RS AGGARWAL CLASS 9 Chapter 1 Number System MCQ

MULTIPLE CHOICE QUESTIONS

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Question 1:

Which of the following is a rational number?

(a) $1+\sqrt{3}$$1+\sqrt{3}$

(b) π

(c) $2\sqrt{3}$$2\sqrt{3}$

(d) 0

Answer 1:

Since, the sum and product of a rational and an irrational is always irrational.

So, $1+\sqrt{3}$$1+\sqrt{3}$ and $2\sqrt{3}$$2\sqrt{3}$ are irrational numbers.

Also, π is an irrational number.

And, 0 is an integer.

So, 0 is a rational number.

Hence, the correct option is (d).

Question 2:

A rational number between –3 and 3 is
(a) 0
(b) –4.3
(c) –3.4
(d) 1.101100110001...

Answer 2:

Since, –4.3 < –3.4 < –3 < 0 < 1.101100110001... < 3

But 1.101100110001... is an irrational number

So, the rational number between –3 and 3 is 0.

Hence, the correct option is (a).

Question 3:

Two rational numbers between $\frac{2}{3}\mathrm{and}\frac{5}{3}$$\frac{2}{3} \mathrm{and} \frac{{\displaystyle 5}}{3}$ are
(a) $\frac{1}{6}\mathrm{and}\frac{2}{6}$$\frac{1}{6} \mathrm{and} \frac{2}{6}$

(b) $\frac{1}{2}\mathrm{and}\frac{2}{1}$$\frac{1}{2} \mathrm{and} \frac{2}{1}$

(c) $\frac{5}{6}\mathrm{and}\frac{7}{6}$$\frac{5}{6} \mathrm{and} \frac{7}{6}$

(d) $\frac{2}{3}\mathrm{and}\frac{4}{3}$$\frac{2}{3} \mathrm{and} \frac{4}{3}$

Answer 3:

We have,

$\frac{2}{3}=\frac{2\times 2}{3\times 2}=\frac{4}{6}\mathrm{and}\frac{5}{3}=\frac{5\times 2}{3\times 2}=\frac{10}{6}$$\frac{2}{3}=\frac{2\times 2}{3\times 2} =\frac{4}{6} \mathrm{and} \frac{{\displaystyle 5}}{3}=\frac{5\times 2}{3\times 2}=\frac{10}{6}$

And, $\frac{1}{2}=\frac{1\times 3}{2\times 3}=\frac{3}{6}\mathrm{and}\frac{2}{1}=\frac{2\times 6}{1\times 6}=\frac{12}{6}$$\frac{1}{2}=\frac{1\times 3}{2\times 3}=\frac{3}{6} \mathrm{and} \frac{2}{1}=\frac{2\times 6}{1\times 6}=\frac{12}{6}$

Also, $\frac{2}{3}=\frac{2\times 2}{3\times 2}=\frac{4}{6}\mathrm{and}\frac{4}{3}=\frac{4\times 2}{3\times 2}=\frac{8}{6}$$\frac{2}{3}=\frac{2\times 2}{3\times 2} =\frac{4}{6} \mathrm{and} \frac{{\displaystyle 4}}{3}=\frac{4\times 2}{3\times 2}=\frac{8}{6}$

Since, $\frac{1}{6}<\frac{2}{6}<\frac{3}{6}\left(\frac{1}{2}\right)<\frac{4}{6}\left(=\frac{2}{3}\right)<\frac{5}{6}<\frac{7}{6}<\frac{8}{6}\left(=\frac{4}{3}\right)<\frac{10}{6}\left(=\frac{5}{3}\right)<\frac{12}{6}\left(=\frac{2}{1}\right)$$\frac{1}{6}<\frac{2}{6}<\frac{{\displaystyle 3}}{{\displaystyle 6}}\left(\frac{1}{2}\right)<\frac{4}{6}\left(=\frac{2}{3}\right)<\frac{5}{6}<\frac{7}{6}<\frac{{\displaystyle 8}}{{\displaystyle 6}}\left(=\frac{4}{3}\right)<\frac{10}{6}\left(=\frac{5}{3}\right)<\frac{{\displaystyle 12}}{{\displaystyle 6}}\left(=\frac{2}{1}\right)$

So, the two rational numbers between $\frac{2}{3}\mathrm{and}\frac{5}{3}$$\frac{2}{3} \mathrm{and} \frac{{\displaystyle 5}}{3}$ are $\frac{5}{6}\mathrm{and}\frac{7}{6}$$\frac{5}{6} \mathrm{and} \frac{7}{6}$.

Hence, the correct opion is (c).

Question 4:

Every point on a number line represents
(a) a rational number
(b) a natural number
(c) an irrational number
(d) a unique number

Answer 4:

As, all rational numbers, all natural numbers and all irrational numbers can be represented on a nuumber line in an unique way.

So, every point on a number line represents a unique number.

Hence, the correct option is (d).

Question 5:

Which of the following is a rational number?
(a) $\sqrt{2}$$\sqrt{2}$
(b) $\sqrt{23}$$\sqrt{23}$
(c) $\sqrt{225}$$\sqrt{225}$
(d) 0.1010010001....

Answer 5:

(c) $\sqrt{225}$$\sqrt{225}$

Because 225 is a square of 15, i.e., $\sqrt{225}$$\sqrt{225}$ = 15, and it can be expressed in the $\frac{p}{q}$$\frac{p}{q}$ form, it is a rational number.

Question 6:

Every rational number is
(a) a natural number
(b) a whole number
(c) an integer
(d) a real number

Answer 6:

(d) a real number

Every rational number is a real number, as every rational number can be easily expressed on the real number line.

Question 7:

Between any two rational numbers there
(a) is no rational number
(b) is exactly one rational numbers
(c) are infinitely many rational numbers
(d) is no irrational number

Answer 7:

(c) are infinitely many rational numbers

Because the range between any two rational numbers can be easily divided into any number of divisions, there can be an infinite number of rational numbers between any two rational numbers.

Question 8:

The decimal representation of a rational number is
(a) always terminating
(b) either terminating or repeating
(c) either terminating or non-repeating
(d) neither terminating nor repeating

Answer 8:

(b) either terminating or repeating

As per the definition of rational numbers, they are either repeating or terminating decimals.

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Question 9:

The decimal representation of an irrational number is
(a) always terminating
(b) either terminating or repeating
(c) either terminating or non-repeating
(d) neither terminating nor repeating

Answer 9:

(d) neither terminating nor repeating

As per the definition of irrational numbers, these are neither terminating nor repeating decimals.

Question 10:

The decimal expansion that a rational number cannot have is
(a) 0.25
(b) $0.25\overline{28}$$0.25\overline{28}$
(c) $0.\overline{2528}$$0.\overline{2528}$
(d) 0.5030030003...

Answer 10:

As, any number which have a terminating or non-terminating recurring decimal expansion is a rational number.

So, 0.5030030003... which is non-termintaing non-recurring decimal expansion is not a rational number.

Hence, the correct option is (d).

Question 11:

Which of the following is an irrational number?
(a) 3.14
(b) 3.141414...
(c) 3.14444...
(d) 3.141141114...

Answer 11:

(d) 3.141141114...

Because 3.141141114... is neither a repeating decimal nor a terminating decimal, it is an irrational number.

Question 12:

A rational number equivalent to $\frac{7}{19}$$\frac{7}{19}$ is
(a) $\frac{17}{119}$$\frac{17}{119}$

(b) $\frac{14}{57}$$\frac{14}{57}$

(c) $\frac{21}{38}$$\frac{21}{38}$

(d) $\frac{21}{57}$$\frac{21}{57}$

Answer 12:

Since, $\frac{7}{19}=\frac{7\times 3}{19\times 3}=\frac{21}{57}$$\frac{7}{19}=\frac{7\times 3}{19\times 3}=\frac{21}{57}$

Hence, the correct option is (d).

