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81. The value of $\frac{{(243)}^{0.13} \times {(243)}^{0.07}}{{(7)}^{0.25} \times {(49)}^{0.075} \times {(343)}^{0.2}}$ is = ?

A. $\frac{3}{7}$

B. $\frac{7}{3}$

C. $1 \frac{3}{7}$

D. $2 \frac{2}{7}$

Discuss

Solution:

\begin{aligned} \frac{{(243)}^{0.13} \times {(243)}^{0.07}}{{(7)}^{0.25} \times {(49)}^{0.075} \times {(343)}^{0.2}} \\ = \frac{{(243)}^{(0.13 + 0.07)}}{{(7)}^{0.25} \times {(7^{2})}^{0.075} \times {(7^{3})}^{0.2}} \\ = \frac{{(243)}^{0.2}}{{(7)}^{0.25} \times {(7)}^{(2 \times 0.075)} \times {(7)}^{(3 \times 0.2)}} \\ = \frac{{(3^{5})}^{0.02}}{{(7)}^{0.25} \times {(7)}^{0.15} \times {(7)}^{0.6}} \\ = \frac{{(3)}^{(5 \times 0.2)}}{{(7)}^{(0.25 + 0.15 + 0.6)}} \\ = \frac{3^{1}}{7^{1}} \\ = \frac{3}{7} \end{aligned}

82. $\frac{2^{n + 4} - 2 \times 2^{n}}{2 \times 2^{(n + 3)}} + 2^{- 3}$ is equal to = ?

A. 2⁽ⁿ⁺¹⁾

B. $(\frac{9}{8} - 2^{n})$

C. $(- 2^{n + 1} + \frac{1}{8})$

D. 1

Discuss

Solution:

\begin{aligned} \frac{2^{n + 4} - 2 \times 2^{n}}{2 \times 2^{(n + 3)}} + 2^{- 3} \\ = \frac{2^{n + 4} - 2^{n + 1}}{2^{(n + 4)}} + \frac{1}{2^{3}} \\ = \frac{2^{n + 1} (2^{3} - 1)}{2^{(n + 4)}} + \frac{1}{2^{3}} \\ = \frac{2^{n + 1} \times 7}{2^{n + 1} \times 2^{3}} + \frac{1}{2^{3}} \\ = (\frac{7}{8} + \frac{1}{8}) \\ = \frac{8}{8} \\ = 1 \end{aligned}

83. $\frac{256 \times 256 - 144 \times 144}{112}$ is equal to = ?

A. 420

B. 400

C. 360

D. 320

Discuss

Solution:

\begin{aligned} \frac{256 \times 256 - 144 \times 144}{112} \\ = \frac{{(256)}^{2} - {(144)}^{2}}{112} \\ = \frac{(112) (400)}{112} \\ = 400 \end{aligned}

84. $\sqrt{3 \sqrt{3 \sqrt{3........}}}$ is equal to = ?

A. $\sqrt{3}$

B. 3

C. $2 \sqrt{3}$

D. $3 \sqrt{3}$

Discuss

Solution:

Shortcut method:
When the question is in form

\begin{aligned} \Rightarrow \sqrt{n \sqrt{n \sqrt{n}}} . . . . . . . . \infty \\ \Rightarrow So n is answer \\ \Rightarrow 3 \end{aligned}

85. The value of $\frac{2^{n - 1} - 2^{n}}{2^{n + 4} + 2^{n + 1}} is = ?$

A. $- \frac{1}{36}$

B. $\frac{2}{3}$

C. $\frac{1}{13}$

D. $\frac{5}{13}$

Discuss

Solution:

\begin{aligned} \frac{2^{n - 1} - 2^{n}}{2^{n + 4} + 2^{n + 1}} \\ = \frac{2^{n - 1} (1 - 2)}{2^{n + 1} (2^{3} + 1)} \\ = (- \frac{1}{9}) {.2}^{(n - 1) - (n + 1)} \\ = (- \frac{1}{9}) {.2}^{- 2} \\ = (- \frac{1}{9}) . \frac{1}{2^{2}} \\ = (- \frac{1}{9}) \times \frac{1}{4} \\ = - \frac{1}{36} \end{aligned}

86. If $x = 5 + 2 \sqrt{6},$ then $\sqrt{x} - \frac{1}{\sqrt{x}}$ = is?

A. $2 \sqrt{2}$

B. $2 \sqrt{3}$

C. $\sqrt{3} + \sqrt{2}$

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

Discuss

Solution:

\begin{aligned} {(\sqrt{x} - \frac{1}{\sqrt{x}})}^{2} \\ = x + \frac{1}{x} - 2 \\ = (5 + 2 \sqrt{6}) + \frac{1}{(5 + 2 \sqrt{6})} - 2 \\ = (5 + 2 \sqrt{6}) + \frac{1}{(5 + 2 \sqrt{6})} \times \frac{(5 - 2 \sqrt{6})}{(5 - 2 \sqrt{6})} - 2 \\ = (5 + 2 \sqrt{6}) + (5 - 2 \sqrt{6}) - 2 \\ = 10 - 2 \\ = 8 \\ ∴ (\sqrt{x} - \frac{1}{\sqrt{x}}) = \sqrt{8} = 2 \sqrt{2} \end{aligned}

87. $(4 + \sqrt{7}),$ expressed as a perfect square, is equal to = ?

