Ch.4_Solution_Manual_Ed.1_v7_

Ch.4_Solution_Manual_Ed.1_v7_ - Exercises 4.1 Exercise 4.1,...

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Exercises 4.1 Exercise 4.1, Solution 1: a. 500: 400: 800 Dividing all terms by their common factor 100, we get, 5 : 4 : 8 Dividing all terms by 4, we get 1.25 : 1 : 2 Therefore, the ratio in the lowest integer is 5 : 4 : 8 and the equivalent ratio having smallest term 1 is 1.25 : 1 : 2. b. 36 : 5 : 9 Dividing all terms by 5, we get 7.2 : 1 : 1.80 Therefore, the ratio in the lowest integer is 36 : 5 : 9 and the equivalent ratio having smallest term 1 is 7.2 : 1 : 1.80. c. 5.1 : 25.5 : 34 Moving decimal by 1 place to the right, we get, 51 : 255 : 340 Dividing all terms by their common factor 17, we get, 3 : 15 : 20 Dividing all terms by 3, we get 1 : 5 : 6.666666… 1 : 5 : 6.67 Therefore, the ratio in the lowest integer is 3 : 15 : 20 and the equivalent ratio having smallest term 1 is 1 : 5 : 6.67. Exercise 4.1, Solution 3: a. 4 : 6 and 6 : 10 Consider 4 : 6 Dividing all terms by their common factor 2 2 : 3 Consider 6 : 10 Dividing all terms by their common factor 2 3 : 5 Therefore, the given ratios are not equal. b. 8 : 10 and 28 : 35 Consider 8 : 10 Dividing all terms by their common factor 2 4 : 5 Consider 28 : 35 Dividing all terms by their common factor 7 4 : 5 Therefore, the given ratios are equal. c. 6 : 8 and 27 : 32 Consider 6 : 8 Dividing all terms by their common factor 2 3 : 4 Consider 27 : 32

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There is no common factor, therefore the lowest ratio is 27 : 32. Therefore, the given ratios are not equal. d. 16 : 22 and 64 : 88 Consider 16 : 22 Dividing all terms by their common factor 2 8 : 11 Consider 64 : 88 Dividing all terms by their common factor 8 8 : 11 Therefore, the given ratios are equal. Exercise 4.1, Solution 5: 30 minutes to 3 hours and 15 minutes 3 hours and 15 minutes = 180 min + 15 min = 195 min 30 : 195 Divide both terms by 15 2 : 13 400 m to 3.5 km 400 : 3500 Divide both terms by 100 4 : 35 750 g to 12 kg 750 : 12,000 Divide both terms by 750 1 : 16 Exercise 4.1, Solution 7: \$1 = 100¢ 10 : 100 Dividing by 10, we get, 1 : 10 Therefore, the ratio of a Canadian dime to a Canadian loonie is 1 : 10. Exercise 4.1, Solution 9: 3000 km : 6 hr Dividing both terms by 6, we get, Therefore, the ratio of the distance to the time taken by the aircraft is 500 km/hr. Exercise 4.1, Solution 11: 3850 : 7000 : 3500 Dividing by common factor 50, we get, 77 : 140 : 70 77 : 140 : 70 Dividing by common factor 7, we get, 11 : 20 : 10 Therefore, the ratio of the distance of the swim to the bike track to the race track in the lowest integer is 11 : 20 : 10. Exercise 4.1, Solution 13: 24 : 4 : 2 : 8 Dividing all terms by 2, we get, 2
12 : 2 : 1 : 4 Therefore, the ratio of the ingredients in her recipe is 12 : 2 : 1 : 4. New ratio = 24 : 4 : 1½ : 8 24 : 4 : : 8 Multiplying every term by 2, we get 48 : 8 : 3 : 16 Therefore, the new ratio of ingredients in her recipe is 48 : 8 : 3 : 16. Exercise 4.1, Solution 15:

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This note was uploaded on 04/04/2012 for the course MATH 1052 taught by Professor Kit during the Winter '12 term at Fanshawe.

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Ch.4_Solution_Manual_Ed.1_v7_ - Exercises 4.1 Exercise 4.1,...

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