Question 13:

Choose the rational number which does not lie between $-\frac{2}{3}\mathrm{and}-\frac{1}{5}$$-\frac{2}{3} \mathrm{and} -\frac{1}{5}$.
(a) $-\frac{3}{10}$$-\frac{3}{10}$

(b) $\frac{3}{10}$$\frac{3}{10}$

(c) $-\frac{1}{4}$$-\frac{1}{4}$

(d) $-\frac{7}{20}$$-\frac{7}{20}$

Answer 13:

We have, $-\frac{2}{3}=-\frac{2\times 20}{3\times 20}=-\frac{40}{60}\mathrm{and}-\frac{1}{5}=-\frac{1\times 12}{5\times 12}=-\frac{12}{60}$$-\frac{2}{3}=-\frac{2\times 20}{3\times 20}=-\frac{40}{60} \mathrm{and} -\frac{1}{5}=-\frac{1\times 12}{5\times 12}=-\frac{12}{60}$

And, $-\frac{3}{10}=-\frac{3\times 6}{10\times 6}=-\frac{18}{60},\frac{3}{10}=\frac{3\times 6}{10\times 6}=\frac{18}{60},-\frac{1}{4}=-\frac{1\times 15}{4\times 15}=-\frac{15}{60},\mathrm{and}-\frac{7}{20}=-\frac{7\times 3}{20\times 3}=-\frac{21}{60}$$-\frac{3}{10}=-\frac{3\times 6}{10\times 6}=-\frac{18}{60}, \frac{3}{10}=\frac{3\times 6}{10\times 6}=\frac{18}{60}, -\frac{1}{4}=-\frac{1\times 15}{4\times 15}=-\frac{15}{60}, \mathrm{and} -\frac{7}{20}=-\frac{7\times 3}{20\times 3}=-\frac{21}{60}$

Since, $-\frac{40}{60}\left(=-\frac{2}{3}\right)<-\frac{21}{60}\left(=-\frac{7}{20}\right)<-\frac{18}{60}\left(=-\frac{3}{10}\right)<-\frac{15}{60}\left(=-\frac{1}{4}\right)<-\frac{12}{60}\left(=-\frac{1}{5}\right)<\frac{18}{60}\left(=\frac{3}{10}\right)$$-\frac{40}{60}\left(=-\frac{2}{3}\right)<-\frac{21}{60}\left(=-\frac{7}{20}\right)<-\frac{18}{60}\left(=-\frac{3}{10}\right)<-\frac{15}{60}\left(=-\frac{1}{4}\right)<-\frac{12}{60}\left(=-\frac{1}{5}\right)<\frac{18}{60}\left(=\frac{3}{10}\right)$

So, the rational number which does not lie between $-\frac{2}{3}\mathrm{and}-\frac{1}{5}$$-\frac{2}{3} \mathrm{and} -\frac{1}{5}$ is $\frac{3}{10}$$\frac{3}{10}$.

Hence, the correct option is (b).

Question 14:

π is
(a) a rational number
(b) an integer
(c) an irrational number
(d) a whole number

Answer 14:

Since, π has a non-terminating non-recurring decimal expansion.

So, π is an irrational number.

Hence, the correct option is (c).

Question 15:

Decimal expansion of $\sqrt{2}$$\sqrt{2}$ is
(a) a finite decimal
(b) 1.4121
(c) non-terminating recurring
(d) non-terminating, non-recurring

Answer 15:

(c) a non-terminating and non-repeating decimal

Because $\sqrt{2}$$\sqrt{2}$ is an irrational number, its decimal expansion is non-terminating and non-repeating.

Question 16:

Which of the following is an irrational number?

(a) $\sqrt{23}$

(b) $\sqrt{225}$

(c) 0.3799

(d) $7.\overline{478}$

Answer 16:

Since, $\sqrt{225}$ = 15, which is an integer,

0.3799 is a number with terminating decimal expansion, and

$7.\overline{478}$ is a number with non-terminating recurring decimal expansion

Also, 23 is a prime number.

So, $\sqrt{23}$ is an irrational number.                $\left[\because \sqrt{n} \mathrm{is} \mathrm{always} \mathrm{an} \mathrm{irrational} \mathrm{number}, \mathrm{if} n \mathrm{is} \mathrm{a} \mathrm{prime} \mathrm{number}.\right]$

Hence, the correct option is (a).

Question 17:

How many digits are there in the repeating block of digits in the decimal expansion of $\frac{17}{7}$?
(a) 16
(b) 6
(c) 26
(d) 7

Answer 17:

$\because \frac{17}{7}=2.\overline{428571}$

So, there are 6 digits in the repeating block of digits in the decimal expansion of $\frac{17}{7}$.

Hence, the correct option is (b).

 

Question 18:

Which of the following numbers is irrational?

(a) $\sqrt{\frac{4}{9}}$

(b) $\frac{\sqrt{1250}}{\sqrt{8}}$

(c) $\sqrt{8}$

(d) $\frac{\sqrt{24}}{\sqrt{6}}$

Answer 18:

Since,

$\sqrt{\frac{4}{9}}=\frac{2}{3}$, which is a rational number,

$\frac{\sqrt{1250}}{\sqrt{8}}=\sqrt{\frac{1250}{8}}=\sqrt{\frac{625}{4}}=\frac{25}{2}$, which is a rational number,

$\sqrt{8}=2\sqrt{2}$, which is an irrational number, and

$\frac{\sqrt{24}}{\sqrt{6}}=\sqrt{\frac{24}{6}}=\sqrt{4}=2$, which is a rational number

Hence, the correct option is (c).

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Question 19:

The product of two irrational number is
(a) always irrational
(b) always rational
(c) always an integer
(d) sometimes rational and sometimes irrational

Answer 19:

(d) sometimes rational and sometimes irrational

For example:
$\sqrt{2}$ is an irrational number, when it is multiplied with itself  it results into 2, which is a rational number.
$\sqrt{2}$  when multiplied with $\sqrt{3}$, which is also an irrational number, results into $\sqrt{6}$, which is an irrational number.

Question 20:

Which of the following is a true statment?
(a) The sum of two irrational numbers is an irrational number
(b) The product of two irrational numbers is an irrational number
(c) Every real number is always rational
(d) Every real number is either rational or irrational

Answer 20:

(d) Every real number is either rational or irrational.

Because a real number can be further categorised into either a rational number or an irrational number, every real number is either rational or irrational.

Question 21:

Which of the following is a true statment?
(a) $\mathrm{\pi } \mathrm{and} \frac{22}{7}$are both rationals
(b) $\mathrm{\pi } \mathrm{and} \frac{22}{7}$are both irrationals
(c) $\mathrm{\pi }$ is rational and $\frac{22}{7}$is irrational
(d) $\mathrm{\pi }$ is irrational and $\frac{22}{7}$is rational

Answer 21:

(d) $\mathrm{\pi }$ is irrational and $\frac{22}{7}$ is rational.
Because the value of $\mathrm{\pi }$ is neither repeating nor terminating, it is an irrational number. $\frac{22}{7}$, on the other hand, is of the form $\frac{p}{q}$, so it is a rational number.

Question 22:

A rational number lying between $\sqrt{2} \mathrm{and} \sqrt{3}$ is
(a) $\frac{\left(\sqrt{2}+\sqrt{3}\right)}{2}$
(b) $\sqrt{6}$
(c) 1.6
(d) 1.9

Answer 22:

Since, $\frac{\left(\sqrt{2}+\sqrt{3}\right)}{2}$ and $\sqrt{6}$ are irrational numbers,

And, $\sqrt{2}=1.414 \mathrm{and} \sqrt{3}=1.732$

So, the rational number lying between $\sqrt{2} \mathrm{and} \sqrt{3}$ is 1.6 .

Hence, the correct option is (c).

 

Question 23:

Which of the following is a rational number?
(a) $\sqrt{5}$
(b) 0.101001000100001...
(c) π
(d) 0.853853853...

Answer 23:

Since, a number whose decimal expansion is terminating or non-terminating recurring is rational number.

So, 0.853853853... is a rational number.

Hence, the correct option is (d).

Question 24:

The product of a nonzero rational number with an irrational number is always a/an
(a) irrational number
(b) rational number
(c) whole number
(d) natural number

Answer 24:

Since, the product of a non-zero rational number with an irrational number is always an irrational number.

Hence, the correct option is (a).

Question 25:

The value of $0.\overline{2}$ in the form $\frac{p}{q}$, where p and q are integers and q ≠ 0, is
(a) $\frac{1}{5}$  

(b) $\frac{2}{9}$

(c) $\frac{2}{5}$

(d) $\frac{1}{8}$

Answer 25:

Let $x=0.\overline{2}=0.222... .....\left(1\right)$
Multiplying both sides by 10, we get
$10x=2.\overline{2} .....\left(2\right)$
Subtracting (1) from (2), we get
$$\begin{gathered} 10x-x=2.\overline{2}-0.\overline{2} \\ \Rightarrow 9x=2 \\ \Rightarrow x=\frac{2}{9} \\ \therefore 0.\overline{2}=\frac{2}{9} \end{gathered}$$
Hence, the correct answer is option (b).

