Explanation for Multiplication Multiply 475 by 9. 475 multiplicand Explanation: Multiplication is merely a short 9 multiplier method for addition. We know that we should have 9X5 units or 45 units in the sum of the units 4275 product if we were adding. We write the 5 in the units place in the product and save the four tens to add to the tens. 9X7 tens=63 tens. 63 tens +4 tens=67 tens. 67 tens make 6 hundreds, with a remainder of 7 tens. Write the 7 in the product and save the 6 hundreds. 9X4 hundreds=36 hundreds. 36 hundreds + 6 hundreds—42 hundreds. Therefore the product is 4275. Exercise 24. 1. 3,672 6. 8,692 11. 6,125 16. 9,934 2. 4,895 7. 3,714 12. 8,309 17. 6,388 3. 2,506 8. 5,206 13. 4,763 18. 2,766 4. 8,179 9. 7,628 14. 2,918 19. 5,177 5. 3,457 10. 49,53 15. 5,046 20. 2,855 Multiply each number in the above list by 2, by 3, by 4, by 5, by 6, by 7, by 8, by 9. Multiply 358 by 42. 358 multiplicand Explanation: We first 42 multiplier multiply by the 2 (the units' figure of the multiplier) as in 2X358=716 1st partial product example one. The 4 in the 4 tens X35851432 2nd partial product multiplier represents 4 tens. 15036 product 358 X4 tens gives 1432 teng. The 2 of this partial product must then be written in tens place, the 3 in hundreds' place, etc. If the multiplier contained a figure in hundreds’ place, where would the first figure of the third partial product be written? The teacher should insist on enough explanations from the pupils to see that they understand the process and then drill for speed. 37. 5437 699 38. 9836 562 39. 5474 875 40. 3869 547 41. 9574 954 42. 7389 875 43. 46. 47. 4156 697 44. 3578 46. 6879 48. 7412 389 Practice problems involving zeros in the multiplier are found on page 24. Exercise 26. Practical Short Methods in Multiplication 1. Multiply 365 by 41. Regular form: A shorter form: 365 365 41 1460 365 14965 1460 14965 2. How many figures are required in the regular form? How many are used in the short form? 3. Using the shorter form, multiply 365 by 31, by 21, by 51, by 71, by 61, by 81, by 91. 4. Multiply 365 by 14. Regular form: A shorter form: 365 365 14 1460 1460 5110 365 5110 In this example, the figure 1 is in tens' place in the multiplier instead of units' place. We may then write down 365 and below it multiply it by 4, placing the 0 in units' place. 5. Using the shorter form, multiply 365 by 18 and by 15. 6. In the same way multiply 866 by 12, by 16 and by 19. 7. Find the cost of a farm of 81 acres at $165 an acre. Suggestion for explanation: If one acre costs $165, 81 acres cost 81 times $165 or ? dollars. 8. A farmer sold a carload of 19 cattle averaging 1,475 pounds in weight. How many pounds did the carload of cattle weigh? (19 X 1475 lb.=? lb.) 9. A truck farmer, on a farm of 16 acres, made an average profit of $85 per acre. What were his total profits on his farm? Exercise 27. Multiplication Problems 1. Margaret raised 45 bushels of tomatoes in her garden last summer and sold them at 75 cents a bushel. How much did she receive for her entire crop? 2. Harold raised 62 bushels of onion sets which he sold for 90 cents a bushel. How much did he receive for them? 3. A bushel of potatoes weighs 60 pounds. How many pounds would a wagon load containing 35 bushels weigh? 4. Frank bought a 6-pound roast for 25 cents per pound. Find the cost of the roast. 5. A farmer sold 64 hogs at $25 per head. How much did he receive for them? 6. Edna bought 5 pounds of sugar at 11 cents a pound; 3 pounds of rice at 12 cents a pound; and 3 pounds of apples at 6 cents a pound. Find the total amount of her purchases. 7. Clark's father bought a farm of 80 acres at $135 per acre. Find the cost of the farm. 8. A tourist drove his car at the rate of 15 miles per hour. How far could he travel at that rate in 12 hours? 9. Ruth's father gives her an allowance of 35 cents a week for spending money. How much does she receive in a year of 52 weeks? 10. A stock feeder sold a herd of 23 cattle for $169 per head. How much did the entire herd bring? 11. A sack of wheat holds 2 bushels. A bushel of wheat weighs 60 pounds. How many pounds are there in a wagon load consisting of 20 sacks of wheat? In problem I it is not necessary to put down the complete row of zeros which would be obtained by multiplying each figure of the multiplicand (8739) by zero. Since the first figure, obtained in multiplying by the 5 in tens' place in the multiplier, will fall in tens' place in the product, it is necessary to fill in units' place with a zero. In the same way, two zeros must be used to fill in the units' and tens' places of the product in problem II and the first figure obtained by multiplying by the 9 must fall in hundreds' place. In problem III it is not necessary to use any zeros because there is already a figure in tens' place in the first partial product. Be careful always to place the first figure, obtained by multiplying by the hundreds' figure of the multiplier, in hundreds' column in the partial products. |