Square Root And Cube Root — Part 1 MCQs with Explanations

$$\eqalign{ & {\left( {.000216} \right)^{\frac{1}{3}}} = {\left( {\frac{{216}}{{{{10}^6}}}} \right)^{\frac{1}{3}}} \cr & = {\left( {\frac{{6 \times 6 \times 6}}{{{{10}^2} \times {{10}^2} \times {{10}^2}}}} \right)^{\frac{1}{3}}} \cr & = \frac{6}{{{{10}^2}}} \cr & = \frac{6}{{100}} \cr & = 0.06 \cr} $$

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

L.C.M. of 21, 36, 66 = 2772. Now, 2772 = 2 x 2 x 3 x 3 x 7 x 11 To make it a perfect square, it must be multiplied by 7 x 11. So, required number = 2 2 x 3 2 x 7 2 x 11 2 = 213444

$$\eqalign{ & \,\,\,\,\,\,1|\overline 1 .\overline {56} \overline {25} \,(\,1.25 \cr & \,\,\,\,\,\,\,\,\,|1 \cr & \,\,\,\,\,\,\,\,\,| - - - - - \cr & \,\,\,22|\,\,\,\,56 \cr & \,\,\,\,\,\,\,\,\,|\,\,\,\,44 \cr & \,\,\,\,\,\,\,\,\,| - - - - - \cr & 245|\,\,\,1225 \cr & \,\,\,\,\,\,\,\,\,|\,\,\,1225 \cr & \,\,\,\,\,\,\,\,\,| - - - - - \cr & \,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,\,\,\,\,\,x \cr & \,\,\,\,\,\,\,\,\,| - - - - - \cr & \therefore \sqrt {1.5625} = 1.25 \cr} $$

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

$$\eqalign{ & \sqrt {4{a^2} - 4a + 1} + 3a \cr & = \sqrt {{{\left( 1 \right)}^2} + {{\left( {2a} \right)}^2} - 2 \times 1 \times 2a} + 3a \cr & = \sqrt {{{\left( {1 - 2a} \right)}^2}} + 3a \cr & = \left( {1 - 2a} \right) + 3a \cr & = \left( {1 + a} \right) \cr & = \left( {1 + 0.1039} \right) \cr & = 1.1039 \cr} $$

$$\eqalign{ & x = \frac{{ {\sqrt 3 + 1} }}{{ {\sqrt 3 - 1} }} \times \frac{{ {\sqrt 3 + 1} }}{{ {\sqrt 3 + 1} }} \cr & \,\,\,\,\,\, = \frac{{{{\left( {\sqrt 3 + 1} \right)}^2}}}{{ {3 - 1} }} \cr & \,\,\,\,\,\, = \frac{{3 + 1 + 2\sqrt 3 }}{2} \cr & \,\,\,\,\,\, = 2 + \sqrt 3 \cr & y = \frac{{ {\sqrt 3 - 1} }}{{ {\sqrt 3 + 1} }} \times \frac{{ {\sqrt 3 - 1} }}{{ {\sqrt 3 - 1} }} \cr & \,\,\,\,\,\, = \frac{{{{\left( {\sqrt 3 - 1} \right)}^2}}}{{ {3 - 1} }} \cr & \,\,\,\,\,\, = \frac{{3 + 1 - 2\sqrt 3 }}{2} \cr & \,\,\,\,\,\, = 2 - \sqrt 3 \cr & \therefore {x^2} + {y^2} = {\left( {2 + \sqrt 3 } \right)^2} + {\left( {2 - \sqrt 3 } \right)^2} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 2\left( {4 + 3} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 14 \cr} $$

Money collected = (59.29 x 100) paise = 5929 paise. ∴ Number of members = $$\sqrt {5929} $$ = 77

$$\eqalign{ & \sqrt {\left( {7 + 3\sqrt 5 } \right)\left( {7 - 3\sqrt 5 } \right)} \cr & = \sqrt {{{\left( 7 \right)}^2} - {{\left( {3\sqrt 5 } \right)}^2}} \cr & = \sqrt {49 - 45} \cr & = \sqrt 4 \cr & = 2 \cr} $$

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

$$\eqalign{ & {\text{Give}}\,{\text{Expression}} \cr & = \frac{{25}}{{11}} \times \frac{{14}}{5} \times \frac{{11}}{{14}} \cr & = 5 \cr} $$

$$\eqalign{ & {\text{Let}}\sqrt {0.0169 \times x} = 1.3 \cr & {\text{Then}},\,0.0169x = {\left( {1.3} \right)^2} = 1.69 \cr & \Rightarrow x = \frac{{1.69}}{{0.0169}} = 100 \cr} $$

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

A number ending in 8 can never be a perfect square.

