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 square root square root of perfect square finding square root of a number square root of a perfect square

Find the square root of $\displaystyle 9\frac { 49 }{ 64 } $.


  1. $\displaystyle 3\frac { 1 }{ 8 } $.
  2. $\displaystyle 4\frac { 3 }{ 8 } $.
  3. $\displaystyle 3\frac { 7 }{ 8 } $.
  4. $\displaystyle 4\frac { 1}{ 8 } $.
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\sqrt { 9\displaystyle\frac { 49 }{ 64 }  } =\sqrt {\displaystyle \frac { 625 }{ 64 }  } =\displaystyle\frac { \sqrt { 625 }  }{ \sqrt { 64 }  } =\displaystyle\frac { 25 }{ 8 } =3\displaystyle\frac { 1 }{ 8 } $.

Multiple choice maths square root square root of perfect square finding square root of a number square root of a perfect square

Find the square root of: $\displaystyle27\frac{9}{16}$

  1. $\displaystyle2\frac{1}{4}$
  2. $\displaystyle7\frac{1}{4}$
  3. $\displaystyle3\frac{1}{4}$
  4. $\displaystyle5\frac{1}{4}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Given number is $ 27\dfrac {9}{16} = \dfrac {441}{16} $

Square root of $ \dfrac {441}{16} =  \dfrac { \sqrt {441}}{\sqrt {16}} = \dfrac {21}{4} = 5 \dfrac {1}{4} $

Multiple choice maths square root square root of perfect square finding square root of a number square root of a perfect square

The square root of $71\, \times\, 72\, \times\, 73\, \times\, 74\, +\, 1$ is :

  1. 9,375

  2. 9,625

  3. 5,625

  4. 5,255
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$\text{Here. consider the fact that the product of 4 consecutive numbers + 1 is perfect square.}$

$So, let\, x = 71$

⇒ $(71)(72)(73)*(74) + 1 = x(x + 1) (x + 2) (x + 3) +1$ 

⇒$ (x² + 3x)(x² + 3x + 2) + 1$

⇒$ (x² + 3x)² + 2(x² + 3x) + 1$

⇒ $(x² + 3x + 1)^2$

⇒ $Square \,root \,of \,(x² + 3x + 1)² = x² + 3x + 1$

$Here, x = 71$

$Therefore, square root is = (71)² + (3*71) + 1$

⇒ $5041 + 213 + 1$ 

= $5255$

Multiple choice maths square root square root of perfect square finding square root of a number square root of a perfect square

The square root of $\displaystyle \frac{441}{961}$ is :

  1. $\displaystyle \frac{21}{39}$
  2. $\displaystyle \frac{37}{21}$
  3. $\displaystyle \frac{21}{31}$
  4. $\displaystyle \frac{11}{13}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

We have to find square root of $\displaystyle \frac {441}{961}$

$\therefore \displaystyle \frac{\sqrt{144}}{\sqrt{961}}=\frac{21}{31}$

Multiple choice maths square root square root of perfect square finding square root of a number square root of a perfect square

The value of $\sqrt{214+\sqrt{130-\sqrt{88-\sqrt{44+\sqrt{25}}}}}$

  1. $14$
  2. $15$
  3. $16$
  4. $17$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

 $\sqrt{214+\sqrt{130-\sqrt{88-\sqrt{44+\sqrt{25}}}}}$
$\Rightarrow \sqrt{214+\sqrt{130-\sqrt{88-\sqrt{44+5}}}}$
$\Rightarrow \sqrt{214+\sqrt{130-\sqrt{88-\sqrt{49}}}}$
$\Rightarrow \sqrt{214+\sqrt{130-\sqrt{88-7}}}$
$\Rightarrow \sqrt{214+\sqrt{130-\sqrt{81}}}$
$\Rightarrow  \sqrt{214+\sqrt{130-9}}$
$\Rightarrow \sqrt{214+\sqrt{121}}$
$\Rightarrow \sqrt{214+11}$
$\Rightarrow \sqrt{225}=15$

Multiple choice maths square root square root of perfect square finding square root of a number square root of a perfect square

Find the square root of the following $\displaystyle\frac{2025}{4900}$

  1. $\displaystyle\frac{55}{80}$
  2. $\displaystyle\frac{55}{70}$
  3. $\displaystyle\frac{45}{80}$
  4. $\displaystyle\frac{45}{70}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let us find the square root of $2025\;and\;4900$ by factorising them.
$3\mid \; \; 2025\ { \overline { 3\mid \; \; 675 }  }\ { \overline { 3\mid \; \; 225 }  }\ { \overline { 3\mid \; \; \; \; 75 }  }\ { \overline { 5\mid \; \; \; \; 25 }  }\ { \overline { 5\mid \; \; \; \; \; 5 }  }\ { \overline { \; \; \mid \; \; \; \; 1 }  }$
$2025=\underline{3\times3}\times\underline{3\times3}\times\underline{5\times5}$
$\sqrt{2025}=3\times3\times5=45$
$2\mid \; \; 4900\ { \overline { 2\mid \; \; 2450 }  }\ { \overline { 5\mid \; \; 1225 }  }\ { \overline { 5\mid \; \; \; \; 245 }  }\ { \overline { 7\mid \; \; \; \; \;49 }  }\ { \overline { 7\mid \; \; \; \; \; \;7 }  }\ { \overline { \; \; \;\mid \; \; \; \; \;1 }  }$
$4900=\underline{2\times2}\times\underline{5\times5}\times\underline{7\times7}$
$\sqrt{4900}=2\times5\times7=70$
So, $\cfrac{\sqrt{2025}}{\sqrt{4900}}=\cfrac{45}{70}$.

