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 complex numbers and linear inequations identities of complex numbers powers of imaginary unit i algebra of complex numbers

$\left(\sqrt[3]{3}+\left(3^\cfrac{5}{6}\right)i\right)^3$ is an integer where $i=\sqrt{-1}$. The value of the integer is equal to.

  1. $24$
  2. $-24$
  3. $-22$
  4. $-21$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
$\rightarrow { \left( \sqrt [ 3 ]{ 3 } +{ 3 }^\cfrac{ 5}{6 }i \right)  }^{ 3 }=3{ \left( 1+\sqrt { 3 } i \right)  }^{ 3 }=3{ \left( 1+\sqrt { 3 } i \right)  }^{ 2 }\left( 1+\sqrt { 3 } i \right) $
$\rightarrow 3{ \left( 1+\sqrt { 3 } i \right)  }^{ 3 }=3\left( 1+3\sqrt { 3 } { i }^{ 3 }+3\sqrt { 3 } i\left( 1+\sqrt { 3 } i \right)  \right) $
$\rightarrow 3\left( 1-3\sqrt { 3 } i+3\sqrt { 3 } i-9 \right) $
$\rightarrow 3\left( -8 \right) =-24$
Multiple choice maths complex numbers and linear inequations identities of complex numbers powers of imaginary unit i algebra of complex numbers

 The value of $\sqrt{i}$ is 

  1. $1-i$
  2. $1+i$
  3. $ \pm \left( {1 + i} \right)$
  4. $i-1$
  5. $\frac{{ \pm 1}}{{\sqrt 2 }}\left( {1 + i} \right)$
Reveal answer Fill a bubble to check yourself
E Correct answer
Explanation
We can write a complex number in the form $2=(a, b)=a+ib$
$z=\sqrt{i}$
Thus
$i=z^2=a^2-b^2+2abi=(a^2-b^2, 2ab)$
$a^2-b^2=0$
$2ab=1$
$2a^2=1$
$a^2=\dfrac{1}{2}$
$a=\pm \dfrac{1}{\sqrt{2}}$
$\sqrt{i}=\dfrac{1}{\sqrt{2}}+\dfrac{i}{\sqrt{2}}=\pm \dfrac{1}{\sqrt{2}}(1+i)$.
Multiple choice maths complex numbers and linear inequations identities of complex numbers powers of imaginary unit i algebra of complex numbers

If ${ \left( \sqrt { 3 } -i \right)  }^{ n }={ 2 }^{ n }, n\in Z$, then $n$ is multiple

  1. $6$
  2. $10$
  3. $9$
  4. $12$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
$(\sqrt {3}-i)^{n}=2^{n}$
$\Rightarrow \quad\left(\dfrac {\sqrt {3}-i}{2}\right) ^{n}=1$
$\Rightarrow \quad i \left(\dfrac {-1}{2}-\dfrac {i\sqrt {3}}{2}\right) ^{n}=1$
$\therefore \quad i n^{2n}=1$
$\therefore \quad i=1$ and $n^{2n}=1$
$’n ’$ is multiple of $’3 ’$ and $’4 ’$ 
$\Rightarrow \quad’n ’$ is multiple of $’12 ’$
Multiple choice maths vedic mathematics square roots using vedic maths square and square roots using vedic mathematics history of mathematics

Predict square root of $3136$ using Vilokanam method.

  1. $54$
  2. $56$
  3. $64$
  4. $66$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

We have to find the square root of $3136$ using Vilokanam method.

Its unit digit is $6$.
Therefore, the unit digit of the square root will be $4$ or $6$. 
Ignoring the last two digits (unit digit and ten’s digit) we get $31$. 
The greatest number whose square is less than or equal to $31$ is $5$.
Adjusting above obtained two unit digits $4$ or $6$ to the right of $5$, we get two numbers $54$ and $56$. 
The unique number with unit digit $5$ which lies between $54$ and $56$ is $55$. 
And  $(55)^2 = 3025$
Since, $3136>3025$, therefore, the required square root is $56$.
Thus $\sqrt{3136}=56$

Multiple choice maths vedic mathematics square roots using vedic maths square and square roots using vedic mathematics history of mathematics

Find square root of $9604$ using Vilokanam method.

  1. $98$
  2. $92$
  3. $88$
  4. $82$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

We have to find the square root of $9604$ using Vilokanam method.

Its unit digit is $4$.
Therefore, the unit digit of the square root will be $2$ or $8$. 
Ignoring the last two digits (unit digit and ten’s digit) we get $96$. 
The greatest number whose square is less than or equal to $96$ is $9$.
Adjusting above obtained two unit digits $2$ or $8$ to the right of $9$, we get two numbers $92$ and $98$. 
The unique number with unit digit $5$ which lies between $92$ and $98$ is $95$. 
And  $(95)^2 = 9025$
Since, $9604>9025$, therefore, the required square root is $98$.
Thus $\sqrt{9604}=98$

Multiple choice maths vedic mathematics square roots using vedic maths square and square roots using vedic mathematics history of mathematics

Find square root of $961$ using Vilokanam method.

  1. $29$
  2. $30$
  3. $31$
  4. $32$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

We have to find the square root of $961$ using Vilokanam method.

Its unit digit is $1$.
Therefore, the unit digit of the square root will be $1$ or $9$. 
Ignoring the last two digits (unit digit and ten’s digit) we get $9$. 
The greatest number whose square is less than or equal to $9$ is $3$.
Adjusting above obtained two unit digits $1$ or $9$ to the right of $2$, we get two numbers $31$ and $39$. 
The unique number with unit digit $5$ which lies between $31$ and $39$ is $35$. 
And  $(35)^2 = 1225$
Since, $961<1225$, therefore, the required square root is $31$.
Thus $\sqrt{961}=31$