Practice MCQ Questions and Answer on Square Root and Cube Root

Solution:

 $$\eqalign{ & {\text{ }}a = \frac{{\sqrt 5 + 1}}{{\sqrt 5 - 1}} \cr & \,\,\,\,\,\, {\text{ = }}\frac{{\left( {\sqrt 5 + 1} \right)}}{{\left( {\sqrt 5 - 1} \right)}} \times \frac{{\left( {\sqrt 5 + 1} \right)}}{{\left( {\sqrt 5 + 1} \right)}} \cr & \,\,\,\,\,\,\, = \frac{{{{\left( {\sqrt 5 + 1} \right)}^2}}}{{\left( {5 - 1} \right)}} \cr & \,\,\,\,\,\,\, = \frac{{5 + 1 + 2\sqrt 5 }}{4} \cr & \,\,\,\,\,\,\, = \left( {\frac{{3 + \sqrt 5 }}{2}} \right) \cr & {\text{ }}b = \frac{{\sqrt 5 - 1}}{{\sqrt 5 + 1}} \cr & \,\,\,\,\,\,\, = \frac{{\left( {\sqrt 5 - 1} \right)}}{{\left( {\sqrt 5 + 1} \right)}} \times \frac{{\left( {\sqrt 5 - 1} \right)}}{{\left( {\sqrt 5 - 1} \right)}} \cr & \,\,\,\,\,\,\, = \frac{{{{\left( {\sqrt 5 - 1} \right)}^2}}}{{\left( {5 - 1} \right)}} \cr & \,\,\,\,\,\,\, = \frac{{5 + 1 - 2\sqrt 5 }}{4} \cr & \,\,\,\,\,\,\, = \left( {\frac{{3 - \sqrt 5 }}{2}} \right) \cr & \,\,\,\,\,\,\, \therefore {a^2} + {b^2} \cr & \,\,\,\,\,\,\, = {\left( {\frac{{3 + \sqrt 5 }}{2}} \right)^2} + {\left( {\frac{{3 - \sqrt 5 }}{2}} \right)^2} \cr & \,\,\,\,\,\,\, = \frac{{{{\left( {3 + \sqrt 5 } \right)}^2}}}{4} + \frac{{{{\left( {3 - \sqrt 5 } \right)}^2}}}{4} \cr & \,\,\,\,\,\,\, = \frac{{{{\left( {3 + \sqrt 5 } \right)}^2} + {{\left( {3 - \sqrt 5 } \right)}^2}}}{4} \cr & \,\,\,\,\,\,\, = \frac{{9 + 2.3.\sqrt 5 + 5 + 9 - 2.3.\sqrt 5 + 5}}{4} \cr & \,\,\,\,\,\,\, = \frac{{2\left( {9 + 5} \right)}}{4} \cr & \,\,\,\,\,\,\, = \frac{{28}}{4} \cr & \,\,\,\,\,\,\, = 7 \cr & {\text{Also, }} \cr & ab = \frac{{\left( {3 + \sqrt 5 } \right)}}{2} \times \frac{{\left( {3 - \sqrt 5 } \right)}}{2} \cr & \,\,\,\,\,\,\, = \frac{{\left( {9 - 5} \right)}}{4} \cr & \,\,\,\,\,\,\, = 1 \cr & \,\,\,\,\,\,\, \therefore \left( {\frac{{{a^2} + ab + {b^2}}}{{{a^2} - ab + {b^2}}}} \right) \cr & \,\,\,\,\,\,\, = \frac{{\left( {{a^2} + {b^2}} \right) + ab}}{{\left( {{a^2} + {b^2}} \right) - ab}} \cr & \,\,\,\,\,\,\, = \frac{{7 + 1}}{{7 - 1}} \cr & \,\,\,\,\,\,\, = \frac{8}{6} \cr & \,\,\,\,\,\,\, = \frac{4}{3} \cr} $$

Solution:

 $$\eqalign{ & {\text{Let}}, \cr & \sqrt {11881} \times \sqrt x = 10137 \cr & {\text{Then}}, \cr & 109 \times \sqrt x = 10137 \cr & \Leftrightarrow \sqrt x = \frac{{10137}}{{109}} = 93 \cr & \Leftrightarrow x = {\left( {93} \right)^2} \cr & \Leftrightarrow x = 8649 \cr} $$

Solution:

