Rational & Irrational Numbers — Multiple Choice Questions

Which of the following is a rational number?

1. π

2. $\sqrt{2}$

3. 3.4

4. 1.010010001..

Answer

3.4 is a terminating decimal, thus it is a rational number.

Hence, Option 3 is the correct option.

Which of the following is an irrational number?

1. 2.7

2. $2.7\overline{2}$

3. $\sqrt{11}$

4. $\dfrac{2}{7}$

Answer

$\sqrt{11}$ is square root of non-perfect square i.e 11, thus it is an irrational number.

Hence, Option 3 is the correct option.

Which of the following is a prime number?

1. 51

2. 57

3. 71

4. 81

Answer

71 has only two factors i.e is 1 and 71 itself, thus it is a prime number.

Hence, Option 3 is the correct option.

When $8.\overline{32}$ is expressed as a vulgar fraction, then it becomes :

1. $\dfrac{208}{824}$

2. $\dfrac{824}{99}$

3. $\dfrac{800}{99}$

4. $\dfrac{416}{45}$

Answer

Let x = $8.\overline{32}$

⇒ x = 8.32323232..     ..........(1)

Multiplying both sides by 100 (since there are two recurring digits)

⇒ 100x = 832.3232..     ..........(2)

Subtracting equation (1) from (2), we get :

⇒ 100x - x = 832.3232.. - 8.3232..

⇒ 99x = 824

⇒ x = $\dfrac{824}{99}$

Hence, Option 2 is the correct option.

0.3 when expressed as a ratio of two integers, becomes :

1. $\dfrac{103}{330}$

2. $\dfrac{52}{165}$

3. $\dfrac{103}{111}$

4. $\dfrac{104}{333}$

Answer

$\dfrac{103}{330} = 0.3\overline{12}$

$\dfrac{52}{165} = 0.3\overline{15}$

$\dfrac{103}{111} = 0.\overline{927}$

$\dfrac{104}{333} = 0.\overline{312}$

Since, $0.3\overline{12}$ is nearest to 0.3

Hence, Option 1 is the correct option.

Only by inspecting the prime factors of the denominator, state which of the following fractions will be a terminating decimal?

1. $\dfrac{7}{12}$

2. $\dfrac{2}{15}$

3. $\dfrac{3}{16}$

4. $\dfrac{4}{21}$

Answer

A rational number will have a terminating decimal expansion if and only if the prime factorization of its denominator contains only powers of 2 and/or 5.

The prime factor of 16 is only 2.

∴ $\dfrac{3}{16}$ will have a terminating decimal.

Hence, Option 3 is the correct option.

Only by inspecting the prime factors of the denominators, state which of the following fractions will be a recurring decimal?

1. $\dfrac{7}{16}$

2. $\dfrac{8}{51}$

3. $\dfrac{3}{25}$

4. $\dfrac{11}{20}$

Answer

A rational number will have a recurring decimal expansion if and only if the prime factorization of its denominator contains any prime factor other than 2 or 5.

The prime factors of 51 are 17 and 3.

Hence, Option 2 is the correct option.

The number which is to be subtracted from $\sqrt{72}$ to get $\sqrt{32}$ is:

1. $2\sqrt{10}$

2. $4\sqrt{2}$

3. $3\sqrt{2}$

4. $2\sqrt{2}$

Answer

Let the number to be subtarcted be x.

$\Rightarrow \sqrt{72} - x = \sqrt{32} \\[1em] \Rightarrow x = \sqrt{72} - \sqrt{32} \\[1em] \Rightarrow x = \sqrt{36 \times 2} - \sqrt{16 \times 2} \\[1em] \Rightarrow x = 6\sqrt{2} - 4\sqrt{2} \\[1em] \Rightarrow x = 2\sqrt{2}.$

Hence, Option 4 is the correct option.

If x = 3 + $2\sqrt{2}$, then $x + \dfrac{1}{x}$ equals to :

1. $4\sqrt{2}$

2. $6\sqrt{2}$

3. 6

4. 4

Answer

Given,

⇒ x = 3 + $2\sqrt{2}$

$\Rightarrow \dfrac{1}{x} = \dfrac{1}{(3 + 2\sqrt{2})}$

Rationalizing,

$\Rightarrow \dfrac{1}{(3 + 2\sqrt{2})} \times \dfrac{(3 - 2\sqrt{2})}{(3 - 2\sqrt{2})} \\[1em] \Rightarrow \dfrac{3 - 2\sqrt{2}}{3^2 - (2\sqrt{2})^2} \\[1em] \Rightarrow \dfrac{3 - 2\sqrt{2}}{9 - 8} \\[1em] \Rightarrow 3 - 2\sqrt{2}$

Adding x and $\dfrac{1}{x}$ we get

$\Rightarrow x + \dfrac{1}{x} = 3 + 2\sqrt{2} + 3 - 2\sqrt{2} \\[1em] \Rightarrow 6$

Hence, Option 3 is the correct option.

