Practice MCQ Questions and Answer on Surds and Indices

Solution:

 $$\eqalign{ & {\left( {\sqrt 8 } \right)^{\frac{1}{3}}} \cr & = {\left( {{8^{\frac{1}{2}}}} \right)^{\frac{1}{3}}} \cr & = {8^{\left( {\frac{1}{2} \times \frac{1}{3}} \right)}} \cr & = {8^{\frac{1}{6}}} \cr & = {\left( {{2^3}} \right)^{\frac{1}{6}}} \cr & = {2^{\left( {3 \times \frac{1}{6}} \right)}} \cr & = {2^{\frac{1}{2}}} \cr & = \sqrt 2 \cr} $$

Solution:

 $$\eqalign{ & {\text{Given,}}\,4x = \sqrt 5 + 2 \cr & \Rightarrow 16x = 4\left( {\sqrt 5 + 2} \right) \cr & \Rightarrow 16x = 4\sqrt 5 + 8 \cr & \therefore \frac{1}{{16x}} = \frac{1}{{4\sqrt 5 + 8}} \cr & \Rightarrow \frac{1}{{16x}} = \frac{{4\sqrt 5 - 8}}{{\left( {4\sqrt 5 + 8} \right)\left( {4\sqrt 5 - 8} \right)}} \cr & \left[ {{\text{Rationalising the denominator}}} \right] \cr & \Rightarrow \frac{1}{{16x}} = \frac{{4\sqrt 5 - 8}}{{80 - 64}} \cr & \Rightarrow \frac{1}{{16x}} = \frac{{4\sqrt 5 - 8}}{{16}} \cr & \Rightarrow \frac{1}{{16x}} = \frac{{4\left( {\sqrt 5 - 2} \right)}}{{16}} \cr & \Rightarrow \frac{1}{{16x}} = \frac{{\sqrt 5 - 2}}{4} \cr & \therefore \left( {x - \frac{1}{{16x}}} \right) \cr & = \frac{{\sqrt 5 + 2}}{4} - \frac{{\sqrt 5 - 2}}{4} \cr & = \frac{{\sqrt 5 + 2 - \sqrt 5 + 2}}{4} \cr & = \frac{4}{4} \cr & = 1 \cr} $$

Solution:

 $$\eqalign{ & \sqrt {\frac{{\left( {\sqrt {12} - \sqrt 8 } \right)\left( {\sqrt 3 + \sqrt 2 } \right)}}{{5 + \sqrt {24} }}} \cr & = \sqrt {\frac{{\sqrt {36} + \sqrt {24} - \sqrt {24} - \sqrt {16} }}{{5 + \sqrt {24} }}} \cr & = \sqrt {\frac{{6 - 4}}{{5 + \sqrt {24} }}} \cr & = \sqrt {\frac{2}{{5 + \sqrt {24} }} \times \frac{{5 - \sqrt {24} }}{{5 - \sqrt {24} }}} \cr & = \sqrt {\frac{{2\left( {5 - \sqrt {24} } \right)}}{{25 - 24 }}} \cr & = \sqrt {2\left( {5 - 2\sqrt 6 } \right)} \cr & = \sqrt {2\left\{ {{{\left( {\sqrt 3 } \right)}^2} + {{\left( {\sqrt 2 } \right)}^2} - 2\sqrt 3 \times \sqrt 2 } \right\}} \cr & = \sqrt {2{{\left( {\sqrt 3 - \sqrt 2 } \right)}^2}} \cr & = \sqrt 2 \left( {\sqrt 3 - \sqrt 2 } \right) \cr & = \sqrt 6 - 2 \cr} $$

Solution:

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

Solution:

 $$\eqalign{ & {\text{let}} \cr & a = 0.73 \cr & b = 0.27 \cr & = \frac{{{a^3} + {b^3}}}{{{a^2} + {b^2} - ab}} \cr & = \frac{{\left( {a + b} \right)\left( {{a^2} + {b^2} - ab} \right)}}{{\left( {{a^2} + {b^2} - ab} \right)}} \cr & = \left( {a + b} \right) \cr & = \left( {0.73 + 0.27} \right) \cr & = 1 \cr} $$

Solution:

 a = bp and b = cq ∴ c = ar = (bp)r = (b)pr = (cq)pr = cpqr ⇒ pqr = 1

Solution:

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

Solution:

 $$\eqalign{ & 5\sqrt 3 + \sqrt {75} = 17.32 \cr & 5\sqrt 3 + 5\sqrt 3 = 17.32 \cr & 10\sqrt 3 = 17.32\,.\,.\,.\,.\,.\,\left( 1 \right) \cr & \therefore 14\sqrt 3 + \sqrt {108} \cr & = 14\sqrt 3 + \sqrt {12 \times 9} \cr & = 14\sqrt 3 + 6\sqrt 3 \cr & = 20\sqrt 3 \cr & = 2 \times 10\sqrt 3 \cr & = 2 \times 17.32 \cr & = 34.64 \cr} $$

Solution:

 $$\frac{{3\sqrt 2 }}{{\sqrt 6 + \sqrt 3 }} - \frac{{2\sqrt 6 }}{{\sqrt 3 + 1}} + \frac{{2\sqrt 3 }}{{\sqrt 6 + 2}}$$ $$ = \left( {\frac{{3\sqrt 2 }}{{\sqrt 6 + \sqrt 3 }} \times \frac{{\sqrt 6 - \sqrt 3 }}{{\sqrt 6 - \sqrt 3 }}{\text{ }}} \right) - $$      $$\left( {\frac{{2\sqrt 6 }}{{\sqrt 3 + 1}} \times \frac{{\sqrt 3 - 1}}{{\sqrt 3 - 1}}{\text{ }}} \right) + $$     $$\left( {\frac{{2\sqrt 3 }}{{\sqrt 6 + 2}}{\text{ }} \times \frac{{\sqrt 6 - 2}}{{\sqrt 6 - 2}}{\text{ }}} \right)$$ $$ = \frac{{3\sqrt 2 \left( {\sqrt 6 - \sqrt 3 } \right)}}{3} - $$    $$\frac{{2\sqrt 6 \left( {\sqrt 3 - 1} \right)}}{2} + $$   $$\frac{{2\sqrt 3 \left( {\sqrt 6 - 2} \right)}}{2}$$ $$\eqalign{ & = \sqrt {12} - \sqrt 6 - \sqrt {18} + \sqrt 6 + \sqrt {18} - 2\sqrt 3 \cr & = \sqrt {12} - 2\sqrt 3 \cr & = 2\sqrt 3 - 2\sqrt 3 \cr & = 0 \cr} $$

Solution:

 I. 333 > 333 (33)11 > 333 (27)11 > 333 It is clear that 333 > 333, this statement is true II. 333 > 333 It is clear that this statement is true III. 333 > 333 This statement is true So, all I, II and III statement are true, option D is correct.