Question 26:

The simplest for of $1.\overline{6}$ is
(a) $\frac{833}{500}$
(b) $\frac{8}{5}$
(c) $\frac{5}{3}$
(d) none of these

Answer 26:

(c) $\frac{5}{3}$
Let x = 1.6666666...       ...(i)
Multiplying by 10 on both sides, we get:
10x = 16.6666666...       ...(ii)
Subtracting (i) from (ii), we get:
9x = 15
$\Rightarrow$x = $\frac{15}{9} =\frac{5}{3}$

Question 27:

The simplest form of $0.\overline{54}$ is
(a) $\frac{27}{50}$
(b) $\frac{6}{11}$
(c) $\frac{4}{7}$
(d) none of these

Answer 27:

(b) $\frac{6}{11}$
Let x = 0.545454...               ...(i)
Multiplying both sides by 100, we get:
100x = 54.5454545...           ...(ii)
Subtracting (i) from (ii), we get:
99x = 540
$\Rightarrow$x = $\frac{54}{99}$ = $\frac{6}{\mathrm{11<br><br>}}$

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Question 28:

The simplest form of $0.3\overline{2}$ is
(a) $\frac{16}{45}$
(b) $\frac{32}{99}$
(c) $\frac{29}{90}$
(d) none of these

Answer 28:

(c) $\frac{29}{90}$
Let x = 0.3222222222...          ...(i)
Multiplying by 10 on both sides, we get:
10x = 3.222222222...              ...(ii)
Again, multiplying by 10 on both sides, we get:
100x = 32.222222222...          ...(iii)
On subtracting (ii) from (iii), we get:
90x = 29
∴ x = $\frac{29}{90}$

Question 29:

The simplest form of $0.12\overline{3}$ is
(a) $\frac{41}{330}$
(b) $\frac{37}{330}$
(c) $\frac{41}{333}$
(d) none of these

Answer 29:

(d) none of these
Let x = 0.12333333333...         ...(i)
Multiplying by 100 on both sides, we get:
100x = 12.33333333...             ...(ii)
Multiplying by 10 on both sides, we get:
1000x = 123.33333333...         ...(iii)
Subtracting (ii) from (iii), we get:
900x = 111

$\Rightarrow$x = $\frac{111}{900}$

Question 30:

An irrational number between 5 and 6 is
(a) $\frac{1}{2}\left(5+6\right)$
(b) $\sqrt{5+6}$
(c) $\sqrt{5\times 6}$
(d) none of these

Answer 30:

(c) $\sqrt{5\times 6}$

An irrational number between a and b is given as $\sqrt{ab}$.

Question 31:

An irrational number between $\sqrt{2} \mathrm{and} \sqrt{3}$is
(a) $\left(\sqrt{2}+\sqrt{3}\right)$
(b) $\sqrt{2}\times \sqrt{3}$
(c) 5^1/4
(d) 6^1/4

Answer 31:

(d) 6^1/4
An irrational number between $$\begin{gathered} \sqrt{2}\mathrm{and} \sqrt{3}: \\ \sqrt{\sqrt{2}\times \sqrt{3}}=6^{\frac{1}{4}} \end{gathered}$$

Question 32:

An irrational number between $\frac{1}{7}\mathrm{and}\frac{2}{7}$is
(a) $\frac{1}{2}\left(\frac{1}{7}+\frac{2}{7}\right)$
(b) $\left(\frac{1}{7}\times \frac{2}{7}\right)$
(c) $\sqrt{\frac{1}{7}\times \frac{2}{7}}$
(d) none of these

Answer 32:

(c) $\sqrt{\frac{1}{7}\times \frac{2}{7}}$

An irrational number between a and b is given as $\sqrt{ab}$.

Question 33:

The sum of $0.\overline{3} \mathrm{and} 0.\overline{4}$ is
(a) $\frac{7}{10}$

(b) $\frac{7}{9}$

(c) $\frac{7}{11}$

(d) $\frac{7}{99}$

Answer 33:

Let $x=0.\overline{3}=0.333... .....\left(1\right)$
Multiplying both sides by 10, we get
$10x=3.\overline{3} .....\left(2\right)$
Subtracting (1) from (2), we get
$$\begin{gathered} 10x-x=3.\overline{3}-0.\overline{3} \\ \Rightarrow 9x=3 \\ \Rightarrow x=\frac{3}{9} \\ \therefore 0.\overline{3}=\frac{3}{9} \end{gathered}$$
Let $y=0.\overline{4}=0.444... .....\left(3\right)$
Multiplying both sides by 10, we get
$10y=4.\overline{4} .....\left(4\right)$
Subtracting (3) from (4), we get
$$\begin{gathered} 10y-y=4.\overline{4}-0.\overline{4} \\ \Rightarrow 9y=4 \\ \Rightarrow y=\frac{4}{9} \\ \therefore 0.\overline{4}=\frac{4}{9} \end{gathered}$$
Sum of $0.\overline{3}$ and $0.\overline{4}$ = $0.\overline{3}+0.\overline{4}=\frac{3}{9}+\frac{4}{9}=\frac{7}{9}$
Hence, the correct answer is option (b).

Question 34:

The value of $2.\overline{45}+0.\overline{36}$ is

(a) $\frac{67}{33}$

(b) $\frac{24}{11}$

(c) $\frac{31}{11}$

(d) $\frac{167}{110}$

Answer 34:

Let $x=2.\overline{45}=2.4545... .....\left(1\right)$
Multiplying both sides by 100, we get
$100x=245.\overline{45} .....\left(2\right)$
Subtracting (1) from (2), we get
$$\begin{gathered} 100x-x=245.\overline{45}-2.\overline{45} \\ \Rightarrow 99x=245-2=243 \\ \Rightarrow x=\frac{243}{99} \\ \therefore 2.\overline{45}=\frac{243}{99} \end{gathered}$$
Let $y=0.\overline{36}=0.3636... .....\left(3\right)$
Multiplying both sides by 100, we get
$100y=36.\overline{36} .....\left(4\right)$
Subtracting (3) from (4), we get
$$\begin{gathered} 100y-y=36.\overline{36}-0.\overline{36} \\ \Rightarrow 99y=36 \\ \Rightarrow y=\frac{36}{99} \\ \therefore 0.\overline{36}=\frac{36}{99} \end{gathered}$$
So, $2.\overline{45}+0.\overline{36}=\frac{243}{99}+\frac{36}{99}=\frac{243+36}{99}=\frac{279}{99}=\frac{31}{11}$
Hence, the correct answer is option (c).

Question 35:

Which of the following is the value of $\left(\sqrt{11}-\sqrt{7}\right) \left(\sqrt{11}+\sqrt{7}\right)$?
(a) –4
(b) 4
(c) $\sqrt{11}$
(d) $\sqrt{7}$

Answer 35:

$$\begin{gathered} \left(\sqrt{11}-\sqrt{7}\right)\left(\sqrt{11}+\sqrt{7}\right) \\ ={\left(\sqrt{11}\right)}^2-{\left(\sqrt{7}\right)}^2 \left[\left(a-b\right)\left(a+b\right)=a^2-b^2\right] \\ =11-7 \\ =4 \end{gathered}$$
Hence, the correct answer is option (b).

Question 36:

$\left(-2-\sqrt{3}\right) \left(-2+\sqrt{3}\right)$ when simplified is
(a) positive and irrational
(b) positive and rational
(c) negative and irrational
(d) negative and rational

Answer 36:

$$\begin{gathered} \left(-2-\sqrt{3}\right)\left(-2+\sqrt{3}\right) \\ =-\left(2+\sqrt{3}\right)\times \left[-\left(2-\sqrt{3}\right)\right] \\ =\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right) \end{gathered}$$
$$\begin{gathered} =2^2-{\left(\sqrt{3}\right)}^2 \left[\left(a+b\right)\left(a-b\right)=a^2-b^2\right] \\ =4-3 \\ =1 \end{gathered}$$
Thus, the given expression when simplified is positive and rational.

Hence, the correct answer is option (b).

Question 37:

$\left(6+\sqrt{27}\right)-\left(3+\sqrt{3}\right)+\left(1-2\sqrt{3}\right)$ when simplified is
(a) positive and irrational
(b) positive and rational
(c) negative and irrational
(d) negative and rational

Answer 37:

$$\begin{gathered} \left(6+\sqrt{27}\right)-\left(3+\sqrt{3}\right)+\left(1-2\sqrt{3}\right) \\ =6+\sqrt{3\times 3\times 3}-\left(3+\sqrt{3}\right)+\left(1-2\sqrt{3}\right) \\ =6+3\sqrt{3}-3-\sqrt{3}+1-2\sqrt{3} \\ =4 \end{gathered}$$
Thus, the given expression when simplified is positive and rational.

Hence, the correct answer is option (b).

Question 38:

When $15\sqrt{15}$ is divided by $3\sqrt{3}$, the quotient is

(a) $5\sqrt{3}$

(b) $3\sqrt{5}$

(c) $5\sqrt{5}$

(d) $3\sqrt{3}$

Answer 38:

$$\begin{gathered} 15\sqrt{15}÷3\sqrt{3} \\ =\frac{15\sqrt{5\times 3}}{3\sqrt{3}} \\ =\frac{15\times \sqrt{5}\times \sqrt{3}}{3\sqrt{3}} \left[\sqrt{ab}=\sqrt{a}\times \sqrt{b}\right] \\ =5\sqrt{5} \end{gathered}$$
Hence, the correct answer is option (c).

