Practice MCQ Questions and Answer on Surds and Indices

111.

$$\frac{2^{n+4}-2\left(2^n\right)}{2\left(2^{n+3}\right)}$$    when simplified is = ?

• (A) $${\text{2}}^{n+1}-\frac{1}{8}$$
• (B) -2(n+1)
• (C) 1 - 2n
• (D) $$\frac{7}{8}$$

Show Answer

 $$\eqalign{ & \frac{{{2^{n + 4}} - 2\left( {{2^n}} \right)}}{{2\left( {{2^{n + 3}}} \right)}}{\text{ }} \cr & = \frac{{{2^{n + 4}} - {2^{n + 1}}}}{{{2^{n + 4}}}}{\text{ }} \cr & = \frac{{{2^{n + 4}}}}{{{2^{n + 4}}}} - \frac{{{2^{n + 1}}}}{{{2^{n + 4}}}}{\text{ }} \cr & = 1 - {2^{(n + 1) - \left( {n + 4} \right)}} \cr & {\text{ = 1}} - {2^{ - 3}} \cr & {\text{ = 1}} - \frac{1}{8}{\text{ }} \cr & {\text{ = }}\frac{7}{8} \cr} $$

112.

If $$N=\frac{\sqrt{7}-\sqrt{3}}{\sqrt{7}+\sqrt{3}},$$    then what is the value of $$N+\frac{1}{N}?$$

• (A) 2√2
• (B) 5
• (C) 10
• (D) 13

Show Answer

 $$\eqalign{ & N = \frac{{\sqrt 7 - \sqrt 3 }}{{\sqrt 7 + \sqrt 3 }} \cr & = \frac{{\sqrt 7 - \sqrt 3 }}{{\sqrt 7 + \sqrt 3 }} \times \frac{{\sqrt 7 - \sqrt 3 }}{{\sqrt 7 - \sqrt 3 }} \cr & = \frac{{7 + 3 - 2\sqrt {21} }}{4} \cr & = \frac{{10 - 2\sqrt {21} }}{4} \cr & = \frac{5}{2} - \frac{{\sqrt {21} }}{2} \cr & {\text{Similarly }}\frac{1}{N} = \frac{5}{2} + \frac{{\sqrt {21} }}{2} \cr & \therefore N + \frac{1}{N} \cr & = \frac{5}{2} - \frac{{\sqrt {21} }}{2} + \frac{5}{2} + \frac{{\sqrt {21} }}{2} \cr & = \frac{5}{2} + \frac{5}{2} \cr & = 5 \cr} $$

113.

If $${\left(\frac{3}{5}\right)}^3{\left(\frac{3}{5}\right)}^{-6}={\left(\frac{3}{5}\right)}^{2x-1}$$     then x is equal to ?

• (A) -2
• (B) -1
• (C) 1
• (D) 2

Show Answer

 $$\eqalign{ & {\text{ }}{\left( {\frac{3}{5}} \right)^3}{\left( {\frac{3}{5}} \right)^{ - 6}} = {\left( {\frac{3}{5}} \right)^{2x - 1}} \cr & \Rightarrow {\text{ }}{\left( {\frac{3}{5}} \right)^{\left( {3 - 6} \right)}} = {\left( {\frac{3}{5}} \right)^{2x - 1}} \cr & \Rightarrow {\text{ }}{\left( {\frac{3}{5}} \right)^{ - 3}} = {\left( {\frac{3}{5}} \right)^{2x - 1}} \cr & \Rightarrow 2x - 1 = - 3 \cr & \Rightarrow 2x = - 2 \cr & \Rightarrow x = - 1 \cr} $$

114.

If m and n are whole numbers such that mⁿ = 121, the value of (m - 1)^{n + 1} is:

• (A) 1
• (B) 10
• (C) 121
• (D) 1000

Show Answer

 We know that 112 = 121. Putting m = 11 and n = 2, we get: (m - 1)n + 1 = (11 - 1)(2 + 1) = 103 = 1000

115.

$$8^{2.4}\times 2^{3.7}÷{\left(16\right)}^{1.3}=2^{?}$$

• (A) 4.8
• (B) 5.7
• (C) 5.8
• (D) 7.1

Show Answer

 $$\eqalign{ & {\text{Let }}{8^{2.4}} \times {2^{3.7}} \div {\left( {16} \right)^{1.3}} = {2^x} \cr & {\text{Then,}}{\left( {{2^3}} \right)^{2.4}} \times {2^{3.7}} \div {\left( {{2^4}} \right)^{1.3}} = {2^x} \cr & \Leftrightarrow {2^{\left( {3 \times 2.4} \right)}} \times {2^{3.7}} \div {2^{\left( {4 \times 1.3} \right)}} = {2^x} \cr & \Leftrightarrow {2^{7.2}} \times {2^{3.7}} \div {2^{5.2}} = {2^x} \cr & \Leftrightarrow {2^x} = {2^{\left( {7.2 + 3.7 - 5.2} \right)}} \cr & \Leftrightarrow {2^x} = {2^{5.7}} \cr & \Leftrightarrow x = 5.7 \cr} $$

116.

