Practice MCQ Questions and Answer on Surds and Indices

11.

The value of $$\sqrt[3]{2^4\sqrt{2^{-5}\sqrt{2^6}}}$$    is = ?

• (A) 1
• (B) 2
• (C) $${\text{2}}^{\frac{5}{3}}$$
• (D) 25

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 $$\eqalign{ & \root 3 \of {{2^4}\sqrt {{2^{ - 5}}\sqrt {{2^6}} } } {\text{ }} \cr & = \root 3 \of {{2^4}\sqrt {{2^{ - 5}}{{\left( {{2^6}} \right)}^{\frac{1}{2}}}} } {\text{ }} \cr & = \root 3 \of {{2^4}\sqrt {{2^{ - 5}}{{\left( 2 \right)}^{\left( {6 \times \frac{1}{2}} \right)}}} } {\text{ }} \cr & = \root 3 \of {{2^4}\sqrt {{2^{ - 5}}{{.2}^3}} } {\text{ }} \cr & = \root 3 \of {{2^4}\sqrt {{2^{\left( { - 5 + 3} \right)}}} } {\text{ }} \cr & = \root 3 \of {{2^4}\sqrt {{2^{ - 2}}} } {\text{ }} \cr & = \root 3 \of {{2^4}.{{\left( {{2^2}} \right)}^{\frac{1}{2}}}} \cr & {\text{ = }}\root 3 \of {{2^4}{{.2}^{\left( { - 2 \times \frac{1}{2}} \right)}}} \cr & = \root 3 \of {{2^4}{{.2}^{\left( { - 1} \right)}}} \cr & = \root 3 \of {{2^{\left( {4 - 1} \right)}}} \cr & {\text{ = }}\root 3 \of {{2^3}} \cr & {\text{ = }}{\left( {{2^3}} \right)^{\frac{1}{3}}} \cr & {\text{ = }}{{\text{2}}^{\left( {3 \times \frac{1}{3}} \right)}} \cr & {\text{ = 2 }} \cr} $$

12.

The value of $$\frac{1}{1+\sqrt{2}+\sqrt{3}}+\frac{1}{1-\sqrt{2}+\sqrt{3}}\text{is:}$$

• (A) √2
• (B) √3
• (C) 1
• (D) 4(√3 + √2)

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

13.

The value of $$\frac{1}{\sqrt{3.25}+\sqrt{2.25}}$$    $$+\frac{1}{\sqrt{4.25}+\sqrt{3.25}}$$    $$+\frac{1}{\sqrt{5.25}+\sqrt{4.25}}$$    $$+\frac{1}{\sqrt{6.25}+\sqrt{5.25}}$$  ÃƒÂƒÃ‚ƒÃ‚ƒÃ‚ƒÃ‚ƒÃ‚ƒÃ‚ƒÃ‚ƒÃ‚ƒÃ‚‚  is = ?

• (A) 1.00
• (B) 1.25
• (C) 1.50
• (D) 2.25

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 $$\eqalign{ & \frac{1}{{\sqrt {3.25} + \sqrt {2.25} }} \times \frac{{\sqrt {3.25} - \sqrt {2.25} }}{{\sqrt {3.25} - \sqrt {2.25} }} \cr & = \frac{{\sqrt {3.25} - \sqrt {2.25} }}{{3.25 - 2.25}} \cr & = \sqrt {3.25} - \sqrt {2.25} \,......(i) \cr & {\text{Similarly}} \cr & \frac{1}{{\sqrt {4.25} + \sqrt {3.25} }} \cr & = \sqrt {4.25} - \sqrt {3.25} \,.......(ii) \cr & \frac{1}{{\sqrt {5.25} + \sqrt {4.25} }} \cr & = \sqrt {5.25} - \sqrt {4.25} \,.......(iii) \cr & \frac{1}{{\sqrt {6.25} + \sqrt {5.25} }} \cr & = \sqrt {6.25} - \sqrt {5.25} \,.......(iv) \cr & {\text{Now}}\,{\text{add}}\,{\text{all}}\,{\text{them}} \cr & \sqrt {3.25} - \sqrt {2.25} + \sqrt {4.25} - \sqrt {3.25} + \sqrt {5.25} - \sqrt {4.25} + \sqrt {6.25} - \sqrt {5.25} \cr & = \sqrt {6.25} - \sqrt {2.25} \cr & = 2.5 - 1.5 \cr & = 1 \cr} $$

14.