Question 39:

The value of $\sqrt{20}\times \sqrt{5}$ is

(a) 10

(b) $2\sqrt{5}$

(c) $20\sqrt{5}$

(d) $4\sqrt{5}$

Answer 39:

$$\begin{gathered} \sqrt{20}\times \sqrt{5} \\ =\sqrt{2\times 2\times 5}\times \sqrt{5} \\ =2\sqrt{5}\times \sqrt{5} \end{gathered}$$
$$\begin{gathered} =2\times 5 \\ =10 \end{gathered}$$
Hence, the correct answer is option (a).

Question 40:

The value of $\frac{4\sqrt{12}}{12\sqrt{27}}$ is
(a) $\frac{1}{9}$

(b) $\frac{2}{9}$

(c) $\frac{4}{9}$

(d) $\frac{8}{9}$

Answer 40:

$$\begin{gathered} \frac{4\sqrt{12}}{12\sqrt{27}} \\ =\frac{4\sqrt{2\times 2\times 3}}{12\sqrt{3\times 3\times 3}} \\ =\frac{4\times 2\sqrt{3}}{12\times 3\sqrt{3}} \\ =\frac{2}{9} \end{gathered}$$
Hence, the correct answer is option (b).

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Question 41:

$\sqrt{10}\times \sqrt{15}=?$
(i) $\sqrt{25}$
(ii) $5\sqrt{6}$
(iii) $6\sqrt{5}$
(iv) None of these

Answer 41:

$$\begin{gathered} \sqrt{10}\times \sqrt{15} \\ =\sqrt{2\times 5}\times \sqrt{3\times 5} \\ =\sqrt{2}\times \sqrt{5}\times \sqrt{3}\times \sqrt{5} \left[\sqrt{ab}=\sqrt{a}\times \sqrt{b}\right] \end{gathered}$$
$$\begin{gathered} =5\times \sqrt{2\times 3} \\ =5\sqrt{6} \end{gathered}$$
Hence, the correct answer is option (ii).
 

Question 42:

$\frac{{\displaystyle \sqrt{32}+\sqrt{48}}}{{\displaystyle \sqrt{8}+\sqrt{12}}}=?$
(a) $\sqrt{2}$
(b) 2
(c) 4
(d) 8

Answer 42:

$$\begin{gathered} \frac{{\displaystyle \sqrt{32}+\sqrt{48}}}{{\displaystyle \sqrt{8}+\sqrt{12}}} \\ =\frac{{\displaystyle \sqrt{16\times 2}+\sqrt{16\times 3}}}{{\displaystyle \sqrt{4\times 2}+\sqrt{4\times 3}}} \\ =\frac{{\displaystyle 4\sqrt{2}+4\sqrt{3}}}{{\displaystyle 2\sqrt{2}+2\sqrt{3}}} \left[\sqrt{ab}=\sqrt{a}\times \sqrt{b}\right] \end{gathered}$$
$$\begin{gathered} =\frac{{\displaystyle 4\left(\sqrt{2}+\sqrt{3}\right)}}{{\displaystyle 2\left(\sqrt{2}+\sqrt{3}\right)}} \\ =\frac{4}{2} \\ =2 \end{gathered}$$
Hence, the correct answer is option (b).

Question 43:

(125)^–1/3 = ?
(a) 5
(b) –5
(c) $\frac{1}{5}$
(d) $-\frac{1}{5}$

Answer 43:

$$\begin{gathered} {\left(125\right)}^{-\frac{1}{3}} \\ ={\left(5^3\right)}^{-\frac{1}{3}} \\ =5^{3\times \left(-\frac{1}{3}\right)} \left[{\left(x^a\right)}^b=x^{ab}\right] \end{gathered}$$
$$\begin{gathered} =5^{-1} \\ =\frac{1}{5} \left(x^{-a}=\frac{1}{x^a}\right) \end{gathered}$$
Hence, the correct answer is option (c).

Question 44:

The value of $7^{\frac{1}{2}}·8^{\frac{1}{2}}$ is
(a) (28)^1/2
(b) (56)^1/2
(c) (14)^1/2
(d) (42)^1/2

Answer 44:

$$\begin{gathered} 7^{\frac{1}{2}}·8^{\frac{1}{2}} \\ ={\left(7\times 8\right)}^{\frac{1}{2}} \left[x^a\times y^a={\left(x\times y\right)}^a\right] \\ ={\left(56\right)}^{\frac{1}{2}} \end{gathered}$$
Hence, the correct answer is option (b).

Question 45:

After simplification, $\frac{{13}^{{\displaystyle \frac{1}{5}}}}{{13}^{{\displaystyle \frac{1}{3}}}}$ is
(a) ${13}^{\frac{2}{15}}$

(b) ${13}^{\frac{8}{15}}$

(c) ${13}^{\frac{1}{3}}$

(d) ${13}^{\frac{-2}{15}}$

Answer 45:

$$\begin{gathered} \frac{{13}^{{\displaystyle \frac{1}{5}}}}{{13}^{{\displaystyle \frac{1}{3}}}} \\ ={13}^{\frac{1}{5}-\frac{1}{3}} \left(\frac{x^a}{x^b}=x^{a-b}\right) \\ ={13}^{\frac{3-5}{15}} \\ ={13}^{\frac{-2}{15}} \end{gathered}$$
Hence, the correct answer is option (d).

Question 46:

The value of $\sqrt[4]{{\left(64\right)}^{-2}}$ is

(a) $\frac{1}{8}$

(b) $\frac{1}{2}$

(c) 8

(d) $\frac{1}{64}$

Answer 46:

$$\begin{gathered} \sqrt[4]{{\left(64\right)}^{-2}}={\left[{\left(64\right)}^{-2}\right]}^{\frac{1}{4}} \\ ={\left[{\left(2^6\right)}^{-2}\right]}^{\frac{1}{4}} \\ ={\left[2^{-12}\right]}^{\frac{1}{4}} \\ =2^{-3} \\ =\frac{1}{2^3} \\ =\frac{1}{8} \end{gathered}$$

$\therefore$ The value of $\sqrt[4]{{\left(64\right)}^{-2}}$ is $\frac{1}{8}$.

Hence, the correct option is (a).

Question 47:

The value of $\frac{2^0+7^0}{5^0}$ is
(a) 0

(b) 2

(c) $\frac{9}{5}$

(d) $\frac{1}{5}$

Answer 47:

$$\begin{gathered} \frac{2^0+7^0}{5^0}=\frac{1+1}{1} \\ =\frac{2}{1} \\ =2 \end{gathered}$$

$\therefore$ The value of $\frac{2^0+7^0}{5^0}$ is 2.

Hence, the correct option is (b).

Question 48:

The value of ${\left(243\right)}^{\frac{1}{5}}$ is

(a) 3

(b) –3

(c) 5

(d) $\frac{1}{3}$

Answer 48:

$$\begin{gathered} {\left(243\right)}^{\frac{1}{5}}={\left(3^5\right)}^{\frac{1}{5}} \\ =3^1 \\ =3 \end{gathered}$$

$\therefore$ The value of ${\left(243\right)}^{\frac{1}{5}}$ is 3.

Hence, the correct option is (a).

Question 49:

$9^3+{\left(-3\right)}^3-6^3$ = ?
(a) 432
(b) 270
(c) 486
(d) 540

Answer 49:

$$\begin{gathered} 9^3+{\left(-3\right)}^3-6^3=\left(729\right)+\left(-27\right)-216 \\ =729-27-216 \\ =729-\left(27+216\right) \\ =729-243 \\ =486 \\ \therefore 9^3+{\left(-3\right)}^3-6^3=486 \end{gathered}$$

Hence, the correct option is (c).

Question 50:

Simplified value of ${\left(16\right)}^{-\frac{1}{4}}\times \sqrt[4]{16}$ is
(a) 0
(b) 1
(c) 4
(d) 16

Answer 50:

$$\begin{gathered} {\left(16\right)}^{-\frac{1}{4}}\times \sqrt[4]{16}={\left(2^4\right)}^{-\frac{1}{4}}\times {\left(2^4\right)}^{\frac{1}{4}} \\ ={\left(2\right)}^{-1}\times {\left(2\right)}^1 \\ =2^{-1+1} \\ =2^0 \\ =1 \end{gathered}$$

$\therefore$Simplified value of ${\left(16\right)}^{-\frac{1}{4}}\times \sqrt[4]{16}$ is 1.

Hence, the correct option is (b).