$$\eqalign{ & \,\,\,\,\,\,\,\,2|\overline 6 \overline {40} \overline {09} \,(\,\,253 \cr & \,\,\,\,\,\,\,\,\,\,\,|4 \cr & \,\,\,\,\,\,\,\,\,\,\,| - - - - - \cr & \,\,\,\,45\,|240 \cr & \,\,\,\,\,\,\,\,\,\,\,|225 \cr & \,\,\,\,\,\,\,\,\,\,\,| - - - - - \cr & 503\,\,|\,\,\,\,\,\,1509 \cr & \,\,\,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,1509 \cr & \,\,\,\,\,\,\,\,\,\,\,| - - - - - \cr & \,\,\,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,\,\,\,\,\,x \cr & \,\,\,\,\,\,\,\,\,\,\,| - - - - - \cr & \therefore \sqrt {64009} = 253 \cr} $$

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

$$\therefore \sqrt {123454321} = 11111$$

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

$$\eqalign{ & \sqrt {\sqrt {17956} + \sqrt {24025} } \cr & = \sqrt {134 + 155} \cr & = \sqrt {289} \cr & = 17 \cr} $$

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

$$\eqalign{ & {\text{Given expression,}} \cr & = \sqrt {176 + 49} \cr & = \sqrt {225} \cr & = 15 \cr} $$

$$\eqalign{ & {\text{Let, }} \cr & {\left( {15} \right)^2} + {\left( {18} \right)^2} - 20 = \sqrt x \cr & {\text{Then, }} \cr & \sqrt x = 225 + 324 - 20 \cr & \Leftrightarrow x = {\left( {529} \right)^2} \cr & \Leftrightarrow x = 279841 \cr} $$

$$\eqalign{ & {\text{Let}}, \cr & \sqrt x \times \sqrt {484} = 1034 \cr & {\text{Then}}, \cr & \sqrt x \times 22 = 1034 \cr & \Leftrightarrow \sqrt x = \frac{{1034}}{{22}} = 47 \cr & \Leftrightarrow x = {\left( {47} \right)^2} \cr & \Leftrightarrow x = 2209 \cr} $$

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

$$\eqalign{ & \frac{{4050}}{{\sqrt x }} = 450 \cr & \Leftrightarrow \sqrt x = \frac{{4050}}{{450}} \cr & \Leftrightarrow \sqrt x = 9 \cr & \Leftrightarrow x = {\left( 9 \right)^2} \cr & \Leftrightarrow x = 81 \cr} $$

$$\eqalign{ & {\text{Let}}, \cr & \sqrt {\frac{{16}}{{25}}} \times \sqrt {\frac{x}{{25}}} \times \frac{{16}}{{25}} = \frac{{256}}{{625}} \cr & {\text{Then}}, \cr & \frac{4}{5} \times \frac{{\sqrt x }}{5} \times \frac{{16}}{{25}} = \frac{{256}}{{625}} \cr & \Leftrightarrow \frac{{64\sqrt x }}{{625}} = \frac{{256}}{{625}} \cr & \Leftrightarrow \sqrt x = \frac{{256}}{{625}} \times \frac{{625}}{{64}} \cr & \Leftrightarrow x = {4^2} \cr & \Leftrightarrow x = 16 \cr} $$

$$\eqalign{ & = \sqrt {\left( {{{272}^2} - {{128}^2}} \right)} \cr & = \sqrt {\left( {272 + 128} \right)\left( {272 - 128} \right)} \cr & = \sqrt {400 \times 144} \cr & = \sqrt {57600} \cr & = 240 \cr} $$

$$\eqalign{ & = 10y\sqrt {{y^3} - {y^2}} \cr & = 10 \times 5\sqrt {{5^3} - {5^2}} \cr & = 50 \times \sqrt {125 - 25} \cr & = 50 \times \sqrt {100} \cr & = 50 \times 10 \cr & = 500 \cr} $$

$$\eqalign{ & = \sqrt {\frac{{25}}{{81}} - \frac{1}{9}} \cr & = \sqrt {\frac{{25 - 9}}{{81}}} \cr & = \sqrt {\frac{{16}}{{81}}} \cr & = \frac{{\sqrt {16} }}{{\sqrt {81} }} \cr & = \frac{4}{9} \cr} $$

$$\eqalign{ & {\text{Let,}} \cr & {\left[ {{{\left( {\sqrt {81} } \right)}^2}} \right]^2} = {\left( x \right)^2} \cr & {\text{Then,}} \cr & \Leftrightarrow {x^2} = {\left( {81} \right)^2} \cr & \Leftrightarrow x = 81 \cr} $$

