1

35. A farmer raised 8288 bushels of corn, averaging 56 bushels to the acre. How many acres did he plant? 148 acres.

36. If a donation of $262275 is divided equally among 269 school libraries, how much will each receive? $975.

37. The earth, at the equator, is about 24899 miles in circumference, and turns on its axis once in 24 hours. How many miles an hour does it turn? 1037 mi.

38. A railroad 238 miles long, cost $3731840. What was the cost per mile? $15680.

39. A fort is 27048 feet distant from the city; the flash of a cannon was seen 24 seconds before the sound was heard. How many feet a second did the sound travel? 1127 feet.

40. Light travels at the rate of 11520000 miles a minute. How many minutes does it require for the light of the sun to reach the earth, the sun being 92160000 miles distant? 8 minutes.

41. Subtract 86247 from 94231 and divide the re

mainder by 16.

499.

42. Divide the sum of 46712 and 6848 by 104.

515.

43. Divide the product of 497 × 583 by 71.

4081.

and divide the sum by 87.

29.

44. To the difference between 2832 and 987 add 678,

45. Multiply the difference between 4896 and 2384 by

49, and divide the product by 112.

1099.

46. Multiply the sum of 228 and 786 by 95, and divide the product by 114.

845.

47. Multiply the sum of 478 and 296 by their difference, and divide the product by 387.

364.

48. A horse dealer received $7560 for horses; he sold a part of them for $3885. If he sold the rest for $175 apiece, how many horses did he sell in the second lot?

21 horses.

49. A farmer expended at one time $7350 for land, and at another, $4655, paying $49 an acre each time. How many acres did he buy in both purchases? 245 acres.

50. A horse dealer bought 58 horses at $77 each, and sold them for $5742. How much did he gain on each horse?

$22.

51. A man bought 240 acres of land, at $26 an acre, giving in payment a house valued at $2820, and horses valued at $180 apiece. How many horses did he give?

19 horses.

52. A speculator bought 25 acres of land for $10625, and after dividing it into 125 village lots, sold each lot for $250. How much did he gain on the whole? On each acre? On each lot? $20625. $825. $165.

CONTRACTIONS IN DIVISION

43. When the divisor can be separated into factors:

1. A man paid $255 for 15 acres of land. How much was that per acre ?

SOLUTION. - 15 acres are 3 times 5 acres; dividing $255 by 3 gives $85, the value of 5 acres; dividing $85 by 5 gives $17, the value of 1

acre.

The solution shows that instead

OPERATION

Dollars. 3 255 the value of 15 acres. 5 85 the value of 5 acres.

=

=

of dividing by the number 15, whose factors are 3 and 5, we may first divide by one factor, then divide the quotient thus obtained by the other factor.

2. Find the quotient of 37, divided by 14.

SOLUTION.-Dividing by 2, the quotient is 18 and 1 unit remaining. That is, 37 = 18 twos and 1 unit remaining. Dividing by 7, the quotient is 2, with a remainder of 4 twos; the whole remainder then is 4 twos plus 1, or 9.

2137

OPERATION

7 18 and 1 over.

2 and 4 twos left.

Rule. 1. Divide the dividend by one of the factors of the divisor; then divide the quotient thus obtained by the other factor:

2. Multiply the last remainder by the first divisor; to the product add the first remainder; the amount will be the true remainder.

NOTE. -When the divisor can be resolved into more than two factors, you may divide by them successively. The true remainder will be found by multiplying each remainder by all the preceding divisors, except that which produced it. To their sum add the remainder from the first divisor.

44. To divide by 1 with ciphers annexed; as, 10, 100, 1000, etc.:

To multiply 6 by 10, annex one cipher; thus, 60. On the principle that division is the reverse of multiplication, to divide 60 by 10, cut off a cipher.

NOTE. Had the dividend been 65, the 5 might have been separated in the same manner as the cipher; 6 being the quotient, 5 the remainder. The same will apply when the divisor is 100, 1000, etc.

Rule. Cut off as many figures from the right of the dividend as there are ciphers in the divisor; the figures cut off will be the remainder, the other figures the quotient.

45. To divide when there are ciphers on the right of the divisor, or on the right of the divisor and dividend:

100, and add 39 to the product (§ 43, Rule); this is the same as annexing the figures cut off to the last remainder.

Rule.-1. Cut off the ciphers at the right of the divisor, and as many figures from the right of the dividend.

2. Divide the remaining figures in the dividend by the remaining figures in the divisor.

3. Annex the figures cut off to the remainder, which gives the true remainder.