If x = 2 - $\sqrt{2}$, then $x - \dfrac{1}{x}$ =

1. 4

2. -4

3. $\dfrac{2 - 3\sqrt{2}}{2}$

4. $\dfrac{2 + 3\sqrt{2}}{2}$

Answer

Given,

⇒ x = 2 - $\sqrt{2}$

$\Rightarrow \dfrac{1}{x} = \dfrac{1}{(2 - \sqrt{2})}$

Rationalizing the denominator, we get :

$\Rightarrow \dfrac{1}{(2 - \sqrt{2})} \times \dfrac{(2 + \sqrt{2})}{(2 + \sqrt{2})} \\[1em] \Rightarrow \dfrac{2 + \sqrt{2}}{2^2 - (\sqrt{2})^2} \\[1em] \Rightarrow \dfrac{2 + \sqrt{2}}{4 - 2} \\[1em] \Rightarrow \dfrac{2 + \sqrt{2}}{2}$

Substituting values in $x - \dfrac{1}{x}$, we get :

$\Rightarrow 2 - \sqrt{2} - \dfrac{2 + \sqrt{2}}{2} \\[1em] \Rightarrow \dfrac{2(2 - \sqrt{2})-(2 + \sqrt{2})}{2} \\[1em] \Rightarrow \dfrac{4 - 2\sqrt{2} - 2 - \sqrt{2}}{2} \\[1em] \Rightarrow \dfrac{2 - 3\sqrt{2}}{2} \\[1em]$

$x - \dfrac{1}{x} = \dfrac{2 - 3\sqrt{2}}{2}$

Hence, Option 3 is the correct option.

If x = 5 + $2\sqrt{6}$, then $x^2 + \dfrac{1}{x^2}$ =

1. 98

2. 142

3. 49

4. 138

Answer

Given,

x = $(5 + 2\sqrt{6})$

$\therefore \dfrac{1}{x} = \dfrac{1}{(5 + 2\sqrt{6})}$

Rationalizing,

$\Rightarrow \dfrac{1}{x} = \dfrac{1}{5 + 2\sqrt{6}} \times \dfrac{(5 - 2\sqrt{6})}{(5 - 2\sqrt{6})} \\[1em] = \dfrac{5 - 2\sqrt{6}}{5^2 - (2\sqrt{6})^2} \\[1em] = \dfrac{5 - 2\sqrt{6}}{25 - 24} \\[1em] = \dfrac{5 - 2\sqrt{6}}{1} \\[1em] = 5 - 2\sqrt{6}.$

By formula,

$x^2 + \dfrac{1}{x^2} = \Big(x + \dfrac{1}{x}\Big)^2 - 2$

Substituting values we get :

$\Rightarrow x^2 + \dfrac{1}{x^2} = (5 + 2\sqrt{6} + 5 - 2\sqrt{6})^2 - 2 \\[1em] = 10^2 - 2 \\[1em] = 100 - 2 \\[1em] = 98.$

Hence, Option 1 is the correct option.

Two rational numbers between $-\dfrac{3}{7} \text{ and } -\dfrac{1}{7}$ is :

1. $\dfrac{4}{14}, \dfrac{3}{14}$

2. $-\dfrac{4}{14}, \dfrac{3}{14}$

3. $\dfrac{4}{14}, -\dfrac{3}{14}$

4. $-\dfrac{4}{14}, -\dfrac{3}{14}$

Answer

Let the first rational number between $-\dfrac{3}{7} \text{ and } -\dfrac{1}{7}$ be x.

$\Rightarrow x = \dfrac{1}{2}\Big[-\dfrac{3}{7} + \Big(-\dfrac{1}{7}\Big)\Big] \\[1em] \Rightarrow x = \dfrac{1}{2}\Big(-\dfrac{4}{7}\Big)\\[1em] \Rightarrow x = -\dfrac{4}{14}$

Let the second rational number be y.

$\Rightarrow y = \dfrac{1}{2}\Big[-\dfrac{4}{14} + \Big(-\dfrac{1}{7}\Big)\Big] \\[1em] \Rightarrow y = \dfrac{1}{2}\Big(\dfrac{-4 - 2}{14}\Big) \\[1em] \Rightarrow y = \dfrac{1}{2}\Big(-\dfrac{6}{14}\Big)\\[1em] \Rightarrow y = -\dfrac{6}{28} = -\dfrac{3}{14}$

Hence, Option 4 is the correct option.

The correct ascending order of $\sqrt{3} , \sqrt[3]{6}, \sqrt[4]{7}$ is :

1. $\sqrt[3]{6}, \sqrt[4]{7}, \sqrt{3}$

2. $\sqrt[4]{7}, \sqrt{3}, \sqrt[3]{6}$

3. $\sqrt{3}, \sqrt[4]{7}, \sqrt[3]{6}$

4. $\sqrt[3]{6}, \sqrt{3}, \sqrt[4]{7}$

Answer

On solving,

⇒ $\sqrt{3}$ = 1.732

⇒ $\sqrt[3]{6}$ = 1.817

⇒ $\sqrt[4]{7}$ = 1.626

∴ $\sqrt[4]{7} \lt \sqrt{3} \lt \sqrt[3]{6}$

Hence, Option 2 is the correct option.

The mixed surd for $\sqrt[3]{432}$ is :

1. $2\sqrt[3]{6}$

2. $6\sqrt[3]{2}$

3. $3\sqrt[3]{6}$

4. $6\sqrt[3]{3}$

Answer

Given,

$\Rightarrow \sqrt[3]{432} \\[1em] \Rightarrow \sqrt[3]{216 \times 2} \\[1em] \Rightarrow \sqrt[3]{216} \times \sqrt[3]{2} \\[1em] \Rightarrow 6\sqrt[3]{2}$

Hence, Option 2 is the correct option.

What is the pure surd for $5\sqrt[3]{2}$ ?

1. $\sqrt[3]{125}$

2. $\sqrt[3]{150}$

3. $\sqrt[3]{250}$

4. $\sqrt[3]{1000}$

Answer

Given,

$\Rightarrow 5\sqrt[3]{2} \\[1em] \Rightarrow \sqrt[3]{5^3 \times 2} \\[1em] \Rightarrow \sqrt[3]{125 \times 2} \\[1em] \Rightarrow \sqrt[3]{250}$

Hence, Option 3 is the correct option.