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 concept of directed numbers and number line addition of directed numbers addition of integers addition of integers on number line

The simplified form of the expression $\left[15-\left{12-\left(7-\overline{6-2}\right)\right}\right]$ will be 

  1. $6$
  2. $3+3$
  3. $8$
  4. $2$ x $4$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$ \left[15-\left{12-\left(7-\overline{6-2}\right)\right}\right]$


$=\left[15-\left{12-\left(7-4\right)\right}\right]$

$=\left[15-\left{12-\left(3\right)\right}\right]$

$= 15-9$

$=6$

$\text{Option A is correct.}$

Multiple choice maths congruency of triangles triangle inequality inequalities in triangle inequalities in triangles

If $\left| {z - 1} \right| + \left| {z + 3} \right| \le 8$ then the range of values of $\left| {z - 4} \right|$

  1. $[1,\,7]$
  2. $[1,\,8]$
  3. $[1,\,9]$
  4. $[2,\,5]$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$|z-1|+|z+3|\le 8$

Using the triangle inequality 
$|z _1\pm z _2|\le |z _1|+|z _2|$
We have $|z-1|+|z+3|\le 8$
$\implies |z-1+z+3|\le 8$
$\implies |z+1|\le 4$
Using triangle inequality again 
$|z|+1\le 4\implies |z|\le 3$
So, the maximum value of $|z _1+ _2|$ is $|z _1|+|z _2|$
And  the minimum value of $|z _1+ _2|$ is $|z _1|-|z _2|$
Hence the maximum value of $|z-4|$ is $|z|+|4|=3+4=7$
 the minimum value of $|z-4|$ is $|z|-|4|=3-4=-1$
Hence the range of $|z-4|$
$1\le|z-4|\le 7$

Multiple choice maths calculating and mental strategies 3 written methods dividing decimals division of decimals

If $\sqrt{.04\times .4\times a} = .004\times .4\times \sqrt{b}$, then $\dfrac{a}{b}$ is 

  1. $16\times 10^{-3}$
  2. $16\times 10^{-4}$
  3. $16\times 10^{-5}$
  4. $16\times 10^{-6}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$\sqrt{0.4\times 0.04\times a}=0.004\times 0.4\times \sqrt{b}$ 

$\Rightarrow \sqrt{\dfrac{a}{b}}=\dfrac{0.004\times 0.4}{\sqrt{0.4\times 0.04}}=\dfrac{16\times 10^{-4}}{4\times 10^{-1-5}}$ 
$\Rightarrow \dfrac{a}{b}=(4\times 10^{25})^{2}=\boxed{16\times 10^{-5}}$

Multiple choice maths squares and square roots finding the square of a number finding square of a number patterns in square numbers

$\sqrt{3\,+\,2\,\sqrt{2}}\,-\,\sqrt{3\,-\,2\,\sqrt{2}}$ is equal to 

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

To find, value of : $\sqrt{3\,+\,2\,\sqrt{2}}\,-\,\sqrt{3\,-\,2\,\sqrt{2}}$ 
Let $x = \sqrt{3\,+\,2\,\sqrt{2}}\,-\,\sqrt{3\,-\,2\,\sqrt{2}}$ 
Squaring both sides:
$x^2$ = $\displaystyle\,\left ( \sqrt{3\,+\,2\sqrt{2}}\,-\,\sqrt{3\,-\,2\sqrt{2}} \right )^{2}$
$\displaystyle\,x^2\,=\,3\,+\,2\sqrt{2}\,+\,3\,-\,2\sqrt{2}\,-\,2(3\,+\,2\sqrt{2})(3\,-\,2\sqrt{2})$
$\Rightarrow x^2\,=\,4$
$\Rightarrow x\,=\,2$
Hence, option 'A' is correct.