Surds And Indices - Aptitude MCQ Questions and Solutions with Explanations

$$\eqalign{ & \sqrt {40 + \sqrt {9\sqrt {81} } } \cr & \Rightarrow \sqrt {40 + \sqrt {9 \times 9} } \cr & \Rightarrow \sqrt {40 + 9} \cr & \Rightarrow \sqrt {49} \cr & \Rightarrow 7 \cr} $$

$$\eqalign{ & \sqrt {8 - 2\sqrt {15} } \cr & = \sqrt {5 + 3 - 2 \times \sqrt 5 \times \sqrt 3 } \cr & = \sqrt {{{\left( {\sqrt 5 } \right)}^2} + {{\left( {\sqrt 3 } \right)}^2} - 2 \times \sqrt 5 \times \sqrt 3 } \cr & = \sqrt {{{\left( {\sqrt 5 - \sqrt 3 } \right)}^2}} \cr & = \left( {\sqrt 5 - \sqrt 3 } \right) \cr} $$

$$\eqalign{ & \sqrt {6 - 4\sqrt 3 + \sqrt {16 - 8\sqrt 3 } } \cr & = \sqrt {6 - 4\sqrt 3 + \sqrt {12 + 4 - 8\sqrt 3 } } \cr & = \sqrt {6 - 4\sqrt 3 + \sqrt {{{\left( {2\sqrt 3 } \right)}^2} + {{\left( 2 \right)}^2} - 2 \times 2\sqrt 3 \times 2} } \cr & = \sqrt {6 - 4\sqrt 3 + \sqrt {{{\left( {2\sqrt 3 - 2} \right)}^2}} } \cr & = \sqrt {6 - 4\sqrt 3 + 2\sqrt 3 - 2} \cr & = \sqrt {{{\left( {\sqrt 3 } \right)}^2} + {{\left( 1 \right)}^2} - 2 \times \sqrt 3 \times 1} \cr & = \sqrt {{{\left( {\sqrt 3 - 1} \right)}^2}} \cr & = \sqrt 3 - 1 \cr} $$

$$\eqalign{ & {\text{Let X = }}\frac{{\sqrt {\sqrt 5 + 2} + \sqrt {\sqrt 5 - 2} }}{{\sqrt {\sqrt 5 + 1} }} \cr & {\text{Then,}} \cr & {{\text{X}}^2} = \frac{{{{\left( {\sqrt {\sqrt 5 + 2} + \sqrt {\sqrt 5 - 2} } \right)}^2}}}{{{{\left( {\sqrt {\sqrt 5 + 1} } \right)}^2}}} \cr} $$
  $${{\text{X}}^2} = {\text{ }}\frac{{\left( {\sqrt 5 + 2} \right) + \left( {\sqrt 5 - 2} \right) + 2\sqrt {\left( {\sqrt 5 + 2} \right)\left( {\sqrt 5 - 2} \right)} }}{{\left( {\sqrt 5 + 1} \right)}}$$
$$\eqalign{ & {{\text{X}}^2} = \frac{{2\sqrt 5 + 2\sqrt {{{\left( {\sqrt 5 } \right)}^2} - {{\left( 2 \right)}^2}} }}{{\sqrt 5 + 1}} \cr & {{\text{X}}^2} = \frac{{2\sqrt 5 + 2}}{{\sqrt 5 + 1}} \cr & {{\text{X}}^2} = \frac{{2\left( {\sqrt 5 + 1} \right)}}{{\left( {\sqrt 5 + 1} \right)}} \cr & {{\text{X}}^2} = 2 \cr & {\text{X = }}\sqrt 2 \cr & \therefore {\text{N}} = \sqrt 2 - \sqrt {3 - 2\sqrt 2 } \cr & {\text{N}} = \sqrt 2 - \sqrt {{{\left( {\sqrt 2 } \right)}^2} + {1^2} - 2 \times \sqrt 2 \times } 1 \cr & {\text{N}} = \sqrt 2 - \sqrt {{{\left( {\sqrt 2 - 1} \right)}^2}} \cr & {\text{N}} = \sqrt 2 - \left( {\sqrt 2 - 1} \right) \cr & {\text{N}} = 1 \cr} $$

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

$$\sqrt {20} + \sqrt {12} + \root 3 \of {729} - \frac{4}{{\sqrt 5 - \sqrt 3 }} - \sqrt {81} $$
$$ = 2\sqrt 5 + 2\sqrt 3 + 9\, - $$   $$\left( {\frac{4}{{\sqrt 5 - \sqrt 3 }} \times \frac{{\sqrt 5 + \sqrt 3 }}{{\sqrt 5 + \sqrt 3 }}} \right)$$     $$\, - \,9$$
$$\eqalign{ & = 2\sqrt 5 + 2\sqrt 3 + 9 - \left( {\frac{{4\left( {\sqrt 5 + \sqrt 3 } \right)}}{2}} \right) - 9 \cr & = 2\sqrt 5 + 2\sqrt 3 + 9 - 2\sqrt 5 - 2\sqrt 3 - 9 \cr & = 0 \cr} $$

