Practice MCQ Questions and Answer on Square Root and Cube Root

41.

The value of $$\sqrt{0.01}\text{ + }$$ $$\sqrt{0.81}\text{ + }$$ $$\sqrt{1.21}\text{ + }$$ $$\sqrt{0.0009}$$   is = ?

• (A) 2.03
• (B) 2.1
• (C) 2.11
• (D) 2.13

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

42.

Solved $$\sqrt[4]{{\left(625\right)}^3}=?$$

• (A) $$\sqrt[3]{1875}$$
• (B) 25
• (C) 125
• (D) None of these

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

43.

The square root of $$\left({272}^2-{128}^2\right)$$   is = ?

• (A) 144
• (B) 200
• (C) 240
• (D) 256

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

44.

The greatest four digit perfect square number is = ?

• (A) 9000
• (B) 9801
• (C) 9900
• (D) 9981

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 $$\eqalign{ & \,\,\,\,\,\,\,9|\overline {99} \,\,\overline {99} \,\,(99 \cr & \,\,\,\,\,\,\,\,\,\,|\,81 \cr & \,\,\,\,\,\,\,\,\,\,| - - - - - - \cr & \,189|\,18\,99 \cr & \,\,\,\,\,\,\,\,\,\,|\,17\,01 \cr & \,\,\,\,\,\,\,\,\,\,| - - - - - - \cr & \,\,\,\,\,\,\,\,\,\,|\,\,\,\,1\,98 \cr & \therefore {\text{Required number}} \cr & = \left( {9999 - 198} \right) \cr & = 9801 \cr} $$

45.

The value of $$\sqrt{2}$$   up to three places of decimal is = ?

• (A) 1.410
• (B) 1.412
• (C) 1.413
• (D) 1.414

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 $$\eqalign{ & \,\,\,\,\,\,\,\,\,\,1|\overline 2 \,.\,\,\overline {00} \,\,\overline {00} \,\,\overline {00} \,(1.414 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,|1 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,| - - - - - - - - \cr & \,\,\,\,\,\,24|\,\,1\,00 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,|\,\,\,\,96 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,| - - - - - - - - \cr & \,\,\,281\,|\,\,\,\,\,\,\,\,\,400 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,\,\,281 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,| - - - - - - - \cr & 2824\,|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,11900 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,11296 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,| - - - - - - - \cr & \therefore \sqrt 2 = 1.414 \cr} $$

46.

If $$2\ast 3=\sqrt{13}$$   and 3 * 4 = 5, then the value of 5 * 12 is ?

• (A) $$\sqrt{17}$$
• (B) $$\sqrt{29}$$
• (C) 12
• (D) 13

Show Answer

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

47.

$$1728÷\sqrt[3]{262144}\times ?-288$$      = 4491

• (A) 148
• (B) 156
• (C) 173
• (D) 177

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 $$\eqalign{ & = 262144 \cr & = 8 \times 8 \times 8 \times 8 \times 8 \times 8 \cr & = {8^6} \cr & \therefore \root 3 \of {262144} \cr & = {8^2} \cr & = 64 \cr & {\text{Let, }} \cr & {\text{1728}} \div \root 3 \of {262144} \times x - 288 = 4491 \cr & {\text{Then,}} \cr & \Leftrightarrow 1728 \div 64 \times x - 288 = 4491 \cr & \Leftrightarrow 27x = 4779 \cr & \Leftrightarrow x = \frac{{4779}}{{27}} \cr & \Leftrightarrow x = 177 \cr} $$ $$\eqalign{ & 8|262144 \cr & - - - - - - - - \cr & 8|\,\,\,32768 \cr & - - - - - - - - \cr & 8|\,\,\,\,4096 \cr & - - - - - - - - \cr & 8|\,\,\,\,\,512 \cr & - - - - - - - - \cr & 8|\,\,\,\,\,\,64 \cr & - - - - - - - - \cr & 8|\,\,\,\,\,\,\,8 \cr & - - - - - - - - \cr} $$

48.

If $$\frac{52}{x}=\sqrt{\frac{169}{289}}\text{,}$$   the value of x is = ?

• (A) 52
• (B) 58
• (C) 62
• (D) 68

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 $$\eqalign{ & \Leftrightarrow \frac{{52}}{x}{\text{ = }}\sqrt {\frac{{169}}{{289}}} \cr & \Leftrightarrow \frac{{52}}{x} = \frac{{13}}{{17}} \cr & \Leftrightarrow x = \left( {\frac{{52 \times 17}}{{13}}} \right) \cr & \Leftrightarrow x = 68 \cr} $$

49.

If $$\sqrt{5}=2.236\text{,}$$   then the value of $$\frac{\sqrt{5}}{2}-$$  $$\frac{10}{\sqrt{5}}+$$  $$\sqrt{125}$$  is equal to = ?

• (A) 5.59
• (B) 7.826
• (C) 8.944
• (D) 10.062

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

50.

1250 oranges were distributed among a group of girls of a class. Each girl got twice as many oranges as the number of girls in that group. The number of girls in the group was = ?

• (A) 25
• (B) 45
• (C) 50
• (D) 100

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