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## S Chand Class 9 ICSE Maths Solutions Chapter 1 Rational and Irrational Numbers Chapter Test

Question 1.

A number is an irrational number if and only if its decimal representation is

(a) non-terminating

(b) non-terminating and repeating

(c) non-terminating and non-repeating

(d) tenninating

Solution:

(c) non-terminating and non-repeating

Question 2.

Which of the following is an irrational number?

(a) \(\sqrt{29}\)

(b) \(\sqrt{441}\)

(c) 0.5948

(d) 5.\(\overline{318}\)

Solution:

(a) \(\sqrt{29}\)

\(\sqrt{29}\) is irrational number as 29 is not a perfect square.

Question 3.

(- 2 – \(\sqrt{3}\)) (- 2 + \(\sqrt{3}\)) when simplified is

(a) positive and irrational

(b) positive and rational

(c) negative and irrational

(d) negative and rational

Solution:

(b) positive and rational

(- 2 – \(\sqrt{3}\)) (- 2 + \(\sqrt{3}\)) = (- 2)² – (\(\sqrt{3}\))²

= 4 – 3 = 1

Which is positive and rational.

Question 4.

If \(\sqrt{6}\) x \(\sqrt{15}\) = x\(\sqrt{10}\) , then the value of x is

(a) 3

(b) ± 3

(c) \(\sqrt{3}\)

(d) \(\sqrt{6}\)

Solution:

(a) 3

\(\sqrt{6}\) x \(\sqrt{15}\) = x\(\sqrt{10}\)

⇒ \(\sqrt{6 \times 15}=x \sqrt{10} \Rightarrow \sqrt{90}=x \sqrt{10}\)

⇒ \(\sqrt{9 \times 10}=x \sqrt{10} \Rightarrow 3 \sqrt{10}=x \sqrt{10}\)

Comparing, we get

∴ x = 3

Question 5.

Two rational numbers between \(\frac { 2 }{ 7 }\) and \(\frac { 2 }{ 14 }\) are

(a) \(\frac { 1 }{ 14 }\) and \(\frac { 2 }{ 14 }\)

(b) \(\frac { 1 }{ 2 }\) and \(\frac { 3 }{ 2 }\)

(c) \(\frac { 3 }{ 14 }\) and \(\frac { 3 }{ 7 }\)

(d) \(\frac { 5 }{ 14 }\) and \(\frac { 8 }{ 14 }\)

Solution:

(d) \(\frac { 5 }{ 14 }\) and \(\frac { 8 }{ 14 }\)

Two rational numbers between \(\frac { 2 }{ 7 }\) and \(\frac { 2 }{ 14 }\) are

\(\frac { 5 }{ 14 }\) and \(\frac { 8 }{ 14 }\)

∵ \(\frac { 2 }{ 7 }\) and \(\frac { 5 }{ 7 }\) = \(\frac { 4 }{ 14 }\) and \(\frac { 10 }{ 14 }\)

and 5 and 8 lie between 4 and 10

Question 6.

An irrational number between \(\frac { 5 }{ 7 }\) and \(\frac { 7 }{ 9 }\) is

(a) 0.75

(b) \(\sqrt{6}\)

(c) 0.7507500075000…

(d) 0.7512

Solution:

(c) 0.7507500075000…

An irrational number between \(\frac { 5 }{ 7 }\) and \(\frac { 7 }{ 9 }\) either \(\sqrt{6}\) or 0.7507500075000…

∵ 0.75 and 0.7512 are rational number

Now \(\sqrt{6}\)

Which does not line between \(\frac { 5 }{ 7 }\) and \(\frac { 7 }{ 9 }\)

∴ 0.7507500075000… is irrational number between \(\frac { 5 }{ 7 }\) and \(\frac { 7 }{ 9 }\)

Question 7.

If \(\sqrt{2}\) = 1.4142, then the value of \(\frac{7}{3+\sqrt{2}}\) correct to two decimal places is

(a) 1.59

(b) 1.60

(c) 2.58

(d) 2.57

Solution:

(a) 1.59

\(\sqrt{2}\) = 1.4142

\(\frac{7}{3+\sqrt{2}}=\frac{7 \times(3-\sqrt{2})}{(3+\sqrt{2})(3-\sqrt{2})}\)

(Rationalising denominator)

\(\frac{7(3-\sqrt{2})}{(9-2)}=\frac{7(3-\sqrt{2})}{7}\) = 3 – \(\sqrt{2}\)

= 3 – 1.4142 = 1.5858 = 1.59

Question 8.

Taking \(\sqrt{3}\) as 1.732 and \(\sqrt{2}\) = 1.414, the value of \(\frac{1}{\sqrt{3}+\sqrt{2}}\) is

(a) 0.064

(b) 0.308

(c) 0.318

(d) 2.146

Solution:

(c) 0.318

\(\sqrt{3}\) = 1.732, \(\sqrt{2}\) = 1.414

\(\frac{1}{\sqrt{3}+\sqrt{2}}=\frac{\sqrt{3}-\sqrt{2}}{(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})}\)

(Rationalising denominator)

\(\frac{\sqrt{3}-\sqrt{2}}{3-2}=\frac{\sqrt{3}-\sqrt{2}}{1}=\sqrt{3}-\sqrt{2}\)

= 1.732 – 1.414 = 0.318

Question 9.

If x = \(\sqrt{3}\) + \(\sqrt{2}\), then the value of (x + \(\frac { 1 }{ x }\)) is

(a) 2

(b) 3

(c) 2\(\sqrt{2}\)

(d) 2\(\sqrt{3}\)

Solution:

(d) 2\(\sqrt{3}\)

x = \(\sqrt{3}\) + \(\sqrt{2}\)

\(\frac{1}{x}=\frac{1}{\sqrt{3}+\sqrt{2}}=\frac{\sqrt{3}-\sqrt{2}}{(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})}\)

(Rationalising denominator)

= \(\frac{\sqrt{3}-\sqrt{2}}{3-2}=\frac{\sqrt{3}-\sqrt{2}}{1}=\sqrt{3}-\sqrt{2}\)

x + \(\frac{1}{x}=\sqrt{3}+\sqrt{2}+\sqrt{3}-\sqrt{2}=2 \sqrt{3}\)

Question 10.

If x = 2 + \(\sqrt{3}\), then the value of \(\sqrt{x}+\frac{1}{\sqrt{x}}\) is

(a) 3 + \(\sqrt{3}\)

(b) \(\sqrt{6}\)

(c) 2\(\sqrt{6}\)

(d) 6

Solution:

(b) \(\sqrt{6}\)

x = 2 + \(\sqrt{3}\)

\(\frac{1}{x}=\frac{1}{2+\sqrt{3}}=\frac{1(2-\sqrt{3})}{(2+\sqrt{3})(2-\sqrt{3})}\)

(Rationalising denominator)

\(\frac{2-\sqrt{3}}{(2)^2-(\sqrt{3})^2}=\frac{2-\sqrt{3}}{4-3}=2-\sqrt{3}\)

∴ x + \(\frac { 1 }{ x }\) = 2 + \(\sqrt{3}\) + 2 – \(\sqrt{3}\) = 4

and (\(\sqrt{x}\) + \(\frac{1}{\sqrt{x}}\))² = x + \(\frac { 1 }{ x }\) + 2

= 4 + 2 = 6

\(\sqrt{x}+\frac{1}{\sqrt{x}}= \pm \sqrt{6}=\sqrt{6}\)