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3. Divide 570 by .005, and give the reason for the

correct placing of the decimal point in the quotient. 4. Extract the cube root of .04 to 3 places of decimals.

Give the reason for the operation by which you divide the number into periods at the commencement of the process.

SECTION I.

1. “Explain each step in the process of subtracting 750 from 850."

805 = 700 + 100 + 5
750 = 700 + 50 + 0
55 =

50 + 5 The result is verified by the resolution of the respective numbers, as seen above; but to exhibit and prove the usual mode of operation, it must be premised that “the difference of two numbers is not affected by adding any equal number to each, and then performing the subtraction ;” thus, 6 - 4 = 16 - 14.

Now in order to make each figure in the subtrahend less than the corresponding one in the minuend, we will add a number of tens to the latter to accomplish the object; but the same must be added to the lower line; and as the numbers increase in a tenfold ratio, let us add ten tens, or one hundred to each. The figures will then stand thus,

h. t.
8 + 10 + 5

5 + 0

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8 +

Hence the rule for "borrowing ten" and "carrying

one."

2. “Explain each step in the process in multiplying 1087 by 5050.”

This question requires us to find the resulting number after the continued addition of 1087 for 5050 times.

1087

5050
54350 = 50 times the multiplicand
5435 = 5000
5489350 = 5050

As a cypher occupies the units' place in the multiplier, it is brought down to hold a similar position in the quotient. Then the operation of multiplying by 5 is commenced ; and this is done in the same manner as though 5 held the units' place. Thus 5 times 7 units are 35 units = 3 tens' + 5 units; the units may be placed under the 5, and the 3 tens carried to the result of the tens. The whole operation will be clearly seen as follows: 5 times 7 units = 35 units = 3 tens + 5 units.

8 tens = 40 tens = 4 hundreds.

1 thousand – 5 thousands. Collecting these results, we obtain 5 thousand + 4 hundred + 3 tens + 5 units, or 5435. Now it was not 5, but 50, by which we were to multiply; consequently, the result is 10 times too small. To multiply by 10, the figures have only to be removed each one place to the left. Hence, by “bringing down the cypher,” we obtain 50 times the multiplicand. The same process could be applied to the other significant figure of the multiplier ; after the result is obtained, multiply by the third power of 10, or, which is the same thing, remove the figures three places towards the left. Add the results, and we obtain the required product.

Note.-It can easily be perceived that whether we multiply 1087 by 5050, or 5050 by 1087, the product would be the same. Now were 5050 some other number, it would, very probably, be more convenient to consider it the multiplicand; thus, were it 5435, the product would be more concisely obtained by considering 5435 the multiplicand and 1087 the multiplier. 3. “ Explain each

the division of £70 10s. 11d. by 820; and express clearly what is the value of the remainder.”

d. s. d.
820)70 10 11(1 81 - 484.

20

step in

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1410
820

590

12 7091 6560 531

4 2124 1640

484

To take the 820th part of £70 10s. 11d., place the number and money in the form of divisor and dividend, as above.

Now, as 70 divided by 820, will not produce a whole number, bring the pounds to their equivalent in shillings, adding in the 10s. The number of shillings will then contain the divisor once with a remainder of 590 shillings. Bring these to pence by multiplying by 12; the quotient is contained in the pence 8 times, with a remainder of 531 pence = 2124 farthings, in which 820 are contained twice, i.e.—the farthings divided into 820 parts will allow of 2 farthings for one such part, and a remainder of 484, which is the number of farthings still undivided, being insufficient to produce one in each of 820 parts : hence there are 10s. 1d. left undistributed over the 820 portions.

SECTION II.

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1. “Find, by practice, the value of 5 cwt. 1 qr. 19 lbs., at £3 15s. per cwt.”

£ d. Price of 1 cwt.

3 15 0

3

18 15 0 1 qr. cwt. 0 18 9 14 lbs. 1 qr.

0 9 41
4 7 를

0 2 89
1 14 lbs. 0 0 873
Total..

£20

6 51

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2 “If 11 articles cost 15s., what would 17 cost? Explain each step of the process of working the sum.'

If the price of 11 articles be 15s., the price of 17, at the same rate, will be the same multiple of the price of 11 as 17 is of 11. Thus, 11 : 17 :: 15 : to the required price, 17

15 x 17 or, 1 : :: 15 :

= £1 3s. 20. 11

11

1

3. “If 5 men receive £18 15s. wages for 12 months, what will be the wages of 16 men for 20 months?" If 5 men for 12 months receive £18 15s.

181 5 men for 1 month would receive £

12 And 1 man

75
4 X 12 X 5

75 x 16 And 16 men

4 X 12 X 5

£ .. 16 men for 20 months would / 75 X 16 X 20

=100 receive

3 4 X 12 X 5 By compound proportion we should have, m. m. 5 : 16

75 X 16 X 20 :: 18 15 :

the same as mo. mo.

4 X 12 X 5 12 : 20 the previous result to be reduced; 75 x 16 x 20

= £100 4 X 12 X 5

£ s.

SECTION III.

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1. (1)

“ Subtract of 1 from 1}, and (2) find the of value i of 16s. 8.1d.” 11 2 1 11 2

11 1 (1.) of

Х 18 3 4 18 3

18 6 11 3 8 18 18

s. d. s. d.

s. d.

16 87 X 3 (2.) 16 87 X X

= 4 61

8 11

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