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91. $\sqrt{6 - 4 \sqrt{3} + \sqrt{16 - 8 \sqrt{3}}}$ is equal to = ?

A. $1 - \sqrt{3}$

B. $\sqrt{3} - 1$

C. $2 (2 - \sqrt{3})$

D. $2 (2 + \sqrt{3})$

Discuss

Solution:

\begin{aligned} \sqrt{6 - 4 \sqrt{3} + \sqrt{16 - 8 \sqrt{3}}} \\ = \sqrt{6 - 4 \sqrt{3} + \sqrt{12 + 4 - 8 \sqrt{3}}} \\ = \sqrt{6 - 4 \sqrt{3} + \sqrt{{(2 \sqrt{3})}^{2} + {(2)}^{2} - 2 \times 2 \sqrt{3} \times 2}} \\ = \sqrt{6 - 4 \sqrt{3} + \sqrt{{(2 \sqrt{3} - 2)}^{2}}} \\ = \sqrt{6 - 4 \sqrt{3} + 2 \sqrt{3} - 2} \\ = \sqrt{{(\sqrt{3})}^{2} + {(1)}^{2} - 2 \times \sqrt{3} \times 1} \\ = \sqrt{{(\sqrt{3} - 1)}^{2}} \\ = \sqrt{3} - 1 \end{aligned}

92. If N = $\frac{\sqrt{\sqrt{5} + 2} + \sqrt{\sqrt{5} - 2}}{\sqrt{\sqrt{5} + 1}} -$ $\sqrt{3 - 2 \sqrt{2}},$ then the value of N is = ?

A. $2 \sqrt{2} - 1$

B. 3

C. 1

D. 2

Discuss

Solution:

\begin{aligned} Let X = \frac{\sqrt{\sqrt{5} + 2} + \sqrt{\sqrt{5} - 2}}{\sqrt{\sqrt{5} + 1}} \\ Then, \\ X^{2} = \frac{{(\sqrt{\sqrt{5} + 2} + \sqrt{\sqrt{5} - 2})}^{2}}{{(\sqrt{\sqrt{5} + 1})}^{2}} \end{aligned}

X^{2} = \frac{(\sqrt{5} + 2) + (\sqrt{5} - 2) + 2 \sqrt{(\sqrt{5} + 2) (\sqrt{5} - 2)}}{(\sqrt{5} + 1)}

\begin{aligned} X^{2} = \frac{2 \sqrt{5} + 2 \sqrt{{(\sqrt{5})}^{2} - {(2)}^{2}}}{\sqrt{5} + 1} \\ X^{2} = \frac{2 \sqrt{5} + 2}{\sqrt{5} + 1} \\ X^{2} = \frac{2 (\sqrt{5} + 1)}{(\sqrt{5} + 1)} \\ X^{2} = 2 \\ X = \sqrt{2} \\ ∴ N = \sqrt{2} - \sqrt{3 - 2 \sqrt{2}} \\ N = \sqrt{2} - \sqrt{{(\sqrt{2})}^{2} + 1^{2} - 2 \times \sqrt{2} \times} 1 \\ N = \sqrt{2} - \sqrt{{(\sqrt{2} - 1)}^{2}} \\ N = \sqrt{2} - (\sqrt{2} - 1) \\ N = 1 \end{aligned}

93. If $\frac{(x - \sqrt{24}) (\sqrt{75} + \sqrt{50})}{\sqrt{75} - \sqrt{50}}$ = 1 then the value of x is = ?

A. $\sqrt{5}$

B. 5

C. $2 \sqrt{5}$

D. $3 \sqrt{5}$

Discuss

Solution:

\begin{aligned} \frac{(x - \sqrt{24}) (\sqrt{75} + \sqrt{50})}{\sqrt{75} - \sqrt{50}} = 1 \\ \Rightarrow (x - \sqrt{24}) = \frac{\sqrt{75} - \sqrt{50}}{\sqrt{75} + \sqrt{50}} \\ \Rightarrow (x - \sqrt{24}) = \frac{{(\sqrt{75} - \sqrt{50})}^{2}}{75 - 50} \\ \Rightarrow (x - \sqrt{24}) = \frac{75 + 50 - 2 \sqrt{75} \sqrt{50}}{25} \\ \Rightarrow (x - \sqrt{24}) = \frac{125 - 2 \times 5 \sqrt{3} \times 5 \sqrt{2}}{25} \\ \Rightarrow (x - \sqrt{24}) = \frac{125 - 50 \sqrt{6}}{25} \\ \Rightarrow (x - \sqrt{24}) = \frac{25 (5 - 2 \sqrt{6})}{25} \\ \Rightarrow x - 2 \sqrt{6} = 5 - 2 \sqrt{6} \\ \Rightarrow x = 5 \end{aligned}

94. Evaluate : $\sqrt{20} + \sqrt{12} + \sqrt[3]{729} -$ $\frac{4}{\sqrt{5} - \sqrt{3}} -$ $\sqrt{81} = ?$

A. $\sqrt{2}$

B. $\sqrt{3}$

C. 0

D. $2 \sqrt{2}$

Discuss

Solution:

\sqrt{20} + \sqrt{12} + \sqrt[3]{729} - \frac{4}{\sqrt{5} - \sqrt{3}} - \sqrt{81}