Multiple choice maths square root square root of perfect square finding square root of a number square root of a perfect square

If the sum $S$ of three consecutive even numbers is a perfect square between $200\;and\;400$, then the square root of $S$ is

  1. $15$
  2. $16$
  3. $18$
  4. $19$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$15^2=225,\;16^2=256$
$17^2=281\;18^2=324$
$19^2=361$
$\;2x-2+2x+2x+2=6(x)$
$\Rightarrow6x=324$ is possible
$\Rightarrow\sqrt{324}=18$

Therefore,square root of $S$ is $18$

Multiple choice maths square root square root of perfect square finding square root of a number square root of a perfect square

Square root of $400$ is?

  1. $40$
  2. $25$
  3. $20$
  4. $100$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$400=2\times 200$

       $=2\times 2\times 100$
       $=2\times 2\times 2\times 50$
       $=2\times 2\times 2\times 2\times 25$
       $=2\times 2\times 2\times 2\times 5\times 5$
       $=2^4\times 5^2$
$\Rightarrow$  $400=2^4\times 5^2$
$\Rightarrow$  $\sqrt{400}=\sqrt{2^4\times 5^2}$
$\Rightarrow$  $\sqrt{400}=2^2\times 5$
$\Rightarrow$  $\sqrt{400}=4\times 5$
$\Rightarrow$  $\sqrt{400}=20$

Multiple choice maths square root square root of perfect square finding square root of a number square root of a perfect square

The value  of  $\sqrt {11 - \sqrt{112} }=  $

  1. $2 + \sqrt{7}$
  2. $2 - \sqrt{7}$
  3. $ \sqrt{7} - 2$

  4. none of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
$11-\sqrt {112}$

$=11-\sqrt {4\times 28}$

$=11-\sqrt {4}\times \sqrt {28}$

$=11-2\times \sqrt {28}$

$=11-2\times \sqrt {7}\times \sqrt {4}$

$=7+4-2\times \sqrt {7}\times \sqrt {4}$

$=(\sqrt {7})^2 +(\sqrt {4})^2 -2\times \sqrt {7}\times \sqrt {4}$

$=(\sqrt {7}-\sqrt {4})^2$

Thus, $11-\surd {112}=(\surd {7} -\surd {4})^2$

Hence,

$\sqrt {11-\sqrt {112}}=\sqrt {(\sqrt {7}-\sqrt {4})^2}=\sqrt {7}-\sqrt {4}=\sqrt {7}-2$
Multiple choice maths square root square root of perfect square finding square root of a number square root of a perfect square

If ${\left( {\dfrac{m}{n}} \right)^{\dfrac{3}{8}}} + {\left( {\dfrac{n}{m}} \right)^{\dfrac{3}{8}}} = 9$ then find the value of ${\left( {\dfrac{m}{n}} \right)^{\dfrac{3}{4}}} + {\left( {\dfrac{n}{m}} \right)^{\dfrac{3}{4}}}$

  1. $79$
  2. $72$
  3. $83$
  4. $84$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
$(\dfrac{m}{n})^\dfrac{3}{8}+(\dfrac{n}{m})^\dfrac{3}{8}=9$
$[(\dfrac{m}{n})^\dfrac{3}{8}+(\dfrac{n}{m})^\dfrac{3}{8}]^{2} =9^{2}$
$ ((\dfrac{m}{n})^\dfrac{3}{8})^{2}+((\dfrac{n}{m})^\dfrac{3}{8})^{2}+2((\dfrac{m}{n})^\dfrac{3}{8})((\dfrac{n}{m})^\dfrac{3}{8})=81$
$ (\dfrac{m}{n})^\dfrac{3}{4}+(\dfrac{n}{m})^\dfrac{3}{4}=81-2=79$
Multiple choice maths square root square root of perfect square finding square root of a number square root of a perfect square

The square root of sum of the digits in the square of $121$ is

  1. $4$
  2. $3$
  3. $6$
  4. $9$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
${ \left( 121 \right)  }^{ 2 }$

$={ \left( 100+21 \right)  }^{ 2 }$     

$={ 100 }^{ 2 }+{ 21 }^{ 2 }+2\left( 100 \right) \left( 21 \right) $      $[\because (a+b)^2= a^2+2ab+b^2]$

$=14641$

Sum of digits $=1+4+6+4+1=16$

Square root$=\sqrt { 16 } =4$
Multiple choice maths square root square root of perfect square finding square root of a number square root of a perfect square

Find the square root of $225$ using "Repeated Subtraction".

  1. $11$
  2. $15$
  3. $5$
  4. $8$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\\225-1=224\\224-3=221\\221-5=216\\216-7=209\\209-9=200\\200-11=189\\189-13=176\\176-15=161\\161-17=144\\144-19=125\\125-21=104\\104-23=81\\81-25=56\\56-27=29\\29-29=0\\\>Total\>steps\>of\>=15\>\\hence\>\sqrt{225}=15$