 Given expression, $$ = \sqrt {\frac{1}{{100}}} + \sqrt {\frac{{81}}{{100}}} + \sqrt {\frac{{121}}{{100}}} + $$      $$\sqrt {\frac{9}{{10000}}} $$ $$\eqalign{ & = \frac{1}{{10}} + \frac{9}{{10}} + \frac{{11}}{{10}} + \frac{3}{{100}} \cr & = 0.1 + 0.9 + 1.1 + 0.03 \cr & = 2.13 \cr} $$

Solution:

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

Solution:

 $$\eqalign{ & \,\,\,\,\,2|4\,\,\overline {12} \,\,\overline {09} \,(203 \cr & \,\,\,\,\,\,\,\,\,|4 \cr & \,\,\,\,\,\,\,\,\,| - - - - - - - - \cr & \,\,40|\,\,\,\,\,\,\,\,12 \cr & \,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,\,\,\,\,0 \cr & \,\,\,\,\,\,\,\,\,| - - - - - - - - \cr & \,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,\,\,\,12\,09 \cr & 403|\,\,\,\,\,\,\,\,\,12\,09 \cr & \,\,\,\,\,\,\,\,\,| - - - - - - - - \cr & \,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,\,\,\,\,\,\,x \cr & \,\,\,\,\,\,\,\,\,| - - - - - - - - \, \cr & \therefore \sqrt {41209} = 203 \cr} $$

Solution:

 Sum of prime numbers greater than 4 but less than 16 $$\eqalign{ & = \left( {5 + 7 + 11 + 13} \right) \cr & = 36 \cr & \therefore \frac{1}{4} \times 36 \cr & = 9 \cr & = {3^2} \cr} $$

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 $$\eqalign{ & = {2^2} \times {2^2} \times {3^2} \times {5^2} \cr & = 3600 \cr} $$

Solution:

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

Solution:

 $$\eqalign{ & {\text{Let,}} \cr & {\text{ }}\frac{x}{{\sqrt {128} }} = \frac{{\sqrt {162} }}{x} \cr & {\text{Then,}} \cr & \Leftrightarrow {x^2} = \sqrt {128 \times 162} \cr & \Leftrightarrow {x^2} = \sqrt {64 \times 2 \times 18 \times 9} \cr & \Leftrightarrow {x^2} = \sqrt {{8^2} \times {6^2} \times {3^2}} \cr & \Leftrightarrow {x^2} = 8 \times 6 \times 3 \cr & \Leftrightarrow {x^2} = 144 \cr & \Leftrightarrow x = \sqrt {144} \cr & \Leftrightarrow x = 12 \cr} $$

Solution:

 $$\eqalign{ & \because a = \frac{{\sqrt 3 + \sqrt 2 }}{{\sqrt 3 - \sqrt 2 }} \cr & = \frac{{\sqrt 3 + \sqrt 2 }}{{\sqrt 3 - \sqrt 2 }} \times \frac{{\sqrt 3 + \sqrt 2 }}{{\sqrt 3 + \sqrt 2 }} \cr & = \frac{{{{\left( {\sqrt 3 + \sqrt 2 } \right)}^2}}}{{{{\left( {\sqrt 3 } \right)}^2} - {{\left( {\sqrt 2 } \right)}^2}}} \cr & = \frac{{3 + 2 + 2\sqrt 6 }}{{3 - 2}} \cr & = 5 + 2\sqrt 6 \cr & \because b = \frac{{\sqrt 3 - \sqrt 2 }}{{\sqrt 3 + \sqrt 2 }} \cr & {\text{ = }}\frac{{\sqrt 3 - \sqrt 2 }}{{\sqrt 3 + \sqrt 2 }} \times \frac{{\sqrt 3 - \sqrt 2 }}{{\sqrt 3 - \sqrt 2 }} \cr & {\text{ = }}\frac{{{{\left( {\sqrt 3 - \sqrt 2 } \right)}^2}}}{{{{\left( {\sqrt 3 } \right)}^2} - {{\left( {\sqrt 2 } \right)}^2}}} \cr & = \frac{{3 + 2 - 2\sqrt 6 }}{{3 - 2}} \cr & = 5 - 2\sqrt 6 \cr & \therefore {\text{ }}{a^2} + {b^2} \cr & = {\left( {5 + 2\sqrt 6 } \right)^2} + {\left( {5 - 2\sqrt 6 } \right)^2} \cr & = 2\left[ {{{\left( 5 \right)}^2} + {{\left( {2\sqrt 6 } \right)}^2}} \right] \cr & = 2\left( {25 + 24} \right) \cr & = 2 \times 49 \cr & = 98 \cr} $$