Surds And Indices - Aptitude MCQ Questions and Solutions with Explanations

3⁸ × 3⁴
= 3^{(8 + 4)}
= 3¹²
= (3⁶)²
= (729)²

$$\eqalign{ & \frac{{343 \times 49}}{{216 \times 16 \times 81}} \cr & = \frac{{{7^3} \times {7^2}}}{{{6^3} \times {2^4} \times {3^4}}} \cr & = \frac{{{7^{\left( {3 + 2} \right)}}}}{{{6^3} \times {{\left( {2 \times 3} \right)}^4}}} \cr & = \frac{{{7^5}}}{{{6^3} \times {6^4}}} \cr & = \frac{{{7^5}}}{{{6^{\left( {3 + 4} \right)}}}} \cr & = \frac{{{7^5}}}{{{6^7}}} \cr} $$

$$\eqalign{ & \frac{{16 \times 32}}{{9 \times 27 \times 81}} \cr & = \frac{{{2^4} \times {2^5}}}{{{3^2} \times {3^3} \times {3^4}}} \cr & = \frac{{{2^{\left( {4 + 5} \right)}}}}{{{3^{\left( {2 + 3 + 4} \right)}}}} \cr & = \frac{{{2^9}}}{{{3^9}}} \cr & = {\left( {\frac{2}{3}} \right)^9} \cr} $$

$$\eqalign{ & \sqrt 2 = 1.414 \cr & \Rightarrow \sqrt 8 {\text{ + 2}}\sqrt {32} - 3\sqrt {128} {\text{ + 4}}\sqrt {50} \cr & \Rightarrow 2\sqrt 2 + 2 \times 4\sqrt 2 - 3 \times 8\sqrt 2 + 4 \times 5\sqrt 2 \cr & \Rightarrow 2\sqrt 2 + 8\sqrt 2 - 24\sqrt 2 + 20\sqrt 2 \cr & \Rightarrow 6\sqrt 2 \cr & \Rightarrow 6 \times 1.414 \cr & \Rightarrow 8.484{\text{ }} \cr} $$

$$\eqalign{ & {\text{Let}}\,{9^3} \times {\left( {81} \right)^2} \div {\left( {27} \right)^3} = {\left( 3 \right)^x}{\text{then}} \cr & \Rightarrow {\left( 3 \right)^x}{\text{ = }}\frac{{{{\left( {{3^2}} \right)}^3} \times {{\left( {{3^4}} \right)}^2}}}{{{{\left( {{3^3}} \right)}^3}}} \cr & \Rightarrow {\left( 3 \right)^x} = \frac{{{3^{\left( {2 \times 3} \right)}} \times {3^{\left( {4 \times 2} \right)}}}}{{{3^{\left( {3 \times 3} \right)}}}} \cr & \Rightarrow {\left( 3 \right)^x} = \frac{{{3^6} \times {3^8}}}{{{3^9}}} \cr & \Rightarrow {\left( 3 \right)^x} = \frac{{{3^{\left( {6 + 8} \right)}}}}{{{3^9}}} \cr & \Rightarrow {\left( 3 \right)^x} = \frac{{{3^{14}}}}{{{3^9}}} \cr & \Rightarrow {\left( 3 \right)^x} = {3^{\left( {14 - 9} \right)}} \cr & \Rightarrow {\left( 3 \right)^x} = {3^5} \cr & \Rightarrow {\left( 3 \right)^x} = 5 \cr} $$

$$\eqalign{ & {\text{Let ,}} \cr & {\left( 6 \right)^4} \div {\left( {36} \right)^3} \times 216 = {6^{\left( {x - 5} \right)}} \cr & {\text{Then,}} \cr & {6^{\left( {x - 5} \right)}} = {\left( 6 \right)^4} \div {\left( {{6^2}} \right)^3} \times {6^3} \cr & \Rightarrow {6^{\left( {x - 5} \right)}} = {6^4} \div {6^{\left( {2 \times 3} \right)}} \times {6^3} \cr & \Rightarrow {6^{\left( {x - 5} \right)}} = {6^4} \div {6^6} \times {6^3} \cr & \Rightarrow {6^{\left( {x - 5} \right)}} = {6^{\left( {4 - 6 + 3} \right)}} \cr & \Rightarrow {6^{\left( {x - 5} \right)}} = 6 \cr & \Rightarrow x - 5 = 1 \cr & \Rightarrow x = 6 \cr} $$

$$\eqalign{ & {\left( {256} \right)^{\frac{5}{4}}} \cr & = {\left( {{4^4}} \right)^{\frac{5}{4}}} \cr & = {4^{\left( {4 \times \frac{5}{4}} \right)}} \cr & = {4^5} \cr & = 1024 \cr} $$

$$\eqalign{ & x = \sqrt {2 + \sqrt {2 + \sqrt {2 + ......} } } \cr & \Rightarrow {x^2} = 2 + \sqrt {2 + \sqrt {2 + .......} } \cr & \Rightarrow {x^2} = 2 + x \cr & \Rightarrow {x^2} - x - 2 = 0 \cr & \Rightarrow x\left( {x - 2} \right) + 1\left( {x - 2} \right) = 0 \cr & \Rightarrow \left( {x + 1} \right)\left( {x - 2} \right) = 0 \cr & \Rightarrow x = 2 \cr} $$

$$\eqalign{ & {\text{2 + }}\sqrt {0.09} - \root 3 \of {0.008} - 75\% \,{\text{of }}2.80 \cr & = 2 + 0.3 - 0.2 - \left( {\frac{3}{4} \times 2.8} \right) \cr & = 2 + 0.3 - 0.2 - 2.10 \cr & = 2.3 - 2.3 \cr & = 0 \cr} $$

$$\eqalign{ & {\left( {3 + 2\sqrt 2 } \right)^{ - 3}} + {\left( {3 - 2\sqrt 2 } \right)^{ - 3}} \cr & = {\left( {\frac{1}{{3 + 2\sqrt 2 }}} \right)^3} + {\left( {\frac{1}{{3 - 2\sqrt 2 }}} \right)^3} \cr} $$
  $$ = {\left( {\frac{1}{{\left( {3 - 2\sqrt 2 } \right)}} \times \frac{{3 + 2\sqrt 2 }}{{3 + 2\sqrt 2 }}} \right)^3} + $$      $${\left( {\frac{1}{{\left( {3 + 2\sqrt 2 } \right)}} \times \frac{{3 - 2\sqrt 2 }}{{3 - 2\sqrt 2 }}} \right)^3}$$
$$\eqalign{ & = {\left( {\frac{{3 - 2\sqrt 2 }}{{9 - 8}}} \right)^3} + {\left( {\frac{{3 + 2\sqrt 2 }}{{9 - 8}}} \right)^3} \cr & = {\left( {3 - 2\sqrt 2 } \right)^3} + {\left( {3 + 2\sqrt 2 } \right)^3} \cr & a = 3 - 2\sqrt 2 \cr & b = 3 + 2\sqrt 2 \cr & \left[ {\because {a^3} + {b^3} = \left( {a + b} \right)\left( {{a^2} + {b^2} - ab} \right)} \right] \cr & = \left( {3 - 2\sqrt 2 + 3 + 2\sqrt 2 } \right)\left( {17 + 17 - 1} \right) \cr & = \left( 6 \right)\left( {33} \right) \cr & = 198 \cr} $$