SChand CLASS 9 Chapter 1 Rational number and Irrational number Test

 Test

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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) terminating

Sol :

(c) non-terminating and non-repeating

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Question 2

Which of the following is an irrational number ?

(a) √29

(b) √441

(c) 0.5948

(d) $5.\sqrt{318}$

Sol :

(a) √29 is irrational number as 29 is not a perfect square.

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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

Sol :

(b) positive and rational

$(-2-\sqrt{3})(-2+\sqrt{3})=(-2)^2-(\sqrt{3})^2$

=4-3=1

Which is positive and rational.

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Question 4

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

(a) 3

(b) ± 3

(c) √3

(d) √6

Sol :

⇒√6×√15=x√10

⇒$\sqrt{6 \times 15}$=x√10

⇒√90=x√10

⇒$\sqrt{9 \times 10}=x\sqrt{10}$

⇒3$\sqrt{10}=x\sqrt{10}$

Comparing, we get

∴ x = 3

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Question 5

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

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

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

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

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

Sol :

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

Two rational numbers between $\frac{2}{7} \text{ and } \frac{2}{14}$ are $\frac{5}{14}\text{ and }\frac{8}{14}$

∵$\frac{2}{7} \text{ and } \frac{5}{7}=\frac{4}{14} \text{ and } \frac{10}{14}$ and 5 and 8 lie between 4 and 10

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Question 6

An irrational number between $\frac{5}{7} \text{ and } \frac{7}{9}$ is

(a) 0.75

(b) √6

(c) 0.7507500075000…

(d) 0.7512

Sol :

(c) 0.7507500075000…

An irrational number between $\frac{5}{7} \text{ and } \frac{7}{9}$ either √6 or 0.7507500075000....

∵ 0.75 and 0.7512 are rational number

Which does not line between $\frac{5}{7} \text{ and } \frac{7}{9}$

∴ 0.7507500075000… is irrational number between $\frac{5}{7} \text{ and } \frac{7}{9}$

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Question 7

If √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

Sol :

(a) 1.59

√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-√2

= 3 – 1.4142 = 1.5858 = 1.59

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Question 8

Taking √3 as i.732 and √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

Sol :

(c) 0.318

√3 = 1.732,  √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

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Question 9

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

(a) 2

(b) 3

(c) $2\sqrt{2}$

 

(d) $2\sqrt{3}$

Sol :

(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)

$=\frac{\sqrt{3}-\sqrt{2}}{3-2}=\frac{\sqrt{3}-\sqrt{2}}{1}=\sqrt{3}-\sqrt{2}$

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

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Question 10

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

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

(b) √6

(c) 2√6

(d) 6

Sol :

(b) √6

x=2+√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 $\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)^2=x+\frac{1}{x}+2$

=4+2=6

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

=√6