 24 gallons at 17s. 6d.; what will be the price of the mixture?

galls. 30 x 15 = 450 = cost of 30

94)1510(164 40 X 16 = 640 = cost of 40

94 24 X 3 = 420 = cost of 24

570 Whole cost=1510 = cost of 94

564

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Def. 1. Lines are measured by considering how often they contain a certain length, called the lineal unit.

Def. 2. Surfaces are measured by considering how often they contain a certain surface, called the superficial or square unit, which is a square, whose side is equal to the lineal unit.

Def. 3. Solids are measured by considering how often they contain a certain solid, called the solid, or cubic unit, which is a cube, whose faces are equal to the square unit.

Def. 4. The dimensions of a magnitude are its length, breadth, and thickness.-A line is of one, a surface of two, a solid of three, dimensions.

Obs. The Rules of Mensuration are Rules which enable us to find, by lineal measurements of the dimensions of a magnitude, the extent of its surface, and its capacity or solid content. These Rules are too numerous for insertion here. The Rules for finding the area of a rectangle, and the content of a rectangular parallelopiped are as follows.

I. TO FIND THE AREA OF A RECTANGLE.

Rule. Multiply the units in the length by the units in the breadth, the result is the number of square units in the area.

This operation may be conducted either (1) by expressing the length and breadth in terms of the same unit, fractional or integral; or (2) by the method commonly called "Cross Multiplication;" or (3) by changing the number of feet and inches in the length and breadth into a number of feet and parts of feet expressed in the dnodenary scale.

EXAMPLE Find the area of a rectangular room 14 ft. 6 in. wide, by 20 ft. 9 in. Jong.

1 29 By first method width = 144ft. =- ft.

2 2

3 83 length = 20— ft.-- ft.

4 4
29 83

2407
.. area=-- Х
2 4

8
= 300 sq. ft. 126 sq. in.

ft. pr. sec. By second method

20 90 14 6 0

sq. ft. =

sq. ft.

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Observations on the 2nd method. In lineal measure inches are also called prines, and twelfths of inches seconds ; in square measure, a rectangle 12 in. by 1 in, i.e. the twelfth of a square foot is called a superficial prime; the twelfth of a prime, i.e. a square inch is called a superficial second; the twelfth of a second, a third; and so on. Hence multiplying the number of feet in a length by the number of inches in the breadth we find the number of primes in the area; or feet multiplied by inches give superficial primes. So inches multiplied by inches give square inches or superficial seconds; inches multiplied by seconds give thirds. Observing these facts, the Rule for the performance of the second method is as follows:

Write the multiplier under the multiplicand, placing feet under feet, &c. Beginning with the highest denomination multiply by each term of the multiplier, taking care to write quantities of the same denomination under each other. Add the products. The result will be in sq. ft., primes, seconds, &c.

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11.

TO FIND THE SOLID CONTENT OF A RECTANGULAR PARALLI LOPIPED.

Rule. Multiply together the units in the length, breadth, and thickness ; the result is the number of cubic units.

This operation (as the last) may be conducted by each of the three methods, already mentioned.

The twelfth part of a solid foot is called a solid or cubic prime; the twelfth part of a prime, a second ; and so on. Since the twelfth part of a solid foot is a solid whose base is a square foot, and height I inch; therefore a number of square feet multiplied into a number of linear inches give a number of cubic primes. Also since the twelfth part of a cubic prime is a solid whose base is a superficial prime and height an inch, or whose base is a square inch or superficial second, and height a foot; therefore a number of superficial primes multiplied into a number of linear inches, or a number of superficial seconds multiplied into a number of linear feet, give a number of cubic seconds. Also since the twelfth part of a cubic second is a cubic inch, therefore a number of superficial seconds multiplied into a number of linear inches give a number of cubic thirds. Observing the above facts, the working of the 2nd method will be understood, the operation being performed as in the previous case.

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EXAMPLE.
How many solid feet and inches are there in a block of stone, whose
dimensions are 6 ft. 7 in., 5 ft. 8 in., 3 ft. 4 in. ?
By the first method.

7 2 1
Solid content = 6- X 54 X 3— cub. ft.

12 3 3
79 x 17 x 10

6715 cub. ft. =

cub. ft.
12 X 3 X 3

54
19
= 124— cub. ft. = 124 cub. ft. 608 cub. ins.

54 By the second method.

ft. pr. sec.

7 0 8 0

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By the third method. The dimensions expressed in the duodenary scale are 6.7 ft., 5.8 ft., 3.4 ft.

ft. 6.7 5.8

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