Ex. 98. (Oral.) 1. How many inches in 1 yd.? in yd.? in ‡ yd. ? 2. How many yards in 180 in.? in 48 in.? in 45 in. ? 3. How many yards in 3 rds.? in 4 rds.? in 5 rds.? 4. How many feet in 2 yds.? in 2 rds.? in 2 rds. 2 yds.? 5. How many rods in 33 ft.? How many yards in 33 ft.? 6. In mi. how many rods? yards? feet? 7. How many rods in 0.4 of a mile? in 0.3? in 0.7? 8. What part of a mile are 160 rds.? 80 rds.? 40 rds.? 9. What part of a foot are 4 in.? 3 in.? 6 in.? 8 in. ? 10. What part of a yard are 2 ft.? 1 ft. 6 in.? 2 ft. 6 in.? REDUCTION DESCENDING. 146. Change 10 mi. 40 rds. to feet. 10 mi. 40 rds. 320 3200 40 3240 16 1620 19440 3240 53460 10 x 320 rds. = 3200 rds., to which the 40 rds. are added. Again, 3240 × 16 ft. = 53,460 ft. The multiplicand and multiplier are interchanged in the operation. Reduce to feet: 1. 3 mi. 5 yds. 2 ft. Ex. 99. Reduce to inches: 4. 18 mi. 252 rds. 2 yds. 147. Change 53,463 ft. to a compound quantity. There are 16 ft., or 33 half-feet, in a rod; so the 53,463 ft. are changed to half-feet, and the half-feet to rods, by dividing by 33. The remainder is 6 half-feet = 3 ft. 3240 rds. are changed to miles by dividing by 320, the number of rods in a mile. The remainder is 40 rds. Theyd. of the 4 and the result, 1 ft. 6 31415 ft. . 5 in. 10471 yds. . 2 ft. 2 20942 half-yards. 1903 rds. 9 half-yards 4 yds. 5 mi... 303 rds. yds. should be reduced to lower denominations, in., added to the 2 ft. 5 in. Thus, mi. rds. yds. ft. in. 5 303 4 2 5 1 6 5 303 5 0 11 5 mi. 303 rds. 5 yds. 0 ft. 11 in. Ans. Write the numbers so that units of the same denomination shall be in the same column. The sum of the inches is 16. Divide the 16 in. by 12 (12 in. 1 ft.). The result is 1 ft. 4 in. Write 4 under the column of inches, and carry 1 to the column of feet. The sum of the feet, including the 1 ft. from the by 3 (3 ft. 1 yd.). The result is 2 yds. 0 ft. column of feet, and carry 2 to the yards. = = 16 in., is 6. Divide Write O under the The sum of the yards, including the 2 yds. from the 6 ft., is 8. Divide by 5 (5 yds. 1 rd.). The result is 1 rd. 24 yds. Write 21 under the column of yards, and carry 1 to the rods. The sum of the rods, including the 1 rd. from the 8 yds., is 351. Divide by 320 (320 rds. = 1 mi.). The result is 1 mi. 31 rds. Write 31 under the column of rods, and carry 1 to the miles. The sum of the miles, including the 1 mi. from the 351 rds., is 28. The yd. is changed to 1 ft. 6 in., and added to 0 ft. 4 in. 149. Take 4 mi. 110 rds. 5 yds. 2 ft. from 6 mi. 25 rds. 1 mi. 234 rds. 4 yds. 2 ft. 6 in. Ans. Write the numbers so that units of the same denomination shall be in the same column. Since 5 yds. are more than 4 yds., 1 rd. is reduced to yards, and the result added to 4 yds., making 9 yds. Then 9 yds. - 5 yds. -4 yds., which is written under the column of yards. = Since the minuend has been increased by 1 rd., 1 rd. must be added to the 110 rds. of the subtrahend. (See & 48, page 37.) Since 111 rds. are more than 25 rds., 1 mi. is reduced to rods, and the result added to 25 rds., making 345 rds. Then 345 rds. - 111 rds. 234 rds. = The 234 is written under the column of rods. The 4 mi. are increased by 1 mi., and the result taken from 6 mi. Theyd. is changed to 1 ft. 6 in. 150. Multiply 37 yds. 2 ft. 11 in. by 4. NOTE. When the multiplier is the product of two factors, multiply by one of the factors, and the resulting product by the other. 151. Divide 121 yds. 2 ft. by 73. yds. ft. 73) 121 2(1 yd. 2 ft. 73 48 3 144 2 The remainder from dividing 121 yds. by 73 is 48 yds., which are reduced to feet by multiplying by 3 (3 ft. 1 yd.). The result with the 2 ft. added is 146 feet. 146 146 73. There is no remainder from dividing 146 ft. by 1.yd. 2 ft. Ans. |