Surds And Indices - Aptitude MCQ Questions and Solutions with Explanations

$$\eqalign{ & {\text{Let }}\,{6^{1.2}} \times {36^x} \times {30^{2.4}} \times {25^{1.3}} = {30^5} \cr & {\text{Then,}}\,{6^{1.2}} \times {({6^2})^x} \times {(6 \times 5)^{2.4}} \times {({5^2})^{1.3}} = {30^5} \cr & \Leftrightarrow {6^{1.2}} \times {6^{2x}} \times {6^{2.4}} \times {5^{2.4}} \times {5^{2.6}} = {(6 \times 5)^5} \cr & \Leftrightarrow {6^{\left( {1.2 + 2x + 2.4} \right)}} \times {5^{\left( {2.4 + 2.6} \right)}} = {6^5} \times {5^5} \cr & \Leftrightarrow {6^{\left( {3.6 + 2x} \right)}} \times {5^5} = {6^5} \times {5^5} \cr & \Leftrightarrow 3.6 + 2x = 5 \cr & \Leftrightarrow 2x = 1.4 \cr & \Leftrightarrow x = 0.7 \cr} $$

$$\left[ {\frac{{\sqrt 3 + \sqrt 2 }}{{\sqrt 3 - \sqrt 2 }} - \frac{{\sqrt 3 - \sqrt 2 }}{{\sqrt 3 + \sqrt 2 }}} \right]$$
$$ = \frac{{\sqrt 3 + \sqrt 2 }}{{\sqrt 3 - \sqrt 2 }} \times $$   $$\frac{{\sqrt 3 + \sqrt 2 }}{{\sqrt 3 + \sqrt 2 }} - $$   $$\frac{{\sqrt 3 - \sqrt 2 }}{{\sqrt 3 + \sqrt 2 }} \times $$  $$\frac{{\sqrt 3 - \sqrt 2 }}{{\sqrt 3 - \sqrt 2 }}$$
$$\eqalign{ & = \frac{{{{\left( {\sqrt 3 + \sqrt 2 } \right)}^2}}}{{3 - 2}} - \frac{{{{\left( {\sqrt 3 - \sqrt 2 } \right)}^2}}}{{3 - 2}} \cr & = \left( {3 + 2 + 2\sqrt 6 } \right) - \left( {3 + 2 - 2\sqrt 6 } \right) \cr & = 4\sqrt 6 {\text{ }} \cr} $$

$$\eqalign{ & \sqrt 3 = 1.732 \cr & \Rightarrow \frac{{2 + \sqrt 3 }}{{2 - \sqrt 3 }} \times \frac{{2 + \sqrt 3 }}{{2 + \sqrt 3 }} \cr & \Rightarrow \frac{{{{\left( {2 + \sqrt 3 } \right)}^2}}}{{4 - 3}} \cr & \Rightarrow 4 + 3 + 4\sqrt 3 \cr & \Rightarrow 7 + 4 \times 1.732 \cr & \Rightarrow 7 + 6.928 \cr & \Rightarrow 13.928 \cr} $$

$$\eqalign{ & 16\sqrt {\frac{3}{4}} - 9\sqrt {\frac{4}{3}} \cr & \Rightarrow 16\sqrt {\frac{{3 \times 4}}{{4 \times 4}}} - 9\sqrt {\frac{{3 \times 4}}{{3 \times 3}}} \cr & \Rightarrow 16 \times \frac{{\sqrt {12} }}{4} - \frac{{9\sqrt {12} }}{3} \cr & \Rightarrow 4\sqrt {12} - 3\sqrt {12} \cr & \Rightarrow \sqrt {12} \cr & \Rightarrow 3.46 \cr} $$

$$\eqalign{ & {\text{Let }}{2^{3.6}} \times {4^{3.6}} \times {4^{3.6}} \times {(32)^{2.3}} = {\left( {32} \right)^x} \cr & {\text{Then,}}{2^{3.6}} \times {\left( {{2^2}} \right)^{3.6}} \times {\left( {{2^2}} \right)^{3.6}} \times {({2^5})^{2.3}} = {\left( {{2^5}} \right)^x} \cr & \Leftrightarrow {2^{3.6}} \times {2^{\left( {2 \times 3.6} \right)}} \times {2^{\left( {2 \times 3.6} \right)}} \times {({2^5})^{2.3}} = {\left( {{2^5}} \right)^x} \cr & \Leftrightarrow {2^{\left( {3.6 + 7.2 + 7.2} \right)}} \times {({2^5})^{2.3}} = {\left( {{2^5}} \right)^x} \cr & \Leftrightarrow {2^{18}} \times {({2^5})^{2.3}} = {\left( {{2^5}} \right)^x} \cr & \Leftrightarrow {\left( {{2^5}} \right)^{3.6}} \times {({2^5})^{2.3}} = {\left( {{2^5}} \right)^x} \cr & \Leftrightarrow {\left( {{2^5}} \right)^{\left( {3.6 + 2.3} \right)}} = {\left( {{2^5}} \right)^x} \cr & \Leftrightarrow {\left( {{2^5}} \right)^{5.9}} = {\left( {{2^5}} \right)^x} \cr & \Leftrightarrow x = 5.9 \cr} $$

$$\eqalign{ & {\text{Let }}{25^{2.7}} \times {5^{4.2}} \div {5^{5.4}} = {25^x} \cr & {\text{Then, }}{25^{2.7}} \times {5^{(4.2 - 5.4)}} = {25^x} \cr & \Leftrightarrow {25^{2.7}} \times {5^{( - 1.2)}} = {25^x} \cr & \Leftrightarrow {25^{2.7}} \times \frac{1}{{{5^{1.2}}}} = {25^x} \cr & \Leftrightarrow \frac{{{{25}^{2.7}}}}{{{{\left( {{5^2}} \right)}^{0.6}}}} = {25^x} \cr & \Leftrightarrow \frac{{{{\left( {25} \right)}^{2.7}}}}{{{{\left( {25} \right)}^{0.6}}}} = {25^x} \cr & \Leftrightarrow {25^x} = {25^{\left( {2.7 - 0.6} \right)}} = {25^{2.1}} \cr & \Leftrightarrow x = 2.1 \cr} $$

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

$$\eqalign{ & {{\text{3}}^{x + y}}{\text{ = 81}} \cr & {{\text{3}}^{x + y}}{\text{ = }}{{\text{3}}^4} \cr & x + y = 4.....(i) \cr & {\text{8}}{{\text{1}}^{x - y}}{\text{ = 3}} \cr & {3^{4x - 4y}}{\text{ = }}{{\text{3}}^1} \cr & 4x - 4y{\text{ = 1}}....{\text{(ii)}} \cr & {\text{From equation (i) and (ii)}} \cr & 4x - 4y = 1 \cr & 4x + 4y = 16 \cr & 8x = 17 \cr & x = \frac{{17}}{8} \cr} $$

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

$$\eqalign{ & \root 4 \of {3x + 1} = 2 \cr & \Rightarrow {\left( {\root 4 \of {3x + 1} } \right)^4} = {2^4} \cr & \Rightarrow {\left( {3x + 1} \right)^{4 \times \frac{1}{4}}} = 16 \cr & \Rightarrow 3x = 15 \cr & \Rightarrow x = 5 \cr} $$