Art. 22. To multiply a number by 15: Multiply the number by 10, and to the product add of the product.

(v) Find the sum of the twelve products.

Art. 23. To multiply a number by 99: Take 100 times the number, from which subtract the number itself. How may

a number be multiplied by 98?

(w) Find the sum of the twelve products..

Art. 24. To multiply a number by 75: Multiply of the number by 100. How may a number be multiplied by 663 ?

(x) Find the sum of the twelve products.

Art. 25. To divide a number by 25; by 331; by 121; by 163; by 20; by 50. (See pp. 72, 73, and 74, of this book.) 850 ÷ 25 = 8 times 4+ 2 = 34 93313319 times 3+1 = 28 6371÷121 = 6 times 8+3=51

750167 times 6+3=45

960209 times 5+ 3 = 48

450 50 = 4 times 2+1= 9

(y) Find the sum of the six quotients.

NOTE. - Without a pencil, tell the integral quotient and the remainder resulting from the incomplete division of 1584 by 25.

Art. 26. To divide a number by 125; by 250: Observe that 125 is contained in each thousand of a number 8 times; that 250 is contained in each thousand of a number 4 times. 71251257 times 8+1=57

85002508 times 4+2=34

NOTE.

(z) Find the sum of the twelve quotients.

- Without a pencil, tell the integral quotient and the remainder resulting from the incomplete division of 15450 by 250.

Art. 27. When the same factor occurs in a dividend and in its divisor, it may be omitted from both without changing their ratio. Hence all the factors that are common to a dividend and its divisor may be stricken out (cancelled) and the quotient (ratio) be unchanged.

NOTE.—In this discussion and in the solution of problems in this connection, all the numbers are regarded as abstract.

70

70

II. Divide 420 by 35.

Operation No. 2.

12

420 2 × 2 × 3 × 5 × 7_12

35

5×7

Observe that if all the factors of one of the numbers are cancelled the number becomes 1 and not 0. The factor 5 is 5 times 1; the facHence in the above problem there really remain in the divisor, after the cancellation, the factors 1 and 1 = 1 x 1 = 1.

tor 7, 7 times 1.

III. Divide 48 x 8 x 4 = 1536 by 8 x 4 x 4 = 128.

Operation No. 1.

128)1536(12

128

256 256

Observe that it is not necessary to obtain

the prime factors of a dividend and its divisor to employ cancellation in finding the quotient. In the above the composite factor 8 is stricken out of the divisor and out of the 48 of the dividend.

IV. Divide 56 x 35 = 1960 by 15 x 8 = 120.

NOTE. - Employ "Short Methods" in the solution of the following problems.

How many cords of wood

*

1. In a pile 32 feet by 8 feet by 4 feet?
2. In a pile 40 feet by 16 feet by 6 feet?
3. In a pile 32 feet by 30 feet by 10 feet?
(aa) Find the sum of the three results.

How many acres of land

4. In a piece 180 rods by 28 rods? †
5. In a piece 64 rods by 96 rods?
6. In a piece 136 rods by 32 rods?

(bb) Find the sum of the three results.

7. Multiply 64 by 96 and divide the product by 16 x 24 x 2.

8. Multiply 250 by 72 and divide the product by 16 × 3 x 24.

(cc) Find the sum of the two results.

*Think of a cord as 8 feet by 4 feet by 4 feet.

+ Think of an acre as 40 rods by 4 rods.

Find the cost·

9. Of 346 acres of land at $50 per acre. 10. Of 346 acres of land at $51 per acre. 11. Of 346 acres of land at $52 per acre. 12. Of 346 acres of land at $49 per acre. 13. Of 254 acres of land at $51 per acre.

(dd) Find the sum of the five results.

14. Of 243 ft. iron pipe at 331 a foot.
15. Of 243 ft. iron pipe at 341¢ a foot.
16. Of 243 ft. iron pipe at 35¢ a foot.
17. Of 243 ft. iron pipe at 323 a foot.
18. Of 156 ft. iron pipe at 35¢ a foot.
(ee) Find the sum of the five results.

19. Of 260 lb. butter at 25¢ a pound.
20. Of 260 lb. butter at 26¢ a pound.
21. Of 260 lb. butter at 27 a pound.
22. Of 260 lb. butter at 24
23. Of 184 lb. butter at 27

a pound.

a pound.

a pound.

a pound.

29. Of 330 lb. coffee at 17
30. Of 330 lb. coffee at 15
31. Of 240 lb. coffee at 25¢ a pound,
32. Of 240 lb. coffee at 26 a pound.
33. Of 240 lb. coffee at 27¢ a pound.
(gg) Find the sum of the ten results.