Question 51:

The value of  $\sqrt[4]{\sqrt[3]{2^2}}$ is
(a) $2^{\frac{-1}{6}}$

(b) $2^{-6}$

(c) $2^{\frac{1}{6}}$

(d) $2^6$

Answer 51:

$$\begin{gathered} \sqrt[4]{\sqrt[3]{2^2}}={\left[{\left(2^2\right)}^{\frac{1}{3}}\right]}^{\frac{1}{4}} \\ ={\left[2^{\frac{2}{3}}\right]}^{\frac{1}{4}} \\ =2^{\frac{2}{12}} \\ =2^{\frac{1}{6}} \end{gathered}$$

$\therefore$The value of  $\sqrt[4]{\sqrt[3]{2^2}}$ is $2^{\frac{1}{6}}$.

Hence, the correct option is (c).

Question 52:

Simplified value of ${\left(25\right)}^{\frac{1}{3}}\times 5^{\frac{1}{3}}$ is
(a) 25
(b) 3
(c) 1
(d) 5

Answer 52:

$$\begin{gathered} {\left(25\right)}^{\frac{1}{3}}\times 5^{\frac{1}{3}}={\left(5^2\right)}^{\frac{1}{3}}\times 5^{\frac{1}{3}} \\ =5^{\frac{2}{3}}\times 5^{\frac{1}{3}} \\ =5^{\frac{2}{3}+\frac{1}{3}} \\ =5^{\frac{2+1}{3}} \\ =5^{\frac{3}{3}} \\ =5 \end{gathered}$$

$\therefore$Simplified value of ${\left(25\right)}^{\frac{1}{3}}\times 5^{\frac{1}{3}}$ is 5.

Hence, the correct option is (d).

Question 53:

The value of ${\left[{\left(81\right)}^{\frac{1}{2}}\right]}^{\frac{1}{2}}$ is
(a) 3
(b) –3
(c) 9
(d) $\frac{1}{3}$

Answer 53:

$$\begin{gathered} {\left[{\left(81\right)}^{\frac{1}{2}}\right]}^{\frac{1}{2}}={\left[{\left(3^4\right)}^{\frac{1}{2}}\right]}^{\frac{1}{2}} \\ ={\left[3^2\right]}^{\frac{1}{2}} \\ =3 \end{gathered}$$

$\therefore$The value of ${\left[{\left(81\right)}^{\frac{1}{2}}\right]}^{\frac{1}{2}}$ is 3.

Hence, the correct option is (a).

Question 54:

There is a number x such that x² is irrational but x⁴ is rational. Then, x can be

(a) $\sqrt{5}$

(b) $\sqrt{2}$

(c) $\sqrt[3]{2}$

(d) $\sqrt[4]{2}$

Answer 54:

$$\begin{gathered} \left(\mathrm{a}\right) \mathrm{Let} x=\sqrt{5}. \\ {x}^2={\left(\sqrt{5}\right)}^2=5, \mathrm{which} \mathrm{is} \mathrm{a} \mathrm{rational} \mathrm{number}. \\ \left(\mathrm{b}\right) \mathrm{Let} x=\sqrt{2}. \\ {x}^2={\left(\sqrt{2}\right)}^2=2, \mathrm{which} \mathrm{is} \mathrm{a} \mathrm{rational} \mathrm{number}. \\ \left(\mathrm{c}\right) \mathrm{Let} x=\sqrt[3]{2}. \\ {x}^2={\left(\sqrt[3]{2}\right)}^2={\left(2\right)}^{\frac{2}{3}}, \mathrm{which} \mathrm{is} \mathrm{an} \mathrm{irrational} \mathrm{number}. \\ {x}^4={\left(\sqrt[3]{2}\right)}^4={\left(2\right)}^{\frac{4}{3}}, \mathrm{which} \mathrm{is} \mathrm{also} \mathrm{an} \mathrm{irrational} \mathrm{number}. \\ \left(\mathrm{d}\right) \mathrm{Let} x=\sqrt[4]{2}. \\ {x}^2={\left(\sqrt[4]{2}\right)}^2={\left(2\right)}^{\frac{2}{4}}={\left(2\right)}^{\frac{1}{2}}, \mathrm{which} \mathrm{is} \mathrm{an} \mathrm{irrational} \mathrm{number}. \\ {x}^4={\left(\sqrt[4]{2}\right)}^4={\left(2\right)}^{\frac{4}{4}}=2, \mathrm{which} \mathrm{is} \mathrm{a} \mathrm{rational} \mathrm{number}. \end{gathered}$$

$\therefore$ x can be $\sqrt[4]{2}$.

Hence, the correct option is (d).

PAGE NO -62

Question 55:

If $x=\frac{\sqrt{7}}{5} \mathrm{and} \frac{5}{x}=p\sqrt{7}$ then the value of p is
(a) $\frac{7}{25}$

(b) $\frac{25}{7}$

(c) $\frac{7}{15}$

(d) $\frac{15}{7}$

Answer 55:

$$\begin{gathered} \frac{5}{x}=p\sqrt{7} \\ \Rightarrow \frac{5}{{\displaystyle \frac{\sqrt{7}}{5}}}=p\sqrt{7} \left(\because x=\frac{\sqrt{7}}{5}\right) \\ \Rightarrow \frac{5\times 5}{{\displaystyle \sqrt{7}}}=p\sqrt{7} \\ \Rightarrow \frac{25}{{\displaystyle \sqrt{7}}}=p\sqrt{7} \\ \Rightarrow \frac{25}{{\displaystyle \sqrt{7}\times \sqrt{7}}}=p \\ \Rightarrow \frac{25}{{\displaystyle 7}}=p \\ \therefore \mathrm{The} \mathrm{value} \mathrm{of} p \mathrm{is} \frac{25}{7}. \end{gathered}$$

Hence, the correct option is (b).

Question 56:

The value of ${\left[\frac{256x^{16}}{81y^4}\right]}^{-\frac{1}{4}}$ is
(a) $\frac{3y}{8x^4}$

(b) $\frac{3y}{4x^4}$

(c) $\frac{4y}{5x^4}$

(d) $\frac{4x^4}{3y}$

Answer 56:

$$\begin{gathered} {\left[\frac{256x^{16}}{81y^4}\right]}^{-\frac{1}{4}}={\left[\frac{2^8 x^{16}}{3^4 y^4}\right]}^{-\frac{1}{4}} \\ ={\left[\frac{3^4 y^4}{2^8 x^{16}}\right]}^{\frac{1}{4}} \\ =\left[\frac{{\left(3^4\right)}^{{\displaystyle \frac{1}{4}}} {\left(y^4\right)}^{{\displaystyle \frac{1}{4}}}}{{\left(2^8\right)}^{{\displaystyle \frac{1}{4}}} {\left(x^{16}\right)}^{{\displaystyle \frac{1}{4}}}}\right] \\ =\left[\frac{3 y}{2^2 x^4}\right] \\ =\frac{3y}{4x^4} \\ \therefore \mathrm{The} \mathrm{value} \mathrm{of} {\left[\frac{256x^{16}}{81y^4}\right]}^{-\frac{1}{4}} \mathrm{is} \frac{3y}{4x^4}. \end{gathered}$$

Hence, the correct option is (b).

Question 57:

The value of $x^{p-q}·x^{q-r}·x^{r-p}$ is equal to
(a) 0

(b) 1

(c) x

(d) x^pqr

Answer 57:

$$\begin{gathered} x^{p-q}·x^{q-r}·x^{r-p}=x^{p-q+q-r+r-p} \\ =x^0 \\ =1 \\ \therefore \mathrm{The} \mathrm{value} \mathrm{of} x^{p-q}·x^{q-r}·x^{r-p} \mathrm{is} \mathrm{equal} \mathrm{to} 1. \end{gathered}$$

Hence, the correct option is (b).

Question 58:

The value of $\sqrt{p^{-1}q}·\sqrt{q^{-1}r}·\sqrt{r^{-1}p}$ is
(a) –1

(b) 0

(c) 1

(d) 2

Answer 58:

$$\begin{gathered} \sqrt{p^{-1}q}·\sqrt{q^{-1}r}·\sqrt{r^{-1}p}={\left(p^{-1}q\right)}^{\frac{1}{2}}·{\left(q^{-1}r\right)}^{\frac{1}{2}}·{\left(r^{-1}p\right)}^{\frac{1}{2}} \\ =\left(p^{-\frac{1}{2}}{q}^{\frac{1}{2}}\right).\left(q^{-\frac{1}{2}}{r}^{\frac{1}{2}}\right)·\left(r^{-\frac{1}{2}}{p}^{\frac{1}{2}}\right) \\ =p^{-\frac{1}{2}+\frac{1}{2}}{q}^{\frac{1}{2}-\frac{1}{2}}{r}^{\frac{1}{2}-\frac{1}{2}} \\ =p^0{q}^0{r}^0 \\ =1 \\ \therefore \mathrm{The} \mathrm{value} \mathrm{of} \sqrt{p^{-1}q}·\sqrt{q^{-1}r}·\sqrt{r^{-1}p} \mathrm{is} \mathrm{equal} \mathrm{to} 1. \end{gathered}$$

Hence, the correct option is (c).