Simplified from of $${\left[{\left(\sqrt[5]{x^{-\frac{3}{5}}}\right)}^{-\frac{5}{3}}\right]}^5$$   is = ?

• (A) $$\frac{1}{x}$$
• (B) x
• (C) x-5
• (D) x5

Show Answer

 $$\eqalign{ & {\left[ {{{\left( {\root 5 \of {{x^{ - \frac{3}{5}}}} } \right)}^{ - \frac{5}{3}}}} \right]^5} \cr & = {\left[ {{{\left\{ {{{\left( {{x^{ - \frac{3}{5}}}} \right)}^{\frac{1}{5}}}} \right\}}^{ - \frac{5}{3}}}} \right]^5} \cr & = {\left[ {{{\left( {{x^{^{\left\{ {\left( { - \frac{3}{5}} \right) \times \frac{1}{5}} \right\}}}}} \right)}^{ - \frac{5}{3}}}} \right]^5} \cr & = {\left[ {{{\left( {{x^{ - \frac{3}{{25}}}}} \right)}^{ - \frac{5}{3}}}} \right]^5} \cr & = {\left[ {{x^{\left\{ {\left( { - \frac{3}{{25}}} \right) \times \left( { - \frac{5}{3}} \right)} \right\}}}} \right]^5} \cr & = {\left( {{x^{\frac{1}{5}}}} \right)^5} \cr & = {x^{\left( {\frac{1}{5} \times 5} \right)}} \cr & = x \cr} $$

117.

$$\frac{6^2+7^2+8^2+9^2+{10}^2}{\sqrt{7+4\sqrt{3}}-\sqrt{4+2\sqrt{3}}}$$     is equal to = ?

• (A) 330
• (B) 355
• (C) 305
• (D) 366

Show Answer

 $$\eqalign{ & \frac{{{6^2} + {7^2} + {8^2} + {9^2} + {{10}^2}}}{{\sqrt {7 + 4\sqrt 3 } - \sqrt {4 + 2\sqrt 3 } }} \cr & \Rightarrow \frac{{{6^2} + {7^2} + {8^2} + {9^2} + {{10}^2}}}{{\sqrt {{{\left( {2 + \sqrt 3 } \right)}^2}} - \sqrt {{{\left( {\sqrt 3 + 1} \right)}^2}} }} \cr & \Rightarrow \frac{{{6^2} + {7^2} + {8^2} + {9^2} + {{10}^2}}}{{2 + \sqrt 3 - \sqrt 3 - 1}} \cr & \Rightarrow {6^2} + {7^2} + {8^2} + {9^2} + {10^2} \cr & \Rightarrow 36 + 49 + 64 + 81 + 100 \cr & \Rightarrow 330 \cr} $$

118.

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

• (A) 420
• (B) 400
• (C) 360
• (D) 320

Show Answer

 $$\eqalign{ & \frac{{256 \times 256 - 144 \times 144}}{{112}} \cr & = \frac{{{{\left( {256} \right)}^2} - {{\left( {144} \right)}^2}}}{{112}} \cr & = \frac{{\left( {112} \right)\left( {400} \right)}}{{112}} \cr & = 400 \cr} $$

119.

Which of the following is true? $$\begin{array}{rl}& \text{I}.\sqrt[3]{11}>\sqrt{7}>\sqrt[4]{45}\\ & \text{II}.\sqrt{7}>\sqrt[3]{11}>\sqrt[4]{45}\\ & \text{III}.\sqrt{7}>\sqrt[4]{45}>\sqrt[3]{11}\\ & \text{IV}.\sqrt[4]{45}>\sqrt{7}>\sqrt[3]{11}\end{array}$$

• (A) Only I
• (B) Only II
• (C) Only III
• (D) Only IV

Show Answer

 $$\eqalign{ & \root 3 \of {11} ,\,\sqrt 7 ,\,\root 4 \of {45} \cr & {11^{\frac{1}{3}}},\,{7^{\frac{1}{2}}},\,{45^{\frac{1}{4}}} \cr & {\text{Take LCM of 3, 2, 4 is 12}} \cr & {11^4},\,{7^6},\,{45^3} \cr & {\text{14641,}}\,{\text{117649,}}\,{\text{91125}} \cr & 117649 > 91125 > 14641 \cr & {\text{Option C is true}} \cr & \sqrt 7 > \root 4 \of {45} > \root 3 \of {11} \cr} $$

120.

The value of $$\sqrt{28+10\sqrt{3}}-\sqrt{7-4\sqrt{3}}$$      is closest to:

• (A) 7.2
• (B) 5.8
• (C) 6.1
• (D) 6.5

Show Answer

 $$\eqalign{ & \sqrt {28 + 10\sqrt 3 } - \sqrt {7 - 4\sqrt 3 } \cr & = \sqrt {{{\left( {5 + \sqrt 3 } \right)}^2}} - \sqrt {{{\left( {2 - \sqrt 3 } \right)}^2}} \cr & = 5 + \sqrt 3 - 2 + \sqrt 3 \cr & = 3 + 2\sqrt 3 \cr & = 3 + 2 \times 1.73 \cr & = 3 + 3.46 \cr & = 6.5 \cr} $$