= 2 \sqrt{5} + 2 \sqrt{3} + 9 -

(\frac{4}{\sqrt{5} - \sqrt{3}} \times \frac{\sqrt{5} + \sqrt{3}}{\sqrt{5} + \sqrt{3}})

- 9

\begin{aligned} = 2 \sqrt{5} + 2 \sqrt{3} + 9 - (\frac{4 (\sqrt{5} + \sqrt{3})}{2}) - 9 \\ = 2 \sqrt{5} + 2 \sqrt{3} + 9 - 2 \sqrt{5} - 2 \sqrt{3} - 9 \\ = 0 \end{aligned}

95. If $\frac{4 + 3 \sqrt{3}}{\sqrt{7 + 4 \sqrt{3}}} = A + \sqrt{B},$ then B - A is = ?

A. -13

B. $2 \sqrt{13}$

C. 13

D. $3 \sqrt{3} - \sqrt{7}$

Discuss

Solution:

\begin{aligned} \frac{4 + 3 \sqrt{3}}{\sqrt{7 + 4 \sqrt{3}}} = A + \sqrt{B} \\ \Rightarrow \sqrt{7 + 4 \sqrt{3}} \\ \Rightarrow \sqrt{2^{2} + {(\sqrt{3})}^{2} + 2 \times 2 \sqrt{3}} \\ \Rightarrow \sqrt{{(2 + \sqrt{3})}^{2}} \\ \Rightarrow (2 + \sqrt{3}) \\ \Rightarrow \frac{4 + 3 \sqrt{3}}{2 + \sqrt{3}} = A + \sqrt{B} \\ \Rightarrow \frac{4 + 3 \sqrt{3}}{2 + \sqrt{3}} \times \frac{2 - \sqrt{3}}{2 - \sqrt{3}} = A + \sqrt{B} \\ \Rightarrow \frac{(4 + 3 \sqrt{3}) (2 - \sqrt{3})}{4 - 3} = A + \sqrt{B} \\ \Rightarrow 8 - 4 \sqrt{3} + 6 \sqrt{3} - 9 = A + \sqrt{B} \\ \Rightarrow 2 \sqrt{3} - 1 = A + \sqrt{B} \\ ∴ A = - 1 & \sqrt{B} = 2 \sqrt{3} \\ ∵ B = 2 \sqrt{3} \times 2 \sqrt{3} = 12 \\ So, B - A = 12 - (- 1) = 13 \end{aligned}

96. Given that 10^0.48 = x, 10^0.70 = y and x^z = y² then the value of z is close to = ?

A. 1.45

B. 1.88

C. 2.9

D. 3.7

Discuss

Solution:

\begin{aligned} x^{z} = y^{2} \\ \Leftrightarrow {(10^{0.48})}^{z} = {(10^{0.70})}^{2} \\ \Leftrightarrow 10^{(0.48 z)} = 10^{(2 \times 0.70)} = 10^{1.40} \\ \Leftrightarrow 0.48 z = 1.40 \\ \Leftrightarrow z = \frac{140}{48} = \frac{35}{12} \\ \Leftrightarrow z = 2.9 (approx) \end{aligned}

97. If m and n are whole numbers such that mⁿ = 121, then the value of (m - 1)ⁿ⁺¹ is = ?

A. 1

B. 10

C. 121

D. 1000

Discuss

Solution:

\begin{aligned} We know that \\ 1 1^{2} = 121 \\ Putting \\ m = 11 & n = 2 \\ we get \\ {(m - 1)}^{n + 1} \\ = {(11 - 1)}^{(2 + 1)} \\ = 10^{3} \\ = 1000 \end{aligned}

98. 1 + (3 + 1)(3² + 1)(3⁴ + 1)(3⁸ + 1)(3¹⁶ + 1)(3³² + 1) is equal to =?

A. $\frac{3^{64} - 1}{2}$

B. $\frac{3^{64} + 1}{2}$

C. 3⁶⁴ - 1

D. 3⁶⁴ + 1

Discuss

Solution:

1 + (3 + 1)

(3^{2} + 1)

(3^{4} + 1)

(3^{8} + 1)

(3^{16} + 1)

(3^{32} + 1)

= 1 + \frac{1}{2} [(3 - 1) (3 + 1) (3^{2} + 1) (3^{4} + 1) (3^{8} + 1) (3^{16} + 1) (3^{32} + 1)]

= 1 + \frac{1}{2} [(3^{2} - 1) (3^{2} + 1) (3^{4} + 1) (3^{8} + 1) (3^{16} + 1) (3^{32} + 1)]

= 1 + \frac{1}{2} [(3^{4} - 1) (3^{4} + 1) (3^{8} + 1) (3^{16} + 1) (3^{32} + 1)]

\begin{aligned} = 1 + \frac{1}{2} [(3^{8} - 1) (3^{8} + 1) (3^{16} + 1) (3^{32} + 1)] \\ = 1 + \frac{1}{2} [(3^{16} - 1) (3^{16} + 1) (3^{32} + 1)] \\ = 1 + \frac{1}{2} [(3^{32} - 1) (3^{32} + 1)] \\ = 1 + \frac{1}{2} [(3^{64} + 1)] \\ = \frac{2 + 3^{64} - 1}{2} \\ = \frac{3^{64} + 1}{2} \end{aligned}

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