Practice MCQ Questions and Answer on Square Root and Cube Root

Solution:

 $$\frac{{{{\left( {0.75} \right)}^3}}}{{1 - 0.75}}$$ $${\text{ + }}$$$$\left[ {0.75 + {{\left( {0.75} \right)}^2} + 1} \right]$$ $$ = \frac{{{{\left( {0.75} \right)}^2} \times 0.75}}{{0.25}}$$   $${\text{ + }}$$ $$\left[ {0.75 + 0.5625 + 1} \right]$$ $$ = 0.5625 \times 3 \,\, + $$   $$\left[ {0.75 + 0.5625 + 1} \right]$$ $$\eqalign{ & = 1.6875 + 2.3125 \cr & = 4 \cr} $$ Square root of 4 = 2

Solution:

 $$\eqalign{ & 3\sqrt 5 + \sqrt {125} = 17.88 \cr & \Rightarrow 3\sqrt 5 + \sqrt {25 \times 5} = 17.88 \cr & \Rightarrow 3\sqrt 5 + 5\sqrt 5 = 17.88 \cr & \Rightarrow 8\sqrt 5 = 17.88 \cr & \Rightarrow \sqrt 5 = 2.235 \cr & \therefore \sqrt {80} + 6\sqrt 5 = \sqrt {16 \times 5} + 6\sqrt 5 \cr & = 4\sqrt 5 + 6\sqrt 5 \cr & = 10\sqrt 5 = \left( {10 \times 2.235} \right) = 22.35 \cr} $$

Solution:

 The number of digits of the square root of a perfect square number of n digits is $$\eqalign{ & {\text{(i)}}\frac{n}{2}{\text{, if n is even}} \cr & {\text{(ii)}}\frac{{n + 1}}{2}{\text{, if n is odd}} \cr & {\text{Here, }}n = 12 \cr & {\text{So, required number of digits}} \cr & = \frac{n}{2} \cr & = \frac{{12}}{2} \cr & = 6{\text{ }} \cr} $$

Solution:

 $$\eqalign{ & {\text{Clearly, }}a*b = \sqrt {{a^2} + {b^2}} \cr & \therefore 5*12 \cr & = \sqrt {{5^2} + {{12}^2}} \cr & = \sqrt {25 + 144} \cr & = \sqrt {169} \cr & = 13 \cr} $$

Solution:

 $$\eqalign{ & \frac{{\sqrt 5 }}{2} - \frac{{10}}{{\sqrt 5 }} + \sqrt {125} \, \cr & = \frac{{{{\left( {\sqrt 5 } \right)}^2} - 20 + 2\sqrt 5 \times 5\sqrt 5 }}{{2\sqrt 5 }} \cr & = \frac{{5 - 20 + 50}}{{2\sqrt 5 }} \cr & = \frac{{35}}{{2\sqrt 5 }} \times \frac{{\sqrt 5 }}{{\sqrt 5 }} \cr & = \frac{{35\sqrt 5 }}{{10}} \cr & = \frac{{7 \times 2.236}}{2} \cr & = 7 \times 1.118 \cr & = 7.826 \cr} $$

Solution:

 $$\eqalign{ & {\text{Given,}} \cr & = \root 4 \of {{{\left( {625} \right)}^3}} \cr & = {\left( {625} \right)^{\frac{3}{4}}} \cr & = {\left( {5 \times 5 \times 5 \times 5} \right)^{\frac{3}{4}}} \cr & = {\left( {{5^4}} \right)^{\frac{3}{4}}} \cr & = {5^3} \cr & = 125 \cr} $$

