Square Root And Cube Root - Aptitude MCQ Questions and Solutions with Explanations

$$\eqalign{ & \,\,\,\,\,\,1|\overline 1 \,\,\overline {10} \,\,\overline {25} \,(105 \cr & \,\,\,\,\,\,\,\,\,|\,\,1 \cr & \,\,\,\,\,\,\,\,\,| - - - - - - \cr & \,\,20|\,\,\,\,\,\,10 \cr & \,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,\,\,0 \cr & \,\,\,\,\,\,\,\,\,| - - - - - - \cr & 205|\,\,\,\,\,\,\,10\,25 \cr & \,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,\,\,10\,25 \cr & \,\,\,\,\,\,\,\,\,| - - - - - - - \cr & \,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{X}} \cr & {\text{Number of rows}} \cr & {\text{ = }}\sqrt {10914 + 111} \cr & = \sqrt {11025} \cr & = 105 \cr} $$

Let the number of girls in the group be x
Then, number of oranges given to each girl = 2x
$$\eqalign{ & \therefore x \times 2x = 1250 \cr & \Leftrightarrow 2{x^2} = 1250 \cr & \Leftrightarrow {x^2} = 625 \cr & \Leftrightarrow x = 25 \cr} $$

Clearly, the required number must be a perfect square. Since a number having 8 as the unit's digit cannot be a perfect square, so 121108 is not a perfect square.

$$\eqalign{ & = 262144 \cr & = 8 \times 8 \times 8 \times 8 \times 8 \times 8 \cr & = {8^6} \cr & \therefore \root 3 \of {262144} \cr & = {8^2} \cr & = 64 \cr & {\text{Let, }} \cr & {\text{1728}} \div \root 3 \of {262144} \times x - 288 = 4491 \cr & {\text{Then,}} \cr & \Leftrightarrow 1728 \div 64 \times x - 288 = 4491 \cr & \Leftrightarrow 27x = 4779 \cr & \Leftrightarrow x = \frac{{4779}}{{27}} \cr & \Leftrightarrow x = 177 \cr} $$

$$\eqalign{ & 8|262144 \cr & - - - - - - - - \cr & 8|\,\,\,32768 \cr & - - - - - - - - \cr & 8|\,\,\,\,4096 \cr & - - - - - - - - \cr & 8|\,\,\,\,\,512 \cr & - - - - - - - - \cr & 8|\,\,\,\,\,\,64 \cr & - - - - - - - - \cr & 8|\,\,\,\,\,\,\,8 \cr & - - - - - - - - \cr} $$

$$\eqalign{ & {\text{Let,}} \cr & {\text{ }}99 \times 21 - \root 3 \of x = 1968 \cr & {\text{Then,}} \cr & \Leftrightarrow 2079 - \root 3 \of x = 1968 \cr & \Leftrightarrow \root 3 \of x = 2079 - 1968 \cr & \Leftrightarrow \root 3 \of x = 111 \cr & \Leftrightarrow x = {\left( {111} \right)^3} \cr & \Leftrightarrow x = 1367631 \cr} $$

$$\eqalign{ & = \sqrt {0.000064} \cr & = \sqrt {\frac{{64}}{{{{10}^6}}}} \cr & = \frac{8}{{{{10}^3}}} \cr & = \frac{8}{{1000}} \cr & = 0.008 \cr & \therefore \root 3 \of {\sqrt {0.000064} } \cr & = \root 3 \of {0.008} \cr & = \root 3 \of {\frac{8}{{1000}}} \cr & = \frac{2}{{10}} \cr & = 0.2 \cr} $$

$$\eqalign{ & = \sqrt {0.01} \times \root 3 \of {0.008} - 0.02{\text{ }} \cr & = \sqrt {{{\left( {0.1} \right)}^2}} \times \root 3 \of {{{\left( {0.2} \right)}^3}} - 0.02 \cr & = 0.1 \times 0.2 - 0.02 \cr & = 0.02 - 0.02 \cr & = 0 \cr} $$

$$\eqalign{ & = \root 3 \of {\frac{{0.2 \times 0.2 \times 0.2 + 0.04 \times 0.04 \times 0.04}}{{0.4 \times 0.4 \times 0.4 + 0.08 \times 0.08 \times 0.08}}} \cr & = \root 3 \of {\frac{{0.008 + 0.000064}}{{0.064 + 0.000512}}} \cr & = \root 3 \of {\frac{{0.008064}}{{0.064512}}} \cr & = \root 3 \of {\frac{{8064}}{{64512}}} \cr & = \root 3 \of {\frac{1}{8}} \cr & = \frac{1}{2} \cr & = 0.5 \cr} $$

$$21600 = {2^5} \times {3^3} \times {5^2}$$
To make it a perfect cube, it must be multiplied by (2 × 5), i.e., 10

$$3600 = {2^3} \times {5^2} \times {3^2} \times 2$$
To make it a perfect cube, it must be divided by   $${5^2} \times {3^2} \times 2,i.e.,450$$