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2. “What part of 5 guineas is 13s. 4d. ?”

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or

13 4 = 40 fourpenny pieces 5 guineas = 315 .. 13 4 is of 5 guineas, 40 times the 315th part ;

40 8 315

63 3. “Show that dividing the numerator of a fraction by any number gives the same result as multiplying the denominator by the same number.” Let it be required to divide the fraction of by 3. 6:3 2

6

6 11 11

;

11 x 3 33 6

2 But

33 The same may be shown by dividing a line representing unity into 11 equal parts, and each of these again into 3 others; when the third part of 6 of the elevenths would, in length, be equal to 6 of the thirty-thirds. (See paper on Algebra, Sec. ii. Ques. 2.)

=

11:

SECTION IV. 1. “Multiply .0017 by 450, and give the reason for the correct placing of the decimal point in the product.”

·0017

450
850
68
765

=

In performing this operation, we are finding the product of 1oooo by 450 : for .0017 Ισσσσ.

Hence, after finding, as above, the product of 17 by 450, we have to divide by 10000; this is done by “pointing

66

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off” four figures. Thus the reason is seen for pointing off as many figures in the quotient as there are decimal places in the multiplier and the multiplicand together. 2. Reduce 4s. 7d. to the decimal of a pound.”

7 7 pence

of a pound. 12 x 20

4 4 shillings

20 7

4 7 48 55
But
+

+
12 x 20 20 240 240

240
5500000

229163

229163; but $, when 240 x 100000 100000 reduced to a decimal, becomes .6°; therefore the 4s. 7d., reduced to the decimal of a pound is . 22916'.

The form generally employed and deduced from the above is as follows:

12) 7d.
204.583

£.22916

3. “Divide 570 by .005, and give the reason for the correct placing of the decimal point in the quotient.”

·005)570

114000

·005 = tood; therefore when we have divided by 5, as seen above, we have divided by a number 1000 times greater than the true divisor: therefore the result must be multiplied by 1000, or “place as many figures after the one obtained, by dividing the units, as there are decimal places in the divisor.

4. “ Extract the cube root of .04 to 3 places of decimals. Give the reason for the operation by which you

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.

divide the number into periods at the commencement of the process."

The cube of 1, 10, 100, is formed of the unit followed by three times as many cyphers as there are in the number taken. A number, therefore, between 10 and 100 has its cube between 1000 and 1000000; consequently its cube is formed of 4, 5, or 6 figures, according to the proximity of its value to 10, or 100. Generally, the cube of any number will contain three times its number of figures, minus one or two, according to the value of the first figure in the root. Hence the reason for dividing a number, whose cube root is to be extracted into periods of three figures each, commencing from the units' place.

Now for the cube root of .04; we must add seven cyphers in order to obtain three decimal places in the root; for the periods commence “ from the units' place.”

•040,000,000(•341

27
(c) 32 X 3 - 27 13000
(d) 3 X 4 X 3
(e)

42 16

3076 X 4 = 12304
376

..696000
342 X 3 = 3468
34 X 1 X 3 =

102
12

1
347821 x1= 347821

348179

9

39 =

= 46

The method here employed, is published in “ Tate's Principles of Arithmetic," a copy of which every teacher

The process for obtaining the trial divisor is original and elegant. It will be comprehended

should possess.

from the following explanation, in which it is assumed that the mode of deriving the rule for the extraction of the cube root is understood. By this method, squaring the root already found, and then multiplying by 3 for a new divisor, are superseded by the simple process of addition. The operation commences after the second figure in the root is obtained ; when the two figures found form that part of the root in the formula, (a + b). = a + 3 a b + 3 a b2 + bs, represented by a. Now in the preceding example (leaving out of consideration that the figures found are decimals, as the reasoning will not be affected thereby), the a is 34, which may be resolved into 30 + 4; squaring these by the binomial and multiplying by 3, we have, 3 (302 + 2 X 30 X 4 + 42) 2700 + 2 X 360 + 3 X 16, the first of which numbers and the greater factor of the others are all previously obtained (marked (c), (d), and (e), respectively); for 27 stands two places to the left, and 36 one. Hence the sum of once (c), twice (d), and thrice (e) will give three times the square of 34. The addition is conveniently performed, as in the example, by placing the sum of (d) and (e) under the obtained sum of the three products, and then adding the three numbers which are directly under each other. The advantage of this artifice becomes immensely great, as the number of figures in the root increases.

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11

MENSURATION.

SECTION I. 1. Prove the rule of cross multiplication. 2. Prove a rule for determining the number of standard

rods of brick-work in a wall. 3. Prove a rule for determining the area of a trapezoid.

SECTION II. 1. How many cubical feet of timber are there in the

flooring of a room in. thick, and 17 ft. 6 in. in

length by 15 ft. 3 in, in breadth ? 2. In a wall 10 ft. high, 15 ft. long, and 27 bricks

thick, there is an arched door-way 4 ft. wide and 6 ft. high to the springing of the arch, which is semi-circular; how many standard rods of brick

work are there in the wall ? 3. What is the weight of a circular iron ring whose

inner diameter is 18 in., and whose section is a circle 2 in. in diameter, the weight of a cubic foot of the iron being 450 lbs. ?

SECTION III. 1. After measuring a piece of cloth, to contain 90 yards,

I find that the yard measure that I have used is too short by 1-30th part; what is the true measure

of the cloth ? 2. How many square inches of tin plate are required

to make an open cylindrical vessel to contain a gallon whose height is equal to one-half of its diameter ?

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