Mathematics · Quantitative Aptitude

Surds and Indices

408 Questions

Surds and indices questions focus on evaluating square roots, cube roots, and fractional exponents. These topics are a core part of the quantitative aptitude section in many competitive exams. Practicing these problems builds speed and accuracy for solving numerical equations.

Square roots evaluationCube roots calculationExponents and powersFractional exponentsSurds multiplication

Surds and Indices Questions

Multiple choice maths calculations and mental strategies 1 equations from statements forming equations from statements writing mathematical statements

If $\left(\dfrac { 3 } { 4 }\right)^{th}$ of $x$ of $\left(\dfrac { 1 } { 4 }\right)^{th}$ of $35600 = 1668.75 ,$ find $x$

  1. $\dfrac { 2 } { 3 }$
  2. $\dfrac { 3 } { 4 }$
  3. $\dfrac { 2 } { 5 }$
  4. $\dfrac { 1 } { 4 }$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$\dfrac { 3 }{ 4 } \times x\times \dfrac { 1 }{ 4 } \times 35600=1668.75$

$\Rightarrow x=\dfrac { 1668.75\times 16 }{ 3\times 35600 } =\dfrac { 1 }{ 4 } $      [D]

Multiple choice comparison of irrational numbers surds and law of surds rational and irrational numbers exponents maths

Compare the following pairs of surds. $\sqrt[8]{80}, \sqrt[4]{40}$    

  1. $\sqrt[8]{80} < \sqrt[4]{40}$
  2. $\sqrt[8]{80} \neq \sqrt[4]{40}$
  3. $\sqrt[8]{80} = \sqrt[4]{40}$
  4. $\sqrt[8]{80} > \sqrt[4]{40}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

   $\sqrt[8]{80}, \sqrt[4]{40}$
$={80}^{\frac{1}{8}}, {40}^{\frac{1}{4}}$
$={80}^{\frac{1}{8}}, {40}^{\frac{2}{8}}$
$={80}^{\frac{1}{8}}, {1600}^{\frac{1}{8}}$
Now,
   $80<1600$
$=>{80}^{\frac{1}{8}}<{1600}^{\frac{1}{8}}$
$=>\sqrt[8]{80}< \sqrt[4]{40}$

Multiple choice comparison of irrational numbers surds and law of surds rational and irrational numbers exponents maths

Compare the following pairs of surds $\sqrt[8]{12}, \sqrt[4]{6}$

  1. $\sqrt[8]{2} < \sqrt[4]{6}$
  2. $\sqrt[8]{8} < \sqrt[4]{6}$
  3. $\sqrt[8]{12} < \sqrt[4]{6}$
  4. $\sqrt[8]{12} < \sqrt[4]{4}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

To compare, convert to a common root: sqrt[8]{12} is 12^(1/8) and sqrt[4]{6} is 6^(1/4) = 6^(2/8) = 36^(1/8). Since 12 < 36, sqrt[8]{12} < sqrt[4]{6}.

Multiple choice comparison of irrational numbers surds and law of surds rational and irrational numbers exponents maths

Compare the following pair of surds:

$\sqrt[3]{6}, \sqrt[4]{8}$

  1. $\sqrt[3]{6} > \sqrt[4]{8}$
  2. $\sqrt[3]{6} > \sqrt[4]{4}$
  3. $\sqrt[3]{6} > \sqrt[3]{8}$
  4. $\sqrt[3]{4} > \sqrt[4]{8}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Convert to a common root (12th root): sqrt[3]{6} = 6^(4/12) = 1296^(1/12). sqrt[4]{8} = 8^(3/12) = 512^(1/12). Since 1296 > 512, the first is larger.

Multiple choice comparison of irrational numbers surds and law of surds rational and irrational numbers exponents maths

Compare the following pairs of surds. $\sqrt[4]{64}, \sqrt[6]{128}$    

  1. $\sqrt[4]{64} > \sqrt[6]{128}$
  2. $\sqrt[4]{64} < \sqrt[6]{128}$
  3. $\sqrt[4]{64} \neq \sqrt[6]{128}$
  4. $\sqrt[4]{64} = \sqrt[6]{128}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\sqrt[4]{64}=\sqrt[4]{2^6}=2\sqrt[4]{2^2}=2\sqrt{2}=2\sqrt[6]{2^3}=2\sqrt[6]{8}$
$\sqrt[6]{128}=\sqrt[6]{2^7}=2\sqrt[6]{2}$
$\sqrt[4]{64}>\sqrt[6]{128}$

Multiple choice comparison of irrational numbers surds and law of surds rational and irrational numbers exponents maths

$\sqrt{11}-\sqrt{10} .... \sqrt{12}-\sqrt{11}$,use appropriate inequality to fill the gap.

