The availability of step-by-step OP Malhotra Class 9 Solutions Chapter 1 Rational and Irrational Numbers Chapter Test can make challenging problems more manageable.

## 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
$$\sqrt{x}+\frac{1}{\sqrt{x}}= \pm \sqrt{6}=\sqrt{6}$$