Express 0 . 0 6 ¯ as a rational number in the form p q , where p , q ∈ Z and q ≠ 0 .

Let x = 0 . 0 6 ⇒ x = 0 . 066666 . . . . ( i ) Multiplying both the sides of equation ( i ) by 10 , we get: 10 x = 0 . 666666 . . . . . ( i i ) Subtracting equation i from equation i i , we get: 9 x = 0 . 6 x = 0 . 6 9 = 6 90 = 1 15 . Hence, the required answer is 1 15 .

Simplify 1 5 + 3 + 1 2 ( 5 - 3 ) .

If the denominator of an expression contains a term with a square root (or a number under a radical sign), then the process of converting it to an equivalent expression whose denominator is a rational number, is called rationalising the denominator. In 1 5 + 3 , the denominator contains square root. Thus it needs to be rationalised. 1 5 + 3 = 1 5 + 3 × 5 - 3 5 - 3 = 5 - 3 5 2 - 3 2 = 5 - 3 5 - 3 = 5 - 3 2 = 1 2 5 - 3 ∴ 1 5 + 3 + 1 2 5 - 3 = 1 2 5 - 3 + 1 2 5 - 3 = 5 - 3

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All in One Mathematics>Chapter 1 - Number Systems>Exam Practice

Brijesh Dwevedi and Jitendra Gupta Solutions for Chapter: Number Systems, Exercise 5: Exam Practice

Author:Brijesh Dwevedi & Jitendra Gupta

Brijesh Dwevedi Mathematics Solutions for Exercise - Brijesh Dwevedi and Jitendra Gupta Solutions for Chapter: Number Systems, Exercise 5: Exam Practice

Attempt the practice questions on Chapter 1: Number Systems, Exercise 5: Exam Practice with hints and solutions to strengthen your understanding. All in One Mathematics solutions are prepared by Experienced Embibe Experts.