  1. <

  2. >
  3. $=$

  4. cannot determined
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

We first consider $\sqrt { 11 } -\sqrt { 10 }$ as follows:


$\sqrt { 11 } -\sqrt { 10 } =3.317-3.162=0.156$

Now we find the value of $\sqrt { 12 } -\sqrt { 11 }$ as follows:

$\sqrt { 12 } -\sqrt { 11 } =3.464-3.317=0.147$


Since $0.156>0.147$

Hence, $\sqrt { 11 } -\sqrt { 10 }>\sqrt {12} -\sqrt {11}$

Multiple choice comparison of irrational numbers surds and law of surds rational and irrational numbers exponents maths

Arrange the following in ascending order of magnitude: $\displaystyle \sqrt[4]{90}, \sqrt[3]{10}, \sqrt{6}$

  1. $\displaystyle \sqrt{3} < \sqrt[4]{10} < \sqrt[3]{6}$
  2. $\displaystyle \sqrt{3} > \sqrt[4]{10} > \sqrt[3]{6}$
  3. $\displaystyle \sqrt{3} > \sqrt[4]{10} < \sqrt[3]{6}$
  4. $\displaystyle \sqrt{3} < \sqrt[4]{10} > \sqrt[3]{6}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice comparison of irrational numbers surds and law of surds rational and irrational numbers exponents maths

Arrange the following surds in ascending order of their magnitudes: $\sqrt{5},\sqrt [ 3 ]{ 11 } ,2\sqrt [ 6 ]{ 3 } $

  1. $\sqrt [ 3 ]{ 11 } > \sqrt{5}< 2\sqrt [ 6 ]{ 3 } $
  2. $\sqrt [ 3 ]{ 11 } < \sqrt{5}< 2\sqrt [ 6 ]{ 3 } $
  3. $\sqrt [ 3 ]{ 11 } > \sqrt{5}> 2\sqrt [ 6 ]{ 3 } $
  4. $\sqrt [ 3 ]{ 11 } < \sqrt{5}> 2\sqrt [ 6 ]{ 3 } $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\sqrt{5} = 5^{1/2}$

$\sqrt[3]{11} = 11^{1/3}$

$2\sqrt[6]{3} = \sqrt[6]{12}= 12^{1/6}$

L.C.M of the denominators of the exponents is 12.

So,

$\sqrt{5} = 5^{\frac{1}{2}\times\frac{6}{6}} = \sqrt [12]{5^6}=\sqrt[12]{15625}$

$\sqrt[3]{11} = 11^{\frac{1}{3}\times\frac{4}{4}} = \sqrt [12]{11^4} =\sqrt[12]{14641}$

$2\sqrt[6]{3} = \sqrt[6]{12}= 12^{\frac{1}{6}\times\frac{2}{2}} = \sqrt[12]{12^2} =\sqrt[12]{144}$

Hence, the Ascending order is $2\sqrt[6]{3}, \sqrt[3]{11},\sqrt{5}$

Multiple choice comparison of irrational numbers surds and law of surds rational and irrational numbers exponents maths

Which is greater?
${ \left( \cfrac { 1 }{ 2 }  \right)  }^{ 1/2 } $ or ${ \left( \cfrac { 2 }{ 3 }  \right)  }^{ 1/3 } $

  1. ${ \left( \cfrac { 2 }{ 3 } \right) }^{ 1/3 } $
  2. ${ \left( \cfrac { 1 }{ 2 } \right) }^{ 1/2 } $

  3. Both are equal

  4. None of the above
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$ \left(\dfrac{1}{2}\right)^{1/2} \; or \; \left(\dfrac{2}{3}\right)^{1/3}$


$= \left(\left(\dfrac{1}{2}\right)^{1/2}\right)^6 \; or \; \left(\left(\dfrac{2}{3}\right)^{1/3}\right)^6$


$= \left(\dfrac{1}{2}\right)^3 \; or \; \left(\dfrac{2}{3}\right)^2$


$= \left(\dfrac{1}{8}\right) \; or \; \left(\dfrac{4}{9}\right)$


= 0.125 or 0.44


Since, 0.44 is greater and so is $ \left(\dfrac{2}{3}\right)^{1/3}$