Practice MCQ Questions and Answer on Surds and Indices

Solution:

 $$\eqalign{ & {\text{Take LCM of }}\left( {2,\,3\,\& \,6} \right) = 6 \cr & {\text{I}}.{\left( {27} \right)^{\frac{1}{3}}} > {\left( {13} \right)^{\frac{1}{2}}} {\left( {47} \right)^{\frac{1}{6}}} \cr & = {\left( {27} \right)^{\frac{2}{{3 \times 2}}}} > {\left( {13} \right)^{\frac{3}{{2 \times 3}}}} {\left( {47} \right)^{\frac{{1 \times 1}}{{6 \times 1}}}} \cr & = {\left( {27} \right)^{\frac{2}{6}}} > {\left( {13} \right)^{\frac{3}{6}}} {\left( {47} \right)^{\frac{1}{6}}} \cr & = {\left( {729} \right)^{\frac{1}{6}}} > {\left( {2197} \right)^{\frac{1}{6}}} {\left( {47} \right)^{\frac{1}{6}}} \cr & {\text{This statement is false}} \cr & {\text{II}}.{\left( {23} \right)^{\frac{1}{3}}} {\left( {49} \right)^{\frac{1}{2}}} {\left( {52} \right)^{\frac{1}{6}}} \cr & = {\left( {23} \right)^{\frac{2}{{3 \times 2}}}} {\left( {49} \right)^{\frac{3}{{2 \times 3}}}} {\left( {52} \right)^{\frac{{1 \times 1}}{{6 \times 1}}}} \cr & = {\left( {529} \right)^{\frac{1}{6}}} {\left( {117649} \right)^{\frac{1}{6}}} {\left( {52} \right)^{\frac{1}{6}}} \cr & {\text{This statement is false}} \cr & {\text{III}}.{\left( {53} \right)^{\frac{1}{6}}} {\left( {41} \right)^{\frac{1}{3}}} {\left( {37} \right)^{\frac{1}{2}}} \cr & = {\left( {53} \right)^{\frac{1}{{6 \times 1}}}} {\left( {41} \right)^{\frac{2}{{3 \times 2}}}} {\left( {37} \right)^{\frac{3}{{2 \times 3}}}} \cr & = {\left( {53} \right)^{\frac{1}{6}}} {\left( {1681} \right)^{\frac{1}{6}}} {\left( {50653} \right)^{\frac{1}{6}}} \cr & {\text{This statement is true}} \cr} $$

Solution:

 $$\eqalign{ & {\text{Expression}} \cr & = \left( {1 - \sqrt 2 } \right) + \left( {\sqrt 2 - \sqrt 3 } \right) + \left( {\sqrt 3 - \sqrt 4 } \right) + \,.\,.\,.\,.\, + \left( {\sqrt {15} - \sqrt {16} } \right) \cr & = 1 - \sqrt 2 + \sqrt 2 - \sqrt 3 + \sqrt 3 - \sqrt 4 + \,.\,.\,.\,.\, + \sqrt {15} - \sqrt {16} \cr & = 1 - \sqrt {16} \cr & = 1 - 4 \cr & = - 3 \cr} $$

Solution:

 $$\eqalign{ & {21^?} \times {21^{6.5}} = {21^{12.4}}.....\left[ {\because {a^m} \times {a^n} = {a^{m + n}}} \right] \cr & \Rightarrow {21^{? + 6.5}} = {21^{12.4}} \cr & \Rightarrow ? + 6.5 = 12.4 \cr & \Rightarrow ? = 12.4 - 6.5 \cr & \Rightarrow ? = 5.9 \cr} $$

Solution:

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

Solution:

 $$\eqalign{ & \frac{{0.41 \times 0.41 \times 0.41 + 0.69 \times 0.69 \times 0.69}}{{0.41 \times 0.41 - 0.41 \times 0.69 + 0.69 \times 0.69}} \cr & = \frac{{{{\left( {0.41} \right)}^3} + {{\left( {0.69} \right)}^3}}}{{{{\left( {0.41} \right)}^2} - 0.41 \times 0.69 + {{\left( {0.69} \right)}^2}}} \cr & .....\left[ {\because {a^3} + {b^3} = \left( {a + b} \right)\left( {{a^2} + {b^2} - ab} \right)} \right] \cr & = 0.41 + 0.69 \cr & = 1.1 \cr} $$

Solution:

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

Solution:

 $$\eqalign{ & \frac{{\left( {x - \sqrt {24} } \right)\left( {\sqrt {75} + \sqrt {50} } \right)}}{{\sqrt {75} - \sqrt {50} }}{\text{ = 1 }} \cr & \Rightarrow \left( {x - \sqrt {24} } \right) = \frac{{\sqrt {75} - \sqrt {50} }}{{\sqrt {75} + \sqrt {50} }}{\text{ }} \cr & \Rightarrow \left( {x - \sqrt {24} } \right) = \frac{{{{\left( {\sqrt {75} - \sqrt {50} } \right)}^2}}}{{75 - 50}} \cr & \Rightarrow \left( {x - \sqrt {24} } \right) = \frac{{75 + 50 - 2\sqrt {75} \sqrt {50} }}{{25}} \cr & \Rightarrow \left( {x - \sqrt {24} } \right) = \frac{{125 - 2 \times 5\sqrt 3 \times 5\sqrt 2 }}{{25}} \cr & \Rightarrow \left( {x - \sqrt {24} } \right) = \frac{{125 - 50\sqrt 6 }}{{25}} \cr & \Rightarrow \left( {x - \sqrt {24} } \right) = \frac{{25\left( {5 - 2\sqrt 6 } \right)}}{{25}} \cr & \Rightarrow x - 2\sqrt 6 = 5 - 2\sqrt 6 \cr & \Rightarrow x = 5{\text{ }} \cr} $$

Solution:

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

Solution:

 $$\frac{{1 + \sqrt 2 }}{{\sqrt 5 + \sqrt 3 }} + \frac{{1 - \sqrt 2 }}{{\sqrt 5 - \sqrt 3 }}$$ $$ \Rightarrow \frac{{\left( {1 + \sqrt 2 } \right)\left( {\sqrt 5 - \sqrt 3 } \right) + \left( {1 - \sqrt 2 } \right)\left( {\sqrt 5 + \sqrt 3 } \right)}}{{\left( {\sqrt 5 + \sqrt 3 } \right)\left( {\sqrt 5 - \sqrt 3 } \right)}}$$ $$ \Rightarrow \frac{{\sqrt 5 - \sqrt 3 + \sqrt {10} - \sqrt 6 + \sqrt 5 + \sqrt 3 - \sqrt {10} - \sqrt 6 }}{{5 - 3}}$$ $$\eqalign{ & \Rightarrow \frac{{2\sqrt 5 - 2\sqrt 6 }}{2} \cr & \Rightarrow \frac{{2\left( {\sqrt 5 - \sqrt 6 } \right)}}{2} \cr & \Rightarrow \sqrt 5 - \sqrt 6 \cr} $$

Solution:

 $$\eqalign{ & \sqrt {121} + \sqrt {12321} + \sqrt {1234321} + \sqrt {123454321} \cr & = 11 + 111 + 1111 + 11111 \cr & = 12344 \cr} $$