Simplify: 3 11 + 8 99 - 44

Given, 3 11 + 8 99 - 44 Convert the surds in simplest form by finding the prime factors of each surd. = 3 11 + 8 3 × 3 × 11 - 2 × 2 × 11 = 3 11 + 8 × 3 11 - 2 11 = 3 + 24 - 2 11 = 25 11 Thus, the required answer is 25 11

Simplify: 3 63 - 2 28 - 112

Given, 3 63 - 2 28 - 112 Convert the surds in simplest form by finding the prime factors of each surd. = 3 3 × 3 × 7 - 2 2 × 2 × 7 - 2 × 2 × 2 × 2 × 7 = 3 7 + 4 7 - 4 7 = 3 + 4 - 4 7 = 3 7 Thus, the required answer is 3 7

Simplify: 5 18 + 7 50 - 10 8

Given, 5 18 + 7 50 - 10 8 Convert the surds in simplest form by finding the prime factors of each surd. = 5 2 × 3 × 3 + 7 2 × 5 × 5 - 10 2 × 2 × 2 = 5 × 3 2 + 7 × 5 2 - 10 × 2 2 = 15 2 + 35 2 - 20 2 = 15 + 35 - 20 2 = 30 2 Thus, the required answer is 30 2

Simplify: 6 72 - 18 - 8 32

Given, 6 72 - 18 - 8 32 Convert the surds in simplest form by finding the prime factors of each surd. = 6 2 × 2 × 2 × 3 × 3 - 2 × 3 × 3 - 8 2 × 2 × 2 × 2 × 2 = 6 × 3 × 2 2 - 3 2 - 8 × 2 × 2 2 = 36 2 - 3 2 - 32 2 = 36 - 3 - 32 2 = 2 Thus, the required answer is 2

Simplify: 48 - 12 + 75 - 27

Given, 48 - 12 + 75 - 27 Convert the surds in simplest form by finding the prime factors of each surd. = 2 × 2 × 2 × 2 × 3 - 2 × 2 × 3 + 5 × 5 × 3 - 3 × 3 × 3 = 2 × 2 3 - 2 3 + 5 3 - 3 3 = 4 3 - 2 3 + 5 3 - 3 3 = 4 - 2 + 5 - 3 3 = 4 3 Thus, the required answer is 4 3

Simplify: 162 - 98 - 8 + 50

Given, 162 - 98 - 8 + 50 Convert the surds in simplest form by finding the prime factors of each surd. = 2 × 3 × 3 × 3 × 3 - 2 × 7 × 7 - 2 × 2 × 2 + 5 × 5 × 2 = 3 × 3 2 - 7 2 - 2 2 + 5 2 = 9 2 - 7 2 - 2 2 + 5 2 = 9 - 7 - 2 + 5 2 = 5 2 Thus, the required answer is 5 2

Expand ( 2 + 5 ) 2 . Thus, we have, ( 2 + 5 ) 2 = 2 2 + 5 2 + 2 × 2 × 5 [Use a + b 2 = a 2 + b 2 + 2 a b ] = 4 + 5 + 4 5 = 9 + 4 5 Therefore, the required answer is 9 + 4 5 .

Expand ( 7 - 3 ) 2 . Thus, we have, ( 7 - 3 ) 2 = 7 2 + 3 2 - 2 × 7 × 3 [Use a - b 2 = a 2 + b 2 - 2 a b ] = 49 + 3 - 14 3 = 52 - 14 3 Therefore, the required answer is 52 - 14 3 .

E M B I B E

ICSE MATHEMATICS CLASS 9>Pure Arithmetic>Chapter 1 - Rational and Irrational Numbers>EXERCISE 1C

B Nirmala Shastry Solutions for Exercise 3: EXERCISE 1C

Author:B Nirmala Shastry

B Nirmala Shastry Mathematics Solutions for Exercise - B Nirmala Shastry Solutions for Exercise 3: EXERCISE 1C

Attempt the practice questions from Exercise 3: EXERCISE 1C with hints and solutions to strengthen your understanding. ICSE MATHEMATICS CLASS 9 solutions are prepared by Experienced Embibe Experts.