SIMPLE DIVISION. Division is the method of finding how often one number is contained in another. Case I. When the Divisor does not exceed 12. Divide 252 by 6. RULE WITH EXAMPLE.-Put the numbers down according to the annexed example. 42 Find how often 6)252 the figure by which you are to divide, viz. 6 is contained in the first, or first and second figures; thus, 6 in 2, there are none, then 6 in 25; there are 4 sixes in 25 and 1 over. Put down the 4 under the 5. Suppose the 1 placed before the 2, which would make it 12. Say 6 in 12. There are 2 sixes in 12. Put the 2 under the 2. The number 6 is called the Divisor; 252 the Dividend; and 42 the Quotient. Case II.-When the Divisor is a Composite number. Divide 6789 by 28. RULE WITH EXAMPLE.-The two factors that produce 28, are 4 and 7; divide them by 4 and by 7 as in the example. The quotient found is 242, but with two remainders, viz., 3 and 1. To obtain the complete remainder, 12 7)1697 remains 1 242 remains 3 multiply the first divisor, viz. 4, by the last remainder, viz. 3, and to the product add the first remainder, viz. 1;-thus, 4X3+1 13 the true remainder. Case III. When the Divisor contains several figures. Divide 431769 by 528. 528)4317,69(817 quotient 4224 936 4089 393 remainder. RULE WITH EXAMPLE.*-Put down the sum in this form. Consider whether the divisor, viz. 528, is contained in the first three figures of the dividend, viz. 431; you see at once that it is not; mark off then four figures, viz. 4317. You are now to find how often 528 is contained in 4317; for this purpose find how often the first figure of the divisor, viz. 5, is contained in the first two figures of the dividend, viz. 43. It is contained 8 times; put the 8 on the opposite side of the dividend from the divisor. Multiply 528 by 8, and put the product under the 4317; subtract, and there remains 93; bring to this the next figure of the dividend, viz. 6. You are now to find how often the divisor, 528, is contained in your new dividend, 936; find, as you did before, how often the first figure of the divisor, 5, is contained in the first figure of the dividend, 9. It is contained once; put the one beside the 8; multiply 528 by 1, and place the product under the 936; subtract and you obtain 408; bring to this the next figure of the dividend, 9. Find, as before, how often 528 is contained in 4089. Because 5 is contained 8 times in 40, you will be inclined to try 8. Do it and you will find that you obtain the product 4224, but this is greater than the 4089 from which you have to subtract it; when this is the case you must try a smaller figure, in this case take 7. This is rather a difficult Rule to understand, and I think your Teacher could explain it to you, by means of a black board and a bit of chalk much better than I can hope to do by any written explanation; yet, if you pay attention, I shall do my best to make you understand it. 92. Divide six millions seven hundred and ninety-four thousand, by four hundred and eighty thousand six hundred and nine. 93. Divide £79648 among 274 persons. 94. What is the ninth of £6037 ? 95. A ship sailed in four weeks 1262 miles; how much is that per day? 96. If a vessel contains 648 gallons of water; how long will it take to discharge it all, at the rate of 18 gallons an hour? 97. The population of Ireland is about eight millions, and there are about 30,000 square miles of surface; how many persons to each mile? 98. The earth is about 93 millions of miles distant from the sun; how many days would a horse take in reaching the sun, supposing he went at the rate of 45 miles per day? 99. The rays of light come from the sun to the earth in 8 minutes, or 495 seconds; at what rate does light move per second, the distance from the sun to the earth being 95173000 miles? 100. The circumference of the earth is about 25000 miles; how long would a man take to walk round it at the rate of 27 miles per day? COMPOUND ADDITION. Add together the following sums of money; £64 12s. 44d., £86 15s. 64d., £14 16s. 53d., £34 17s. 93d, £ S. 6 201 221 RULE WITH EXAMPLE.-Place pounds under pounds, shillings under shillings, &c. and draw a line under the row of figures; first add the farthings together; thus, 3 farthings and 3 farthings make 6 farthings, 6 and 2 make 8, and 1 makes 9; but are equal to 24. Put the under the farthings, and add the 2 pence to the pence column. Then 2 pence and 9 pence makell, and 5 make 16, and 6 make 22, and 4 make 26; but 26 pence are equal to 2 shillings and 2 pence. Put the 2 pence under the pence column and add the 2 shillings to the shilling column; then 2 shillings and 7 shillings make 9, and 6 make 15, and 5 make 20, and 2 make 22; now come down the column adding the tens, 22 and 10 (of the 12) make 32, and 10 (of the 15) make 42, and 10 make 52, and 10 make 62. 62 shillings are equal to 3 pounds 2 shillings; set the two shillings under the shilling column, and carry the three pounds to the pound column. Proceed as in Simple Addition. The principles on which the operation depend are the same as for Simple Addition; only that the columns here do not differ from each other in a tenfold degree. |