$$\eqalign{ & \frac{{4 + 3\sqrt 3 }}{{\sqrt {7 + 4\sqrt 3 } }}{\text{ = }}A + \sqrt B \cr & \Rightarrow \sqrt {7 + 4\sqrt 3 } \cr & \Rightarrow \sqrt {{2^2} + {{\left( {\sqrt 3 } \right)}^2} + 2 \times 2\sqrt 3 } \cr & \Rightarrow \sqrt {{{\left( {2 + \sqrt 3 } \right)}^2}} \cr & \Rightarrow \left( {2 + \sqrt 3 } \right) \cr & \Rightarrow \frac{{4 + 3\sqrt 3 }}{{2 + \sqrt 3 }}{\text{ = }}A + \sqrt B \cr & \Rightarrow \frac{{4 + 3\sqrt 3 }}{{2 + \sqrt 3 }} \times \frac{{2 - \sqrt 3 }}{{2 - \sqrt 3 }} = A + \sqrt B \cr & \Rightarrow \frac{{\left( {4 + 3\sqrt 3 } \right)\left( {2 - \sqrt 3 } \right)}}{{4 - 3}} = A + \sqrt B \cr & \Rightarrow 8 - 4\sqrt 3 + 6\sqrt 3 - 9 = A + \sqrt B \cr & \Rightarrow 2\sqrt 3 - 1 = A + \sqrt B \cr & \therefore A = - 1\,\,\& \,\,\sqrt B = 2\sqrt 3 \cr & \because B = 2\sqrt 3 \times 2\sqrt 3 = 12 \cr & {\text{So}},B - A = 12 - \left( { - 1} \right) = 13 \cr} $$

$$\eqalign{ & {x^z}{\text{ = }}{y^2}{\text{ }} \cr & \Leftrightarrow {\left( {{{10}^{0.48}}} \right)^z} = {\left( {{{10}^{0.70}}} \right)^2} \cr & \Leftrightarrow {10^{\left( {0.48z} \right)}} = {10^{\left( {2 \times 0.70} \right)}} = {10^{1.40}} \cr & \Leftrightarrow 0.48z = 1.40 \cr & \Leftrightarrow z = \frac{{140}}{{48}} = \frac{{35}}{{12}} \cr & \Leftrightarrow z = 2.9\left( {{\text{approx}}} \right) \cr} $$

$$\eqalign{ & {\text{We know that}} \cr & {\text{1}}{{\text{1}}^2} = 121 \cr & {\text{Putting}} \cr & m = 11\& n = 2 \cr & {\text{we get}} \cr & {\left( {m - 1} \right)^{n + 1}} \cr & = {\left( {11 - 1} \right)^{\left( {2 + 1} \right)}} \cr & = {10^3} \cr & = 1000 \cr} $$

$$1 + \left( {3 + 1} \right)$$  $$\left( {{3^2} + 1} \right)$$ $$\left( {{3^4} + 1} \right)$$ $$\left( {{3^8} + 1} \right)$$ $$\left( {{3^{16}} + 1} \right)$$ $$\left( {{3^{32}} + 1} \right)$$
$$ = 1 + \frac{1}{2}\left[ {\left( {3 - 1} \right)\left( {3 + 1} \right)\left( {{3^2} + 1} \right)\left( {{3^4} + 1} \right)\left( {{3^8} + 1} \right)\left( {{3^{16}} + 1} \right)\left( {{3^{32}} + 1} \right)} \right]$$
$$ = 1 + \frac{1}{2}\left[ {\left( {{3^2} - 1} \right)\left( {{3^2} + 1} \right)\left( {{3^4} + 1} \right)\left( {{3^8} + 1} \right)\left( {{3^{16}} + 1} \right)\left( {{3^{32}} + 1} \right)} \right]$$
$$ = 1 + \frac{1}{2}\left[ {\left( {{3^4} - 1} \right)\left( {{3^4} + 1} \right)\left( {{3^8} + 1} \right)\left( {{3^{16}} + 1} \right)\left( {{3^{32}} + 1} \right)} \right]$$
$$\eqalign{ & = 1 + \frac{1}{2}\left[ {\left( {{3^8} - 1} \right)\left( {{3^8} + 1} \right)\left( {{3^{16}} + 1} \right)\left( {{3^{32}} + 1} \right)} \right] \cr & = 1 + \frac{1}{2}\left[ {\left( {{3^{16}} - 1} \right)\left( {{3^{16}} + 1} \right)\left( {{3^{32}} + 1} \right)} \right] \cr & = 1 + \frac{1}{2}\left[ {\left( {{3^{32}} - 1} \right)\left( {{3^{32}} + 1} \right)} \right] \cr & = 1 + \frac{1}{2}\left[ {\left( {{3^{64}} + 1} \right)} \right] \cr & = \frac{{2 + {3^{64}} - 1}}{2} \cr & = \frac{{{3^{64}} + 1}}{2} \cr} $$