Solution:

 $$\eqalign{ & \Leftrightarrow {{\text{n}}^2} = {\left( {{\text{25}}} \right)^{64}} \times {\left( {64} \right)^{25}} \cr & \Leftrightarrow {{\text{n}}^2} = {\left( {{{\text{5}}^2}} \right)^{64}} \times {\left( {{2^6}} \right)^{25}} \cr & \Leftrightarrow {{\text{n}}^2} = {5^{128}} \times {2^{150}} \cr & \Leftrightarrow {{\text{n}}^2} = {5^{128}} \times {2^{128}} \times {2^{22}} \cr & \Leftrightarrow n = {5^{64}} \times {2^{64}} \times {2^{11}} \cr & \Leftrightarrow n = {\left( {5 \times 2} \right)^{64}} \times {2^{11}} \cr & \Leftrightarrow n = {10^{64}} \times 2048 \cr} $$ ∴ Sum of digits of n = 2 + 0 + 4 + 8 = 14

Solution:

 $$\eqalign{ & = \sqrt {\frac{3}{{11}}} \cr & = \sqrt {\frac{{3 \times 11}}{{11 \times 11}}} \cr & = \frac{{\sqrt {33} }}{{11}} \cr & = \frac{{5.745}}{{11}} \cr & = 0.5223 \cr} $$

Solution:

 Given expression, $$ = \frac{1}{{\left( {\sqrt 9 - \sqrt 8 } \right)}} \times \frac{{\left( {\sqrt 9 + \sqrt 8 } \right)}}{{\left( {\sqrt 9 + \sqrt 8 } \right)}}$$     $$ - \frac{1}{{\left( {\sqrt 8 - \sqrt 7 } \right)}}$$   $$ \times \frac{{\left( {\sqrt 8 + \sqrt 7 } \right)}}{{\left( {\sqrt 8 + \sqrt 7 } \right)}}$$   $$ + \frac{1}{{\left( {\sqrt 7 - \sqrt 6 } \right)}}$$   $$ \times \frac{{\left( {\sqrt 7 + \sqrt 6 } \right)}}{{\left( {\sqrt 7 + \sqrt 6 } \right)}}$$   $$ - \frac{1}{{\left( {\sqrt 6 - \sqrt 5 } \right)}}$$   $$ \times \frac{{\left( {\sqrt 6 + \sqrt 5 } \right)}}{{\left( {\sqrt 6 + \sqrt 5 } \right)}}$$   $$ + \frac{1}{{\left( {\sqrt 5 - \sqrt 4 } \right)}}$$   $$ \times \frac{{\left( {\sqrt 5 + \sqrt 4 } \right)}}{{\left( {\sqrt 5 + \sqrt 4 } \right)}}$$ $$ = \frac{{\left( {\sqrt 9 + \sqrt 8 } \right)}}{{\left( {9 - 8} \right)}} - \frac{{\left( {\sqrt 8 + \sqrt 7 } \right)}}{{\left( {8 - 7} \right)}}$$     $$ + \frac{{\left( {\sqrt 7 + \sqrt 6 } \right)}}{{\left( {7 - 6} \right)}}$$   $$ - \frac{{\left( {\sqrt 6 + \sqrt 5 } \right)}}{{\left( {6 - 5} \right)}}$$   $$ + \frac{{\left( {\sqrt 5 + \sqrt 4 } \right)}}{{\left( {5 - 4} \right)}}$$ $$ = \left( {\sqrt 9 + \sqrt 8 } \right) - \left( {\sqrt 8 + \sqrt 7 } \right)$$     $$ + \left( {\sqrt 7 + \sqrt 6 } \right)$$   $$ - \left( {\sqrt 6 + \sqrt 5 } \right)$$   $$ + \left( {\sqrt 5 + \sqrt 4 } \right)$$ $$ = \left( {\sqrt 9 + \sqrt 4 } \right)$$ $$ = 3 + 2$$ $$ = 5$$

Solution:

 L.C.M. of 3, 4, 5, 6, 8 is 120 Now 120 = 2 × 2 × 2 × 3 × 5 To make it a perfect square, it must be multiplied by 2 × 3 × 5 So, required number $$\eqalign{ & = {2^2} \times {2^2} \times {3^2} \times {5^2} \cr & = 3600 \cr} $$