254. Written Problems. 1. Find the number of square feet in a walk 50 ft. long and 4 ft. wide. 2. How much will it cost at $.12 a square foot to lay a cement walk 50 ft. long and 5 ft. wide ? 3. Find the area of a walk 60 ft. long and 5 ft. wide. 4. How many square inches are there in 3 sq. ft.? In 3 sq. ft.? 5. How many square inches are there in 3 sq. ft. and 32 sq. in.? 6. Find the number of square feet there are on the floor of your schoolroom. 7. Find the number of square rods there are in your school yard. 8. Find the number of square yards of blackboard there are in your schoolroom. 9. How much did the blackboard in your schoolroom cost at 18¢ a square foot? 255. In your drawings, let 1 in. represent 2 ft. 1. A square that will contain 16 sq. ft. ft. BILLS AND ACCOUNTS 256. 1. Study the bill given in Sec. 78, p. 82. 2. A bill must always show the date of the transaction. What is the date of the transaction referred to in the bill in Sec. 78? 3. The debtor is the party who buys the goods. Who is the debtor in the bill referred to above? 4. The creditor is the party who sells the goods. Who is the creditor in the bill referred to above? 5. An item is a separate debit or credit made in a bill. How How many items are there in the bill referred to above? 6. How many items are there in the bill in Sec. 257? 7. Name the debtor and the creditor in the bill in Sec. 257. LONG DIVISION 258. When all the steps in division are written, the process is called long division. Divide 173 by 3. MODEL: 57/ 3)173 15 23 21 2 3 is contained in 17 five times. Write 5 in the quotient above the 7, as in short division. Multiply 3 by 5 and write the product under 17, and subtract. The remainder is 2. Bring down the 3 of the dividend and find how many times 3 is contained in 23. This is 7 times. Multiply 3 by 7 and write the product under 23, and subtract. Treat the remainder as in short division. CASE ONE 259. When the second figure* of the divisor is the same as or less than the first figure, as in 44, 63, 978, 658, etc.. 1. Divide 2292 by 43. First, find how many places at the left of 2292 it will take to contain 43 at least one time. It will take three places. The first figure of the quotient will be in tens' place. MODEL: 43)2292 215 142 129 13 Step 1. 4 is contained in 22 five times with 2 remainder. This remainder, with the next figure of the dividend, is the dividend of 3, the second figure of the divisor. The dividend of * In 63 regard 6 as the first figure and 3 as the second figure. 3 is 29. Is 3 contained in 29 as many as 5 times? If it is, 5 is the trial quotient figure. 3 is contained in 29. as many as 5 times. Write 5 in the quotient above 9. Step 2. Multiply 43 by 5, and write the product under 229. Step 3. Subtract 215 from 229. The remainder is 14. As this remainder is less than the divisor, the trial quotient figure is the true quotient figure. Step 4. Bring down the next figure of the dividend. The new dividend is 142. Repeat Step 1. 4 is contained in 14 three times with 2 remainder. As 3 is contained in 22 as many as 3 times, 3 is the trial quotient figure. Write 3 in the quotient. Repeat Step 2. Multiply 43 by 3 and write the product under 142. Repeat Step 3. is 13. Subtract 129 from 142. The remainder 7877 by 96; 7877 by 82. 2. Divide 1806 by 43; 3. Divide 2806 by 65; 5927 by 97; 5927 by 51. 4. Divide 16,108 by 72; 18,191 by 85; 45,960 by 71. 5. Divide 2115 by 43. First, find how many places at the left of 2115 it will take to contain 43 at least one time. MODEL: 49,8 43)2115 172 395 387 Step 1. 4 is contained in 21 five times with 1 remainder. As 3 is not contained in 11 as many as 5 times, write 4 as the trial quotient figure, and continue as in the preceding exercise. 260. When the second figure of the divisor is the same as or less than the first figure, the trial quotient figure may be found as follows: 1. If the second figure of the divisor is contained in its dividend as many times as the first figure is contained in its dividend, use this quotient figure as the trial quotient figure. 2. If the second figure of the divisor is not contained in its dividend as many times as the first figure is contained in its dividend, use as a trial quotient figure one less than the quotient figure obtained by dividing by the first figure of the divisor. This will be found to be the true quotient figure. 3. If the first figure of the divisor is contained in its dividend 10 times, use 9 as the trial quotient figure.† * The number formed by annexing the next figure of the dividend to the remainder left after dividing by the first figure of the divisor is the dividend of the second figure of the divisor. †The pupils should become perfectly familiar with these facts through illustration. |