Question 59:

$\sqrt[3]{2}\times \sqrt[4]{2}\times \sqrt[12]{32}=?$

(a) 2

(b) $\sqrt{2}$

(c) $2\sqrt{2}$

(d) $4\sqrt{2}$

Answer 59:

$$\begin{gathered} \sqrt[3]{2}\times \sqrt[4]{2}\times \sqrt[12]{32}={\left(2\right)}^{\frac{1}{3}}\times {\left(2\right)}^{\frac{1}{4}}\times {\left(32\right)}^{\frac{1}{12}} \\ ={\left(2\right)}^{\frac{1}{3}}\times {\left(2\right)}^{\frac{1}{4}}\times {\left(2^5\right)}^{\frac{1}{12}} \\ ={\left(2\right)}^{\frac{1}{3}}\times {\left(2\right)}^{\frac{1}{4}}\times {\left(2\right)}^{\frac{5}{12}} \\ ={\left(2\right)}^{\frac{1}{3}+\frac{1}{4}+\frac{5}{12}} \\ ={\left(2\right)}^{\frac{4+3+5}{12}} \\ ={\left(2\right)}^{\frac{12}{12}} \\ =2 \\ \therefore \sqrt[3]{2}\times \sqrt[4]{2}\times \sqrt[12]{32}=2. \end{gathered}$$

Hence, the correct option is (a).

Question 60:

If ${\left(\frac{2}{3}\right)}^x {\left(\frac{3}{2}\right)}^{2x}=\frac{81}{16} \mathrm{then} x=?$
(a) 1
(b) 2
(c) 3
(d) 4

Answer 60:

$$\begin{gathered} {\left(\frac{2}{3}\right)}^x {\left(\frac{3}{2}\right)}^{2x}=\frac{81}{16} \\ \Rightarrow {\left(\frac{2}{3}\right)}^x{\left(\frac{2}{3}\right)}^{-2x}=\frac{81}{16} \\ \Rightarrow {\left(\frac{2}{3}\right)}^{x-2x}=\frac{81}{16} \\ \Rightarrow {\left(\frac{2}{3}\right)}^{-x}=\frac{3^4}{2^4} \\ \Rightarrow {\left(\frac{3}{2}\right)}^x=\frac{3^4}{2^4} \\ \Rightarrow x=4 \end{gathered}$$
Hence, the correct answer is option (d).

Question 61:

If ${\left(3^3\right)}^2=9^x \mathrm{then} 5^x=?$
(a) 1
(b) 5
(c) 25
(d) 125

Answer 61:

$$\begin{gathered} {\left(3^3\right)}^2=9^x \\ \Rightarrow {\left(3^2\right)}^3={\left(3^2\right)}^x \\ \Rightarrow x=3 \\ \mathrm{Now} \\ {5}^x=5^3=125 \end{gathered}$$
Hence, the correct answer is option (d).

Question 62:

On simplification, the expression $\frac{5^{n+2}-6\times 5^{n+1}}{13\times 5^n-2\times 5^{n+1}}$ equals
(a) $\frac{5}{3}$

(b) $-\frac{5}{3}$

(c) $\frac{3}{5}$

(d) $-\frac{3}{5}$

Answer 62:

$$\begin{gathered} \frac{5^{n+2}-6\times 5^{n+1}}{13\times 5^n-2\times 5^{n+1}} \\ =\frac{5^n\times 5^2-6\times 5^n\times 5}{13\times 5^n-2\times 5^n\times 5} \\ =\frac{5^n\times 5\left[5-6\right]}{5^n\left[13-2\times 5\right]} \\ =\frac{-5}{3} \end{gathered}$$
Hence, the correct answer is option (b). 

Question 63:

The simplest rationalisation factor of $\sqrt[3]{500}$ is
(a) $\sqrt{5}$
(b) $\sqrt{3}$
(c) $\sqrt[3]{5}$
(d) $\sqrt[3]{2}$

Answer 63:

$\sqrt[3]{500}=\sqrt[3]{5\times 5\times 2\times 5\times 2}=\sqrt[3]{5^3\times 2^2}=5\sqrt[3]{4}$  
So, the simplest rationalisation factor of $\sqrt[3]{500}$ is $\sqrt[3]{2}$.
Hence, the correct answer is option (d). 

Question 64:

The simplest rationalisation ractor of $\left(2\sqrt{2}-\sqrt{3}\right)$ is
(a) $2\sqrt{2}+3$
(b) $2\sqrt{2}+\sqrt{3}$
(c) $\sqrt{2}+\sqrt{3}$
(d) $\sqrt{2}-\sqrt{3}$

Answer 64:

Simplest rationalisation ractor of $\left(2\sqrt{2}-\sqrt{3}\right)$ is $2\sqrt{2}+\sqrt{3}$.
Hence, the correct answer is option (b). 

Question 65:

The rationalisation factor of $\frac{1}{2\sqrt{3}-\sqrt{5}}$ is
(a) $\sqrt{5}-2\sqrt{3}$
(b) $\sqrt{3}+2\sqrt{5}$
(c) $\left(\sqrt{3}+\sqrt{5}\right)$
(d) $\sqrt{12}+\sqrt{5}$

Answer 65:

Rationalisation factor of $\frac{1}{2\sqrt{3}-\sqrt{5}}$ will be $2\sqrt{3}+\sqrt{5}=\sqrt{4\times 3}+\sqrt{5}=\sqrt{12}+\sqrt{5}$.
Hence, the correct answer is option (d). 

Question 66:

Rationalisation of the denominator of $\frac{1}{\sqrt{5}+\sqrt{2}}$ gives
(a) $\frac{1}{\sqrt{10}}$
(b) $\sqrt{5}+\sqrt{2}$
(c) $\sqrt{5}-\sqrt{2}$
(d) $\frac{\sqrt{5}{\displaystyle -}{\displaystyle \sqrt{2}}}{3}$

Answer 66:

$\frac{1}{\sqrt{5}+\sqrt{2}}$
Rationalisation of denominator gives
$\frac{1}{\sqrt{5}+\sqrt{2}}\times \frac{\sqrt{5}-\sqrt{2}}{\sqrt{5}-\sqrt{2}}=\frac{\sqrt{5}-\sqrt{2}}{5-2}=\frac{\sqrt{5}-\sqrt{2}}{3}$
Hence, the correct answer is option (d). 

Question 67:

If $x=2+\sqrt{3} \mathrm{then} \left(x+\frac{1}{x}\right)$ equals
(a) $-2\sqrt{3}$
(b) 2
(c) 4
(d) $4-2\sqrt{3}$

Answer 67:

$$\begin{gathered} x=2+\sqrt{3} \\ \frac{1}{x}=\frac{1}{2+\sqrt{3}}=\frac{1}{2+\sqrt{3}}\times \frac{2-\sqrt{3}}{2-\sqrt{3}}=\frac{2-\sqrt{3}}{4-3}=2-\sqrt{3} \end{gathered}$$
$x+\frac{1}{x}=2+\sqrt{3}+2-\sqrt{3}=4$
Hence, the correct answer is option (c). 

PAGE NO -63

Question 68:

$\frac{1}{\left(3+2\sqrt{2}\right)}=?$

(a) $\frac{{\displaystyle 3-2\sqrt{2}}}{{\displaystyle 17}}$

(b) $\frac{\left(3-2\sqrt{2}\right)}{13}$

(c) $\left(3-2\sqrt{2}\right)$

(d) None of these

Answer 68:

$\frac{1}{\left(3+2\sqrt{2}\right)}=\frac{1}{\left(3+2\sqrt{2}\right)}\times \frac{\left(3-2\sqrt{2}\right)}{\left(3-2\sqrt{2}\right)}=\frac{\left(3-2\sqrt{2}\right)}{9-8}=\left(3-2\sqrt{2}\right)$
Hence, the correct answer is option (c). 

Question 69:

If $x=\left(7+4\sqrt{3}\right) \mathrm{then} \left(x+\frac{1}{x}\right)=?$
(a) $8\sqrt{3}$
(b) 14
(c) 49
(d) 48

Answer 69:

Given: $x=\left(7+4\sqrt{3}\right)$
$\frac{1}{x}=\frac{1}{7+4\sqrt{3}}=\frac{1}{7+4\sqrt{3}}\times \frac{7-4\sqrt{3}}{7-4\sqrt{3}}=\frac{7-4\sqrt{3}}{49-48}=7-4\sqrt{3}$
$\left(x+\frac{1}{x}\right)=7+4\sqrt{3}+\left(7-4\sqrt{3}\right)=14$
Hence, the correct answer is option (b). 