Which of the following relation is/are true? $$\begin{array}{rl}& \text{I}.{\left(27\right)}^{\frac{1}{3}}>{\left(13\right)}^{\frac{1}{2}}{\left(47\right)}^{\frac{1}{6}}\\ & \text{II}.{\left(23\right)}^{\frac{1}{3}}{\left(49\right)}^{\frac{1}{2}}{\left(52\right)}^{\frac{1}{6}}\\ & \text{III}.{\left(53\right)}^{\frac{1}{6}}{\left(41\right)}^{\frac{1}{3}}{\left(37\right)}^{\frac{1}{2}}\end{array}$$

• (A) Only III
• (B) Both I and II
• (C) II and III only
• (D) All relations are false

Show Answer

 $$\eqalign{ & {\text{Take LCM of }}\left( {2,\,3\,\& \,6} \right) = 6 \cr & {\text{I}}.{\left( {27} \right)^{\frac{1}{3}}} > {\left( {13} \right)^{\frac{1}{2}}} {\left( {47} \right)^{\frac{1}{6}}} \cr & = {\left( {27} \right)^{\frac{2}{{3 \times 2}}}} > {\left( {13} \right)^{\frac{3}{{2 \times 3}}}} {\left( {47} \right)^{\frac{{1 \times 1}}{{6 \times 1}}}} \cr & = {\left( {27} \right)^{\frac{2}{6}}} > {\left( {13} \right)^{\frac{3}{6}}} {\left( {47} \right)^{\frac{1}{6}}} \cr & = {\left( {729} \right)^{\frac{1}{6}}} > {\left( {2197} \right)^{\frac{1}{6}}} {\left( {47} \right)^{\frac{1}{6}}} \cr & {\text{This statement is false}} \cr & {\text{II}}.{\left( {23} \right)^{\frac{1}{3}}} {\left( {49} \right)^{\frac{1}{2}}} {\left( {52} \right)^{\frac{1}{6}}} \cr & = {\left( {23} \right)^{\frac{2}{{3 \times 2}}}} {\left( {49} \right)^{\frac{3}{{2 \times 3}}}} {\left( {52} \right)^{\frac{{1 \times 1}}{{6 \times 1}}}} \cr & = {\left( {529} \right)^{\frac{1}{6}}} {\left( {117649} \right)^{\frac{1}{6}}} {\left( {52} \right)^{\frac{1}{6}}} \cr & {\text{This statement is false}} \cr & {\text{III}}.{\left( {53} \right)^{\frac{1}{6}}} {\left( {41} \right)^{\frac{1}{3}}} {\left( {37} \right)^{\frac{1}{2}}} \cr & = {\left( {53} \right)^{\frac{1}{{6 \times 1}}}} {\left( {41} \right)^{\frac{2}{{3 \times 2}}}} {\left( {37} \right)^{\frac{3}{{2 \times 3}}}} \cr & = {\left( {53} \right)^{\frac{1}{6}}} {\left( {1681} \right)^{\frac{1}{6}}} {\left( {50653} \right)^{\frac{1}{6}}} \cr & {\text{This statement is true}} \cr} $$

15.

The quotient when 10¹⁰⁰ is divided by 5⁷⁵ is

• (A) 225 × 1075
• (B) 1025
• (C) 275
• (D) 275 × 1025

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 $$\eqalign{ & {\text{Expression,}} \cr & {\text{ = }}\frac{{{{\left( {10} \right)}^{100}}}}{{{{\left( 5 \right)}^{75}}}} \cr & = \frac{{{{\left( {2 \times 5} \right)}^{100}}}}{{{{\left( 5 \right)}^{75}}}} \cr & = \frac{{{{\left( 2 \right)}^{100}} \times {{\left( 5 \right)}^{100}}}}{{{{\left( 5 \right)}^{75}}}} \cr & = {2^{100}} \times \frac{{{5^{100}}}}{{{5^{75}}}} \cr & = {2^{100}} \times {5^{\left( {100 - 75} \right)}}.....\left[ {\because \frac{{{a^m}}}{{{a^n}}} = {a^{m - n}}} \right] \cr & = {2^{100}} \times {5^{25}} \cr & = {2^{25}} \times {5^{25}} \times {2^{75}}.....\left[ {\because {a^m} \times {a^n} = {a^{m + n}}} \right] \cr & = {\left( {10} \right)^{25}} \times {2^{75}}.....\left[ {\because {a^m} \times {b^m} = a{b^m}} \right] \cr} $$

16.

