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