A. ${(2 + \sqrt{7})}^{2}$

B. ${(\frac{\sqrt{7}}{2} + \frac{1}{2})}^{2}$

C. ${\frac{1}{2} {(\sqrt{7} + 1)}^{2}}$

D. $(\sqrt{3} + \sqrt{4})$

Discuss

Solution:

\begin{aligned} (4 + \sqrt{7}) \\ = \frac{7}{2} + \frac{1}{2} + 2 \times \frac{\sqrt{7}}{\sqrt{2}} \times \frac{1}{\sqrt{2}} \\ = {(\frac{\sqrt{7}}{\sqrt{2}})}^{2} + {(\frac{1}{\sqrt{2}})}^{2} + 2 \times \frac{\sqrt{7}}{\sqrt{2}} \times \frac{1}{\sqrt{2}} \\ = {(\frac{\sqrt{7}}{\sqrt{2}} + \frac{1}{\sqrt{2}})}^{2} \\ = \frac{1}{2} {(\sqrt{7} + 1)}^{2} \end{aligned}

88. If 3^x+y = 81 and 81^x-y = 3, then the value of $\frac{x}{y}$ is = ?

A. $\frac{15}{17}$

B. $\frac{17}{30}$

C. $\frac{15}{34}$

D. $\frac{17}{15}$

Discuss

Solution:

\begin{aligned} According to question, \\ \Rightarrow 3^{x + y} = 81 and 8 1^{x - y} = 3 \\ \Rightarrow 3^{x + y} = (3)^{4} and {(3)}^{4 (}^{x - y)} = 3^{1} \\ \Rightarrow x + y = 4 and x - y = \frac{1}{4} \\ x + y = 4...... (i) \\ x - y = \frac{1}{4} . . . . . (i i) \\ Solve the equation of (i) and (ii) \\ x = \frac{17}{8}, \\ y = \frac{15}{8}, \\ \Rightarrow \frac{x}{y} = \frac{17}{15} \end{aligned}

89. $(\frac{1 + \sqrt{2}}{\sqrt{5} + \sqrt{3}} + \frac{1 - \sqrt{2}}{\sqrt{5} - \sqrt{3}})$ simplifies to = ?

A. $\sqrt{5} + \sqrt{6}$

B. $2 \sqrt{5} + \sqrt{6}$

C. $\sqrt{5} - \sqrt{6}$

D. $2 \sqrt{5} - 3 \sqrt{6}$

Discuss

Solution:

\frac{1 + \sqrt{2}}{\sqrt{5} + \sqrt{3}} + \frac{1 - \sqrt{2}}{\sqrt{5} - \sqrt{3}}

\Rightarrow \frac{(1 + \sqrt{2}) (\sqrt{5} - \sqrt{3}) + (1 - \sqrt{2}) (\sqrt{5} + \sqrt{3})}{(\sqrt{5} + \sqrt{3}) (\sqrt{5} - \sqrt{3})}

\Rightarrow \frac{\sqrt{5} - \sqrt{3} + \sqrt{10} - \sqrt{6} + \sqrt{5} + \sqrt{3} - \sqrt{10} - \sqrt{6}}{5 - 3}

\begin{aligned} \Rightarrow \frac{2 \sqrt{5} - 2 \sqrt{6}}{2} \\ \Rightarrow \frac{2 (\sqrt{5} - \sqrt{6})}{2} \\ \Rightarrow \sqrt{5} - \sqrt{6} \end{aligned}

90. $\sqrt{8 - 2 \sqrt{15}}$ is equal to = ?

A. $3 - \sqrt{5}$

B. $\sqrt{5} - \sqrt{3}$

C. $5 - \sqrt{3}$

D. $\sqrt{5} + \sqrt{3}$

Discuss

Solution:

\begin{aligned} \sqrt{8 - 2 \sqrt{15}} \\ = \sqrt{5 + 3 - 2 \times \sqrt{5} \times \sqrt{3}} \\ = \sqrt{{(\sqrt{5})}^{2} + {(\sqrt{3})}^{2} - 2 \times \sqrt{5} \times \sqrt{3}} \\ = \sqrt{{(\sqrt{5} - \sqrt{3})}^{2}} \\ = (\sqrt{5} - \sqrt{3}) \end{aligned}

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