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# Surds and Indices Questions and Answers for Competitive Exams

9 months ago 8.6K Views Surds and indices is a popular topic, which is often asked in the examinations. Some students have to face problems while solving these questions. If given the questions along with the answers, their solutions also are given, with the help of which you can easily understand these questions.

If you have to do good your performance then you must practice these surds and indices in Hindi questions.

## Important Surds and Indices Questions:

$$Q.1.\ 17^{3.5}×17^{7.3}÷17^{4.2}=17^?$$

$$(A) \ 8.4$$

$$(B) \ 8$$

$$(C) \ 6.6$$

$$(D) \ 6.4$$

Ans .  C

Solution .  $$\ 17^{3.5}×17^{7.3}÷17^{4.2}=17^?$$
$$→ {17^{3.5+7.3} \over17^{4.2}}=17^?$$
$$→ 17^{10.8-4.2}=17^?$$
$$17^6.6 = 17^?$$
$$Since, base \ are \ same$$
$$∴ ? =6.6$$

$$Q.2.Simplify\ \left( ^3\sqrt {^6\sqrt { 2^9} \ } \ \right)^4 × \left( ^6\sqrt {^3\sqrt { 2^9} \ } \ \right)^4$$

$$(A) \ 2^4$$

$$(B) \ 2^9$$

$$(C) \ 2^3$$

$$(D) \ 2^{16}$$

Ans .   A

Solution .   $$= \left(2^{9×{1\over6}×{1\over3}} \right)^4 ×\left(2^{9×{1\over3}×{1\over6}} \right)^4$$
$$= (2^{1/2})^4×(2^{1/2})^4$$
$$= 2^2×2^2$$
$$= 2^{2+2}=2^4$$

$$Q.3.\ If \ a = {\sqrt {3} \ \over2}, then \ the \ value \ of \ \sqrt { 1+a} \ +\sqrt {1-a} \ is$$

$$(A) \ 2-\sqrt {3} \$$

$$(B) \ 2+\sqrt {3} \$$

$$(C) \ {\sqrt {3} \over2}$$

$$(D) \ \sqrt {3} \$$

Ans .   D

Solution .   $$\ Given \ a = {\sqrt {3} \ \over2}$$
$$∴ \ \sqrt {1+a } \ + \sqrt {1-a } \ ^2 = (1+a)+(1-a)+2\sqrt { (1+a)(1-a)} \$$
$$[∴ (a+b)^2= a^2+b^2+2ab]$$
$$= 2+2\sqrt {1-a^2 } \$$
$$=\left( 1+\sqrt {1-\left({\sqrt {3 } \ \over2} \right)^2 } \ \right)$$
$$= 2\left(1+\sqrt {1-{3\over4} } \ \right)$$
$$= 2\left(1+\sqrt {{4-3\over4} } \ \right)= 2 \left(1+{1\over2} \right)$$
$$= 2×{3\over2}$$
$$∴ (\sqrt {1+a} \ + \sqrt {1+a} \ )^2 =3$$
$$∴ (\sqrt {1+a} \ + \sqrt {1-a} \ )= \sqrt {3} \$$

$$Q.4.\ \sqrt {2^n }=64, then\ n \ is \ equal \ to$$

$$(A) \ 2$$

$$(B) \ 4$$

$$(C) \ 6$$

$$(D) \ 12$$

Ans .  D

Solution .  $$\sqrt {2^n } \ = 64 → 2^{n/2}= 2^6$$
$$Since, base \ are \ same$$
$$∴ {n\over2} =6$$
$$→ n = 6×2 = 12$$

$$Q.5.\ {1\over(216)^{-2/3}}+{1\over(256)^{-2/3}}+{1\over(243)^{-1/5}}$$
$$(A) \ 103$$
$$(B) \ 105$$
$$(C) \ 107$$
$$(D) \ 109$$

Ans .  A

Solution .  $$\ {1\over(216)^{-2/3}}+{1\over(256)^{-2/3}}+{1\over(243)^{-1/5}}$$
$$= (216)^{2/3}+(256)^{3/4}+(243)^{1/5}$$
$$= (6^2)^{2/3}+(4^4)^{3/4}+(3^5)^{1/5}$$
$$= 36+64+3$$
$$103$$

$$Q.6.\ 2^{x-1}+2^{x+1}=2560 \ , then\ find\ the \ value \ of \ x$$

$$(A) \ 10$$

$$(B) \ 12$$

$$(C) \ 9$$

$$(D) \ 8$$

Ans .   A

Solution .   $$\ 2^{x-1}+2^{x+1}=2560 \$$
$$→ 2^{x-1}+2^2.2^{x-1}=2560$$
$$→ 2^{x-1}(1+4)= 2560$$
$$2^{x-1}={2560\over5}=512$$
$$→ 2^{x-1}=2^9$$
$$∴ \ x-1= 9$$
$$x=10$$

$$Q.7. \ If \ \left(\sqrt {2} \ ^\sqrt {2 } \ \right)^\sqrt {2 } \ = 2^x, then \ x \ is \ equal\ to$$

$$(A) \ 4$$

$$(B)\ 2$$

$$(C) 1$$

$$(D) \sqrt {2} \$$

Ans .   C

Solution .   $$\ Given, \left(\sqrt {2} \ ^\sqrt {2 } \ \right)^\sqrt {2 } \ = 2^x$$
$$→\sqrt {2}^{\sqrt {2} × \sqrt {2} \ \ } \ = 2^x$$
$$→ \left(\sqrt {2} \ \right)^2=2^x$$
$$(2)^1= 2^x$$
$$∴ x = 1$$

$$Q.8.\ 5\sqrt {5 }×5^3 ÷ 5^{-3/2}=5^{a+2}\ , then\ a\ is \ equal \ to$$

$$(A) \ 4$$

$$(B) \ 5$$

$$(C) \ 6$$

$$(D) \ 8$$

Ans .  A

Solution .  $$\ 5\sqrt {5 }×5^3 ÷ 5^{-3/2}=5^{a+2}\$$
$$→ {5.5^{1/2}×5^3\over5^{-3/2}}=5^{a+2}$$
$$→ 5^{1+{1\over2}+3}.5^{3/2}=5^{a+2}$$
$$∴ a+2 = 6$$
$$a=6-2=4$$

$$Q.9.\ \left( {32\over243}\right)^{-4/5}$$

$$(A) \ {4\over9}$$

$$(B) \ {9\over4}$$

$$(C) \ {16\over81}$$

$$(D) \ {81\over16}$$

Ans .  D

Solution .  $$\ \left( {32\over243}\right)^{-4/5} = \left( {2^5\over3^5}\right)^{-4/5}$$
$$=\left( {2\over3}\right)^{5×-{-4\over3}}$$
$$=\left({2\over3}\right)^{-4}=\left({3\over4}\right)^4 = {81\over16}$$

$$Q.10.\ 81^{2.5}×8^{4.5}÷3^{4.8}=9^?$$

$$(A) \ 7.1$$

$$(B) \ 9.4$$

$$(C) \ 4.7$$

$$(D) \ 4.5$$

Ans .   A

Solution .   $$81^{2.5}×9^{4.5}÷3^{4.8}=9^?$$
$$→ {(3^4)^{2.5}×(3^2)^{4.5} \over 3^{4.8}}=9^{?}$$
$$→ {3^{10}×3^9\over3^{4.8}}=(3^2)^?$$
$$→ 3^{19-4.8}=(3^2)^?$$
$$2×?=14.2$$
$$→ ?={14.2\over2}=7.1$$