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 be my multiple, i'll be your factor co-prime numbers lcm lowest common multiple (l.c.m.)

Solve the given exponent:
$\sqrt[4]{12} \times \sqrt[7]{6}$  

  1. $2^{\frac{9}{14}}\times 3^{\frac{11}{28}}$
  2. $3^{\frac{9}{14}}\times 2^{\frac{11}{28}}$
  3. $2^{\frac{1}{14}}\times 3^{\frac{1}{28}}$
  4. $3^{\frac{1}{14}}\times 2^{\frac{1}{28}}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\sqrt[4]{12} \ \times \ \sqrt[7]{6}$


$=(12)^{\frac{1}{4}} \ \times \ (6)^{\frac{1}{7}}$

$=(2\times2\times3)^{\frac{1}{4}} \ \times \ (2\times3)^{\frac{1}{7}}$

$=(2^2\times3)^{\frac{1}{4}} \ \times \ (2\times3)^{\frac{1}{7}}$

$=2^{2\times(\frac{1}{4})}\times 3^{\frac{1}{4}} \ \times2^\frac{1}{7}\times3^\frac{1}{7}$

$=2^{\frac{1}{2}}\times 3^{\frac{1}{4}} \ \times2^\frac{1}{7}\times3^\frac{1}{7}$

$=2^{(\frac{1}{2}+\frac{1}{7})}\times 3^{(\frac{1}{4}+\frac{1}{7})}$-----If base is same, then their powers can be added, by product law.

$=2^{\frac{9}{14}}\times 3^{\frac{11}{28}}$

Option A.

Multiple choice properties of proportion ratio and proportions ratio and proportion maths

If $\left( {{p^2} + {q^2}} \right)/\left( {{r^2} + {s^2}} \right) = \left( {pq} \right)/\left( {rs} \right)$, then what is the value of $\left( {p - q} \right)/\left( {p + q} \right)$ in terms of $r$ and $s$?

  1. $\left( {r + s} \right)/\left( {r - s} \right)$
  2. $\left( {r - s} \right)/\left( {r + s} \right)$
  3. $\left( {r + s} \right)/\left( {r s} \right)$
  4. $\left( {r s} \right)/\left( {r - s} \right)$
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice maths numbers and place value many forms of ten thousand standard form of numbers number names, numerals and place values

The value of ${\left( {{{27}^{\tfrac{{ - 2}}{3}}}} \right)^{\tfrac{1}{2}}} \times {\left( {{{64}^{\tfrac{1}{3}}}} \right)^2} \times {\left( {{{81}^{\tfrac{{ - 3}}{2}}}} \right)^{\tfrac{1}{6}}}$

  1. $\dfrac{1}{9}$
  2. $\dfrac{16}{9}$
  3. $\dfrac{2}{9}$
  4. $-\dfrac{16}{9}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
${\left( {{{27}^{\tfrac{{ - 2}}{3}}}} \right)^{\tfrac{1}{2}}} \times {\left( {{{64}^{\tfrac{1}{3}}}} \right)^2} \times {\left( {{{81}^{\tfrac{{ - 3}}{2}}}} \right)^{\tfrac{1}{6}}}$

$\displaystyle =\left(\dfrac{1}{(27)^{\tfrac{2}{3}}}\right)^{\tfrac{1}{2}}\times (4^{3\times \tfrac{1}{3}})^{2}\times \left(\dfrac{1}{(81)^{\frac{3}{2}}}\right)^{\tfrac{1}{6}}$

$\displaystyle =\left(\dfrac{1}{27}\right)^{\tfrac{2}{3}\times \tfrac{1}{2}}\times (4)^{2}\times \left(\dfrac{1}{81}\right)^{\tfrac{3}{2}\times \tfrac{1}{6}}$

$\displaystyle =\left(\dfrac{1}{3^{3}}\right)^{\tfrac{1}{3}}\times (4)^{2}\times \left(\dfrac{1}{3^{4}}\right)^{\tfrac{1}{4}}$

$\displaystyle =\frac{1}{3}\times 16\times \frac{1}{3}$

$=\dfrac{16}{9}$
Multiple choice maths square and square root perfect square or square number squares and triangles powers and roots

Find the value of each of the following, using the column method.
$(23)^2$
$(52)^2$

  1. 549, 2724

  2. 549, 2704

  3. 529, 2724

  4. 529, 2704
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$(23)^2$
$a=2, b=3$

$i$ $ii$ $iii$
$a^2$ $2ab$ $b^2$
$4$$1$           $\underline { 5 }$ $12$$+0$           $1\underline { 2 }$ $\underline { 9 }$

$\therefore (23)^2=529$

$(52)^2$
$a=5, b=2$

$i$ $ii$ $iii$
$a^2$ $2ab$ $b^2$
$25$$+2$          $\underline { 27 }$ $20$$+0$          $2\underline { 0 }$ $\underline { 4 }$

$\therefore (52)^2=2704$

Multiple choice maths square and square root perfect square or square number squares and triangles powers and roots

Find the number whose square root is twice of its cubic root.