$$\eqalign{ & \,\,\,\,\,\,1|\overline 1 \,\,\overline {58} \,\,\overline {76} \,(126 \cr & \,\,\,\,\,\,\,\,\,|1 \cr & \,\,\,\,\,\,\,\,\,| - - - - - - - - \cr & \,\,22|\,\,\,\,\,\,58 \cr & \,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,44 \cr & \,\,\,\,\,\,\,\,\,| - - - - - - - - \cr & 246\,|\,\,\,\,\,\,\,14\,76 \cr & \,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,\,\,14\,76 \cr & \,\,\,\,\,\,\,\,\,| - - - - - - - \cr & \,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,\,\,\,\,\,\,\,x \cr & \,\,\,\,\,\,\,\,\,| - - - - - - - \cr & \therefore \sqrt {15876} = 126 \cr} $$

$$\eqalign{ & \,\,\,\,\,\,\,\,\,\,1|\overline 3 \,.\,\,\overline {00} \,\,\overline {00} \,\,\overline {00} \,(1.732 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,|1 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,| - - - - - - - - \cr & \,\,\,\,\,\,27|\,\,2\,\,\,00 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,|\,\,1\,\,\,89 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,| - - - - - - - - \cr & \,\,\,343\,|\,\,\,\,\,\,\,\,\,11\,\,00 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,\,\,\,10\,\,29 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,| - - - - - - - \cr & 3492\,|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,71\,\,00 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,69\,\,84 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,| - - - - - - - \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1\,\,16 \cr} $$ $$\eqalign{ & \therefore \sqrt 3 \cr & = 1.73 \cr & = \frac{{173}}{{100}} \cr} $$

The squares of numbers having 1 and 9 as the unit's digit end in the digit 1. $$\eqalign{ & {\text{Such numbers are,}} \cr & 1,9,11,19,21,29,31,39,41,49{\text{ i}}{\text{.e}}{\text{.,}} \cr & {\text{There are 10 such numbers}}{\text{.}} \cr & \therefore {\text{Required percentage}} \cr & = \left( {\frac{{10}}{{50}} \times 100} \right)\% \cr & = 20\% \cr} $$

$$\eqalign{ & {\text{Let the given number be }}x \cr & {\text{Then,}} \cr & \Leftrightarrow {x^2} - 25 = {\left( {x - 25} \right)^2}{\text{ }} \cr & \Leftrightarrow {x^2} - 25 = {x^2} + 625 - 50x \cr & \Leftrightarrow 50x = 650 \cr & \Leftrightarrow x = 13 \cr} $$

$$\eqalign{ & {\left( {11} \right)^2} = 121{\text{ }} \cr & {\text{And }} \cr & {\left( {17} \right)^2} = 289 \cr} $$ So, the perfect squares between 120 and 300 are the squares of numbers from 11 to 17. Clearly, these are 7 in number.

The first perfect square number after 50 is 64 $$\left( {64 = {8^2}} \right)$$ and the last perfect square number before 1000 is 961 $$\left[ {961 = {{\left( {31} \right)}^2}} \right]$$ So, the perfect squares between 50 and 1000 are the squares of numbers from 8 to 31. (31 - 8) + 1 = 24 Clearly, these are 24 in number.

Clearly, the man was born between 1800 and 1850 is 1849. $${\text{And, }}1849 = {\left( {43} \right)^2}$$ So, the man was 43 years old in 1849 $$\eqalign{ & {\text{Thus, he was born in }} \cr & = \left( {1849 - 43} \right) \cr & = 1806 \cr} $$

$$\eqalign{ & \Rightarrow {\text{Q}} = \sqrt {\frac{{{{\left( {8{\text{R}}} \right)}^2}}}{4}} \cr & \Rightarrow {\text{Q}} = \frac{{\sqrt {{{\left( {8{\text{R}}} \right)}^2}} }}{{\sqrt 4 }} \cr & \Rightarrow {\text{Q}} = \frac{{8{\text{R}}}}{2} \cr & \Rightarrow {\text{Q}} = 4{\text{R}} \cr} $$

The smallest such number is 1444 $$\left[ {1444 = {{\left( {38} \right)}^2}} \right]$$ It lies between 1000 and 2000.

$$\eqalign{ & {\text{We have,}} \cr & 1 \times 2 \times 3 \times 4 = 24 \cr & {\text{And }}24 + 1 = 25\left[ {25 = {5^2}} \right] \cr & 2 \times 3 \times 4 \times 5 = 120{\text{ }} \cr & {\text{And 1}}20 + 1 = 121\left[ {121 = {{11}^2}} \right] \cr & 3 \times 4 \times 5 \times 6 = 360{\text{ }} \cr & {\text{And }}360 + 1 = 361\left[ {361 = {{19}^2}} \right] \cr & 4 \times 5 \times 6 \times 7 = 840{\text{ }} \cr & {\text{And }}840 + 1 = 841\left[ {841 = {{29}^2}} \right] \cr & \therefore p = 1 \cr} $$