Question 70:

If $\sqrt{2}=1.41 \mathrm{then} \frac{1}{\sqrt{2}}=?$
(a) 0.075
(b) 0.75
(c) 0.705
(d) 7.05

Answer 70:

$$\begin{gathered} \frac{1}{\sqrt{2}}=\frac{{\displaystyle 1}}{{\displaystyle \sqrt{2}}}\times \frac{\sqrt{2}}{\sqrt{2}}=\frac{\sqrt{2}}{2} \\ \mathrm{Given}: \sqrt{2}=1.41 \\ \mathrm{So}, \frac{{\displaystyle \sqrt{2}}}{{\displaystyle 2}}=\frac{1.41}{2}=0.705 \end{gathered}$$
Hence, the correct answer is option (c). 

Question 71:

If $\sqrt{7}=2.646 \mathrm{then} \frac{1}{\sqrt{7}}$ =?
(a) 0.375
(b) 0.378
(c) 0.441
(d) None of these

Answer 71:

$\frac{1}{\sqrt{7}}=\frac{1}{\sqrt{7}}\times \frac{\sqrt{7}}{\sqrt{7}}=\frac{\sqrt{7}}{7}$
Given that 
$$\begin{gathered} \sqrt{7}=2.646 \\ \mathrm{So}, \frac{\sqrt{7}}{7}=\frac{2.646}{7}=0.378 \end{gathered}$$
Hence, the correct answer is option (b). 

Question 72:

The value of $\sqrt{3-2\sqrt{2}}$ is
(a) $\sqrt{3}+\sqrt{2}$
(b) $\sqrt{3}-\sqrt{2}$
(c) $\sqrt{2}+1$
(d) $\sqrt{2}-1$

Answer 72:

$$\begin{gathered} 3-2\sqrt{2}=2+1-2\times \sqrt{2}\times 1 \\ ={\left(\sqrt{2}\right)}^2+1^2-2\times \sqrt{2}\times 1 \end{gathered}$$
This is of the form
$a^2+b^2-2ab={\left(a-b\right)}^2$
$$\begin{gathered} {\left(\sqrt{2}\right)}^2+1^2-2\times \sqrt{2}\times 1={\left(\left(\sqrt{2}\right)-1\right)}^2 \\ \mathrm{So}, \\ \sqrt{3-2\sqrt{2}}=\sqrt{{\left(\left(\sqrt{2}\right)-1\right)}^2}=\sqrt{2}-1 \end{gathered}$$
Hence, the correct answer is option (d). 
 

Question 73:

The value of $\sqrt{5+2\sqrt{6}}$ is
(a) $\sqrt{5}+\sqrt{6}$
(b) $\sqrt{5}-\sqrt{6}$
(c) $\sqrt{3}+\sqrt{2}$
(d) $\sqrt{3}-\sqrt{2}$

Answer 73:

$$\begin{gathered} 5+2\sqrt{6}=2+3+2\times \sqrt{3}\times \sqrt{2} \\ ={\left(\sqrt{2}\right)}^2+{\left(\sqrt{3}\right)}^2+2\times \sqrt{3}\times \sqrt{2} \end{gathered}$$
This is in the form
$a^2+b^2+2ab={\left(a+b\right)}^2$
So, we have 
${\left(\sqrt{2}\right)}^2+{\left(\sqrt{3}\right)}^2+2\times \sqrt{3}\times \sqrt{2}={\left(\sqrt{2}+\sqrt{3}\right)}^2$
Thus, $\sqrt{5+2\sqrt{6}}=\sqrt{{\left(\sqrt{2}+\sqrt{3}\right)}^2}=\sqrt{2}+\sqrt{3}$
Hence, the correct answer is option (c). 

Question 74:

If $\sqrt{2}=1.414 \mathrm{then} \sqrt{\frac{\left(\sqrt{2}-1\right)}{\left(\sqrt{2}+1\right)}}=?$
(a) 0.207
(b) 2.414
(c) 0.414
(d) 0.621

Answer 74:

$$\begin{gathered} \sqrt{\frac{\left(\sqrt{2}-1\right)}{\left(\sqrt{2}+1\right)}}=\sqrt{\frac{\left(\sqrt{2}-1\right)}{\left(\sqrt{2}+1\right)}\times \frac{\left(\sqrt{2}-1\right)}{\left(\sqrt{2}-1\right)}}=\sqrt{\frac{{\left(\sqrt{2}-1\right)}^2}{2-1}} \\ =\sqrt{\frac{2+1-2\sqrt{2}}{1}} \\ =\sqrt{3-2\sqrt{2}} \\ =\sqrt{3-2\times 1.414} \\ =\sqrt{3-2.828} \\ =\sqrt{0.172} \\ =0.414 \end{gathered}$$
Hence, the correct answer is option (c).

Question 75:

If $x=3+\sqrt{8} \mathrm{then} \left(x^2+\frac{1}{x^2}\right)=?$
(a) 34
(b) 56
(c) 28
(d) 63

Answer 75:

Given: $x=3+\sqrt{8}$
$\frac{1}{x}=\frac{1}{3+\sqrt{8}}=\frac{1}{3+\sqrt{8}}\times \frac{3-\sqrt{8}}{3-\sqrt{8}}=\frac{3-\sqrt{8}}{9-8}=\frac{3-\sqrt{8}}{1}=3-\sqrt{8}$
$x+\frac{1}{x}=\left(3+\sqrt{8}\right)+\left(3-\sqrt{8}\right)=6$
$$\begin{gathered} {\left(x+\frac{1}{x}\right)}^2=x^2+\frac{1}{x^2}+2\times x\times \frac{1}{x}=x^2+\frac{1}{x^2}+2 \\ \Rightarrow 6^2=x^2+\frac{1}{x^2}+2 \\ \Rightarrow 36=x^2+\frac{1}{x^2}+2 \\ \Rightarrow x^2+\frac{1}{x^2}=36-2=34 \end{gathered}$$
Hence, the correct answer is option (a). 

 

Question 76:

Assertion: Three rational numbers between $\frac{2}{5} \mathrm{and} \frac{3}{5}\mathrm{are}\frac{9}{20}, \frac{10}{20}\mathrm{and} \frac{11}{20}$.
Reason: A rational number between two rational numbers p and q is $\frac{1}{2}\left(p+q\right)$.
(a) Both Assertion and Reason are true and Reasom is a correct explanation of Assertion.
(b) Both Assertion and Reason and Reasom are true but Reasom is not a correct explanation of Assertion.
(c) Assertion is true and Reasom is false.
(d) Assertion is false and Reasom is true.

Answer 76:

(a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
$$\begin{gathered} \mathrm{Rational} \mathrm{number} \mathrm{between} \frac{2}{5} \mathrm{and} \frac{3}{5}: \\ \frac{\frac{2}{5}+\frac{3}{5}}{2}=\frac{1}{2}=\frac{10}{20} \\ \mathrm{Rational} \mathrm{number} \mathrm{between} \frac{2}{5} \mathrm{and} \frac{10}{20}: \\ \frac{\frac{2}{5}+\frac{10}{20}}{2}=\frac{18}{40}=\frac{9}{20} \\ \mathrm{Rational} \mathrm{number} \mathrm{between} \frac{3}{5} \mathrm{and} \frac{10}{20}: \\ \frac{\frac{3}{5}+\frac{10}{20}}{2}=\frac{22}{40}=\frac{11}{20} \end{gathered}$$

So, Assertion and Reason are correct (property of rational numbers). Also, Reason is the correct explanation of Assertion.

PAGE NO-64

Question 77:

Assertion: $\sqrt{3}$is an irrational number.
Reason: Square root a positive integer which is not a perfect square is an irrational number.
(a) Both Assertion and Reason are true and Reasom is a correct explanation of Assertion.
(b) Both Assertion and Reason and Reasom are true but Reasom is not a correct explanation of Assertion.
(c) Assertion is true and Reasom is false.
(d) Assertion is false and Reasom is true.

Answer 77:

(a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

$\sqrt{3}\mathrm{is} \mathrm{not} \mathrm{a} \mathrm{perfect} \mathrm{square}; \mathrm{hence}, \mathrm{it} \mathrm{is} \mathrm{irrational}.$ 3√ 3√$\sqrt{3} is not a perfect square and hence is irrational and the reason is correct explanation for the assertion thus \left(a\right) is correct$

Question 78:

Assertion: e is irrational number.
Reason: $\mathrm{\pi }$ is an irrational number.
(a) Both Assertion and Reason are true and Reasom is a correct explanation of Assertion.
(b) Both Assertion and Reason and Reasom are true but Reasom is not a correct explanation of Assertion.
(c) Assertion is true and Reasom is false.
(d) Assertion is false and Reasom is true.