  1. $128$
  2. $64$
  3. $16$
  4. $4$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let the number be $x.$
As per the problem $\sqrt {x}=2\times \sqrt [ 3 ]{x  } $
Raising both sides by $6$ times
$=(x^{1/2})^6 = 2^6(x^{1/3})^6$
$= x^{1/2\times 6} = 2^6 x^{1/3\times 6}$
or $ x^3 = 64 x^2$
or $x=64$

Multiple choice mathematics and statistics angle and their measurement angles and sides naming the sides in a right angled triangle understanding ratios

For a
positive integer n,
let
${f _n}\left( \theta  \right) = \left( {\tan \frac{\theta }{2}} \right)\left( {1 + \sec \theta } \right)\left( {1 + \sec 2\theta } \right)\left( {1 + \sec {2^2}\theta } \right)...\left( {1 + \sec {2^n}\theta } \right),then$

  1. ${f _2}\left( {\frac{\pi }{{16}}} \right) = 1$
  2. ${f _3}\left( {\frac{\pi }{{32}}} \right) = 1$
  3. ${f _4}\left( {\frac{\pi }{{64}}} \right) = 1$
  4. ${f _5}\left( {\frac{\pi }{{128}}} \right) = 1$
Reveal answer Fill a bubble to check yourself
A,B,C,D Correct answer
Explanation

$f _n(\theta)=(\tan \frac{\theta}{2})(1+\text{sec}\theta)(1+\text{sec} 2\theta)(1+\text{sec}2^2 \theta)\cdots(1+\text{sec}2^n \theta)$

           $=\dfrac{\sin \frac{\theta}{2}}{\cos \frac{\theta}{2}}\times \dfrac{1+\cos \theta}{\cos \theta}\times \dfrac{1+\cos 2\theta}{\cos 2\theta}\times \dfrac{1+\cos 2^2 \theta}{\cos 2^2 \theta}\cdots\times \dfrac{1+\cos 2^n \theta}{\cos 2^n \theta}$
           $=\dfrac{\sin \frac{\theta}{2}}{\cos \frac{\theta}{2}}\times \dfrac{2\cos ^2 \frac{\theta}{2}}{\cos \theta}\times \dfrac{2\cos ^2 \theta}{\cos 2 \theta}\times \dfrac{2\cos^2 2\theta}{\cos 2^2 \theta}\times \cdots\times \dfrac{2\cos 2^{n-1}\theta}{\cos^2 2^n\theta}$
           $=(2\sin \frac{\theta}{2}\cos \frac{\theta}{2})\times (2\cos \theta)\times (2\cos 2 \theta)\times \cdots\times \dfrac{2\cos 2^{n-1}\theta}{\cos 2^n \theta}$
           $=(2\sin \theta\cos \theta)\times (2\cos 2 \theta)\times \cdots\times \dfrac{2\cos 2^{n-1}\theta}{\cos 2^n \theta}$
           $=\dfrac{\sin 2^n \theta}{\cos 2^n \theta}=\tan 2^n \theta$

$f _2\bigg(\dfrac{\pi}{16}\bigg)=\tan (2^2\times \dfrac{\pi}{16})=\tan \frac{\pi}{4}=1$
$f _3\bigg(\dfrac{\pi }{32}\bigg)=\tan (2^3\times \dfrac{\pi}{32})=\tan \frac{\pi}{4}=1$
$f _4\bigg(\dfrac{\pi}{64}\bigg)=\tan (2^4\times \dfrac{\pi}{64})=\tan \frac{\pi}{4}=1$
$f _5\bigg(\dfrac{\pi}{128}\bigg)=\tan (2^5\times \dfrac{\pi}{128})=\tan \frac{\pi}{4}=1$

Multiple choice maths square and square root scientific notation use of exponents power of 10

Express $(3000)^2\times (20)^3$ in scientific notation:

  1. $7.2\times 10^{10}$
  2. $3.6\times 10^{8}$
  3. $6\times 10^9$
  4. $4.8\times 10^{12}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

${ 3000 }^{ 2 }\times { 20 }^{ 3 }=(9\times { 10 }^{ 6 })\times (8\times { 10 }^{ 3 })\ =72\times { 10 }^{ 9 }=7.2\times { 10 }^{ 10 }$

So correct answer will be option A