Arrange the following fractions in descending order. a. 3 4 , 13 16 , 7 12

The given fractions are 3 4 , 13 16 , 7 12 Fractions can be compared by converting them into like fractions and then arranging them in ascending or descending order. LCM of 4 , 16 , 12 is 48 . . Thus we convert all the denominators equivalent to the LCM of denominators. Now, 3 4 = 3 × 12 4 × 12 = 36 48 13 16 = 13 × 3 16 × 3 = 39 48 7 12 = 7 × 4 12 × 4 = 28 48 We know, 28 < 36 < 39 39 48 , 36 48 , 28 48 Hence, the following fractions in descending order is 39 48 , 36 48 , 28 48 . .

Arrange the following fractions in descending order: 2 3 , 8 15 , 5 12 , 13 16

The given fractions are 2 3 , 8 15 , 5 12 , 13 16 Fractions can be compared by converting them into like fractions and then arranging them in ascending or descending order. LCM of 3 , 15 , 12 , 16 is 240 . Thus we convert all the denominators equivalent to the LCM of denominators. Now, 2 3 = 2 × 80 3 × 80 = 160 240 8 15 = 8 × 16 15 × 16 = 128 240 5 12 = 5 × 20 12 × 20 = 100 240 13 16 = 13 × 5 16 × 15 = 195 240 We know, 100 < 128 < 160 < 195 ⇒ 195 240 , 160 240 , 128 240 , 100 240 Hence, the following fractions in descending order is 195 240 , 160 240 , 128 240 , 100 240 .

Arrange the following fractions in descending order: 5 6 , 7 8 , 4 15 , 8 13

The given fractions are 5 6 , 7 8 , 4 15 , 8 13 Fractions can be compared by converting them into like fractions and then arranging them in ascending or descending order. LCM of 6 , 8 , 15 , 13 is 1690 . Thus we convert all the denominators equivalent to the LCM of denominators. Now, 5 6 = 2 × 80 3 × 80 = 160 240 8 15 = 8 × 16 15 × 16 = 128 240 5 12 = 5 × 20 12 × 20 = 100 240 13 16 = 13 × 5 16 × 15 = 195 240 We know, 100 < 128 < 160 < 195 ⇒ 195 240 , 160 240 , 128 240 , 100 240 Hence, the following fractions in descending order is 195 240 , 160 240 , 128 240 , 100 240 .

Solve the following: a. 5 1 4 + 1 2 5 - 2 1 8

Given, 5 1 4 + 1 2 5 - 2 1 8 , We can solve this by, = 21 4 + 7 5 - 17 8 = 210 + 56 - 85 40 = 181 40 = 4 21 40 Therefore, the solution is 4 21 40 .

Solve the following: b. 3 1 5 + 2 5 7 - 5 2 3

Given, 3 1 5 + 2 5 7 - 5 2 3 , We can solve this by, = 16 5 + 19 7 - 17 3 = 336 + 285 - 595 105 = 26 105 Therefore, the solution is 26 105 .

E M B I B E

Cambridge connection Mathematics>Chapter 4 - Fractions>Assessment

Prakash Tiwari and Ishita Joshi Solutions for Chapter: Fractions, Exercise 11: Assessment

Author:Prakash Tiwari & Ishita Joshi

Prakash Tiwari Mathematics Solutions for Exercise - Prakash Tiwari and Ishita Joshi Solutions for Chapter: Fractions, Exercise 11: Assessment

Attempt the free practice questions on Chapter 4: Fractions, Exercise 11: Assessment with hints and solutions to strengthen your understanding. Cambridge connection Mathematics solutions are prepared by Experienced Embibe Experts.