Answer 78:

(b) Both Assertion and Reason are true, but Reason is not a correct explanation of Assertion.

It is known that e and $\mathrm{\pi }$ are irrational numbers, but Reason is not the correct explanation.

Question 79:

Assertion: $\sqrt{3}$is an irrational number.
Reason: The sum of rational number and an irrational number is an irrational number.
(a) Both Assertion and Reason are true and Reasom is a correct explanation of Assertion.
(b) Both Assertion and Reason and Reasom are true but Reasom is not a correct explanation of Assertion.
(c) Assertion is true and Reasom is false.
(d) Assertion is false and Reasom is true.

Answer 79:

(b) Both Assertion and Reason are true, but Reason is not a correct explanation of Assertion.
$$\begin{gathered} \sqrt{3} \mathrm{is} \mathrm{not} \mathrm{a} \mathrm{perfect} \mathrm{square} \mathrm{and} \mathrm{is} \mathrm{irrational}. \\ \mathrm{Reason}: \mathrm{Let} \mathrm{the} \mathrm{sum} \mathrm{of} \mathrm{a} \mathrm{rational} \mathrm{numbe}r a \mathrm{and} \mathrm{an} \mathrm{irrational} \mathrm{number} \sqrt{b} \mathrm{be} \mathrm{a} \mathrm{rational} \mathrm{number} c. \\ \mathrm{Thus}, \mathrm{we} \mathrm{have}: \\ a +\sqrt{b}=c \\ \Rightarrow \sqrt{b}=c-a \\ \mathrm{Now}, c-a \mathrm{is} \mathrm{rational} \mathrm{because} \mathrm{both} c \mathrm{and} a \mathrm{are} \mathrm{rational}, \mathrm{but} \sqrt{b} \mathrm{is} \mathrm{irrational}; \mathrm{thus}, \mathrm{we} \mathrm{arrive} \mathrm{at} \mathrm{a} \mathrm{contradiction}. \\ \mathrm{Hence}, \mathrm{the} \mathrm{sum} \mathrm{of} \mathrm{a} \mathrm{rational} \mathrm{number} \mathrm{and} \mathrm{an} \mathrm{irrational} \mathrm{number} \mathrm{is} \mathrm{an} \mathrm{irrational} \mathrm{number}. \\ \mathrm{Thus}, \mathrm{Reason} \mathrm{R} \mathrm{is} \mathrm{not} \mathrm{a} \mathrm{correct} \mathrm{explanation}. \end{gathered}$$

Question 80:

Match the following columns:

Column I		Column II
(a)	$6.\overline{54}$ is ....... .	(p)	14
(b)	$\mathrm{\pi }$ is ...... .	(q)	6
(c)	The length of period of $\frac{1}{7}$=...... .	(r)	a rational number
(d)	If $x=\left(2-\sqrt{3}\right)$, then $\left(x^2+\frac{1}{x^2}\right)$....... .	(s)	an irrational number

(a) ........
(b) ........
(c) ........
(d) ........

Answer 80:

(a) Because it is a non-terminating and repeating decimal, it is a rational number.

(b) $\mathrm{\pi }$ is an irrational number.

(c) $\frac{1}{7}=.142857142857$...
Hence, its period is 6.
                                                     
(d)
$$\begin{gathered} x^2+\frac{1}{x^2} \\ ={\left(2-\sqrt{3}\right)}^2+\frac{1}{{\left(2-\sqrt{3}\right)}^2} \\ =\left(2^2+{\left(\sqrt{3}\right)}^3-2\times 2\times \sqrt{3}\right)+\frac{1}{\left(2^2+{\left(\sqrt{3}\right)}^3-2\times 2\times \sqrt{3}\right)} \\ =\left(4+3-4\times \sqrt{3}\right)+\frac{1}{\left(4+3-4\times \sqrt{3}\right)} \\ =\left(7-4\times \sqrt{3}\right)+\frac{1}{\left(7-4\times \sqrt{3}\right)} \\ =\frac{{\left(7-4\times \sqrt{3}\right)}^2}{\left(7-4\times \sqrt{3}\right)}+\frac{1}{\left(7-4\times \sqrt{3}\right)} \\ =\frac{7^2+{\left(4\sqrt{3}\right)}^2-2\times 7\times 4\sqrt{3}+1}{\left(7-4\times \sqrt{3}\right)} \\ =\frac{49+48-56\sqrt{3}+1}{\left(7-4\times \sqrt{3}\right)} \\ =\frac{98 -56\sqrt{3}}{\left(7-4\times \sqrt{3}\right)} \\ =14\times \frac{\left(\left(7-4\times \sqrt{3}\right)\right)}{\left(7-4\times \sqrt{3}\right)} \\ =14 \end{gathered}$$

Question 81:

Match the following columns:

Column I		Column II
(a)	$\sqrt[4]{{\left(81\right)}^{-2}}=...... .$	(p)	4
(b)	If ${\left(\frac{a}{b}\right)}^{x-2}={\left(\frac{b}{a}\right)}^{x-4}$, then x = ........ .	(q)	$\frac{2}{9}$
(c)	If $x=\left(9+4\sqrt{5}\right)$, then $\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)$= ...... .	(r)	$\frac{1}{9}$
(d)	${\left(\frac{81}{16}\right)}^{-3/4}\times {\left(\frac{64}{27}\right)}^{-1/3}=?$	(s)	3

(a) ......
(b) ......
(c) ......
(d) ......

Answer 81:

(a)
$$\begin{gathered} {\left({\left(81\right)}^{-2}\right)}^{\frac{1}{4}} \\ ={\left({\left(9\right)}^{-4}\right)}^{\frac{1}{4}}={\left(9\right)}^{-4\times \frac{1}{4}}={\left(9\right)}^{-1} \\ =\frac{1}{9} \end{gathered}$$     

(b)

$$\begin{gathered} {\left(\frac{a}{b}\right)}^{x-2}={\left(\frac{b}{a}\right)}^{x-4} \\ \Rightarrow {\left(\frac{b}{a}\right)}^{2-x}={\left(\frac{b}{a}\right)}^{x-4} \\ \Rightarrow 2-x=x-4 \\ \Rightarrow 2x=6 \\ \Rightarrow x=3 \end{gathered}$$

(c)

$$\begin{gathered} x=9+4\sqrt{5} \\ \mathrm{and} \\ \frac{1}{x}=\frac{1}{9+4\sqrt{5}}\times \frac{9-4\sqrt{5}}{9-4\sqrt{5}} \\ =\frac{9-4\sqrt{5}}{81-80} \\ =9-4\sqrt{5} \end{gathered}$$
$$\begin{gathered} \mathrm{Now}, \\ x+\frac{1}{x}=9+4\sqrt{5}+9-4\sqrt{5}=18 \\ \mathrm{Thus}, \mathrm{we} \mathrm{have}: \\ x+\frac{1}{x}=18 \\ \mathrm{We} \mathrm{know}: \\ {\left(\sqrt{\mathrm{x}}-\frac{1}{\sqrt{\mathrm{x}}}\right)}^2=\mathrm{x}+\frac{1}{\mathrm{x}}-2\times \mathrm{x}\times \frac{1}{\mathrm{x}} \\ \Rightarrow {\left(\sqrt{\mathrm{x}}-\frac{1}{\sqrt{\mathrm{x}}}\right)}^2=18-2 \\ \Rightarrow {\left(\sqrt{\mathrm{x}}-\frac{1}{\sqrt{\mathrm{x}}}\right)}^2=16 \\ \mathrm{Taking} \mathrm{the} \mathrm{square} \mathrm{root} \mathrm{of} \mathrm{both} \mathrm{sides}, \mathrm{we} \mathrm{get}: \\ \sqrt{\mathrm{x}}-\frac{1}{\sqrt{\mathrm{x}}}=4 \end{gathered}$$

(d)

$$\begin{gathered} {\left(\frac{3}{2}\right)}^{4\times \frac{-3}{4}}\times {\left(\frac{4}{3}\right)}^{3\times \frac{-1}{3}} \\ ={\left(\frac{3}{2}\right)}^{-3}\times {\left(\frac{4}{3}\right)}^{-1} \\ ={\left({\displaystyle \frac{3}{2}}\right)}^{-3}\times \frac{3}{4} \\ =\left({\displaystyle \frac{3^{-3}}{2^{-3}}}\right)\times \frac{3}{2^2} \\ =\left({\displaystyle \frac{3^{-3}\times 3}{2^{-3}\times 2^2}}\right) \\ =\frac{3^{-2}}{2^{-1}} \\ =\frac{2}{9} \end{gathered}$$