Find the simplest value of $$\text{2}\sqrt{50}$$  + $$\sqrt{18}$$  - $$\sqrt{72}$$ = ?(given $$\sqrt{2}$$ = 1.414)

• (A) 4.242
• (B) 9.898
• (C) 10.6312
• (D) 8.484

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 $$\eqalign{ & {\text{2}}\sqrt {50} {\text{ + }}\sqrt {18} - \sqrt {72} \cr & \Rightarrow {\text{2}} \times {\text{5}}\sqrt 2 {\text{ + 3}}\sqrt 2 - 6\sqrt 2 \cr & \Rightarrow 13\sqrt 2 - 6\sqrt 2 \cr & \Rightarrow 7\sqrt 2 \cr & \Rightarrow 7 \times 1.414 \cr & \Rightarrow 9.898 \cr} $$

17.

$${\left(\frac{x^b}{x^c}\right)}^{\left(b+c-a\right)}.$$   $${\left(\frac{x^c}{x^a}\right)}^{\left(c+a-b\right)}.$$   $${\left(\frac{x^a}{x^b}\right)}^{\left(a+b-c\right)}=?$$

• (A) xabc
• (B) 1
• (C) xab+bc+ca
• (D) xa+b+c

Show Answer

 $${x^{\left( {b - c} \right)\left( {b + c - a} \right)}}.{x^{\left( {c - a} \right)\left( {c + a - b} \right)}}.{x^{\left( {a - b} \right)\left( {a + b - c} \right)}}$$   $$ = {x^{\left( {b - c} \right)\left( {b + c} \right) - a\left( {b - c} \right)}}.$$    $${x^{\left( {c - a} \right)\left( {c + a} \right) - b\left( {c - a} \right)}}.$$   $${x^{\left( {a - b} \right)\left( {a + b} \right) - c\left( {a - b} \right)}}$$ $$\eqalign{ & = {x^{\left( {{b^2} - {c^2} + {c^2} - {a^2} + {a^2} - {b^2}} \right)}}.{x^{ - a\left( {b - c} \right) - b\left( {c - a} \right) - c\left( {a - b} \right)}} \cr & = \left( {{x^0} \times {x^0}} \right) \cr & = \left( {1 \times 1} \right) \cr & = 1 \cr} $$

18.

If $$\sqrt{33}$$  = 5.745, then the value of the following is approximately:

• (A) 0.5223
• (B) 6.32
• (C) 2.035
• (D) 1

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Answer & Solution Answer: Option A No explanation is given for this question Let's Discuss on Board

19.

The least one among $$\text{2}\sqrt{3}\text{,}$$  $$\text{2}\sqrt[4]{5}\text{,}$$  $$\sqrt{8}\text{,}$$  $$\text{3}\sqrt{2}$$  is = ?

• (A) $$\text{2}\sqrt{3}$$
• (B) $$\text{2}\sqrt[4]{5}$$
• (C) $$\sqrt{8}$$
• (D) $$\text{3}\sqrt{2}$$

Show Answer

 $$2\sqrt 3 = {\left( {4 \times 3} \right)^{\frac{1}{2}}} \to {12^{\frac{1}{2}}} \to {12^{\frac{2}{4}}} \to \root 4 \of {144} $$ $$2\root 4 \of 5 = \root 4 \of {\left( {5 \times 16} \right)} \to {80^{\frac{1}{4}}} \to \root 4 \of {80} $$ $$\sqrt 8 = {8^{\frac{1}{2}}} \to {8^{\frac{1}{2}}} \to \boxed{\root 4 \of {64} }\,{\text{smallest}}$$ $$3\sqrt 2 = \sqrt {18} \to {18^{\frac{1}{2}}} \to {18^{\frac{2}{4}}} \to \root 4 \of {324} $$ $$\sqrt 8 \,{\text{ is answer}}$$

20.

If x = $$\sqrt{a\sqrt[3]{ab\sqrt{a\sqrt[3]{ab}}}}.....\propto ,$$     then the value of x is:

• (A) $$\sqrt[5]{a^{2}b}$$
• (B) $$\sqrt[5]{a^4{b}^4}$$
• (C) $$\sqrt[6]{a^{5}b}$$
• (D) $$\sqrt[5]{a^{4}b}$$

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 $$\eqalign{ & x = \sqrt {a\root 3 \of {ab\sqrt {a\root 3 \of {ab} } } } ..... \propto \cr & {\text{Square both side}} \cr & {x^2} = a\,\root 3 \of {ab\,x} \,\,\,\,\,\left( {\because \sqrt {a\root 3 \of {ab} } ..... \propto } \right) \cr & {\text{Again cube both sides}} \cr & {x^6} = {a^3}ab\,x \cr & {x^5} = {a^4}b \cr & x = \root 5 \of {{a^4}b} \cr} $$