Mathematics · Quantitative Aptitude

Surds and Indices

362 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 squares and square roots approximation of square roots estimating square roots square root of non perfect squares

Estiamate the square root of $850$ 

  1. $29.15$
  2. $30.21$
  3. $98.23$
  4. $23.11$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The square root of $850$ is $\sqrt {850}=\sqrt {25 \times 34}=5\sqrt {34}$

Square root of $34$ lie between $5$ and $6$.
 square of $5.5$ is $30.25$ 
Now, we can say that square root of $34$ lie between $5.5$ and $6$.
Now, square of $5.75$ is $33.06$
So, square root of $34$ lie between $5.75$ and $6$.
Now, we have to choose the number $5.85$
$(5.85)^2=34.225$ which is greater than $34$ and close to $34$
So, assume a number $5.84$.
$(5.84)^2=34.1056$
$(5.83)^2=33.9889$
Hence, we can say that square root of 34 lie between $5.83$ and $5.84$.
So, $5\times 5.83=29.15$.

Multiple choice maths rational numbers proof of irrationality of numbers proofs of irrationality introduction to irrational numbers

The equation $\sqrt{x+4}$- $\sqrt{x-3}$+ 1=0 has:

  1. no root

  2. one real root

  3. one real root and one imaginary root

  4. two imaginary roots

  5. two real roots
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\Longrightarrow \sqrt { x+4 } -\sqrt { x-3 } +1=0\ \Longrightarrow \sqrt { x+4 } +1=\sqrt { x-3 } \ \Longrightarrow x+4+1+2\sqrt { x+4 } =x-3\ \Longrightarrow 2\sqrt { x+4 } =-8\ \Longrightarrow x+4=16\ \therefore x=12$

But x = 12 will not satisfy given equation.
$\therefore$ No roots for given equation.

Multiple choice finding nth roots of a complex number n th root of unity demoivre's theorem complex numbers maths

$\begin{array} { l } { 1 , a _ { 1 } , \ldots , a _ { 4 } \text { are the } 5 ^ { \text { th } } \text { roots of unity. The value } } \ { \text { of } \left( 1 + a _ { 1 } \right) \dots \left( 1 + a _ { 4 } \right) \text { is } } \end{array}$ ?

  1. $-16$
  2. $16$
  3. $-1$
  4. $1$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The 5th roots of unity are the roots of the equation x^5 - 1 = 0, which can be factored as (x - 1)(x^4 + x^3 + x^2 + x + 1) = 0. For the non-unity roots a_1, a_2, a_3, a_4, the polynomial can also be expressed as (x - 1)(x - a_1)(x - a_2)(x - a_3)(x - a_4) = x^5 - 1. Substituting x = -1 into both sides gives (-1 - a_1)(-1 - a_2)(-1 - a_3)(-1 - a_4) = (-1)^5 - 1 = -2, which factors out to give the product (1 + a_1)...(1 + a_4) = 1.

Multiple choice finding nth roots of a complex number n th root of unity demoivre's theorem complex numbers maths

The value of ${ \left( 16 \right)  }^{ 1/4 }$ are

  1. $\pm 2,\pm 2i$
  2. $\pm 4,\pm 4i$
  3. $\pm 1,\pm i$

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

Let $x={ \left( 16 \right)  }^{ { 1 }/{ 4 } }$

${ x }^{ 4 }=16$

${ x }^{ 4 }-16=0$

${ x }^{ 4 }-{ 2 }^{ 4 }=0$

$\left( { x }^{ 2 }-{ 2 }^{ 2 } \right) \left( { x }^{ 2 }+{ 2 }^{ 2 } \right) =0$

$\left( x-2 \right) \left( x+2 \right) \left( { x }^{ 2 }+4 \right) =0$

$x-2=0,x+2=0,{ x }^{ 2 }+4=0$

$x=2,-2$ or ${ x }^{ 2 }=-4\Longrightarrow x=\pm \sqrt { -4 } =\pm 2i$

$\therefore x=\pm 2,\pm 2i$

Multiple choice maths applications of matrices and determinants elementary transformations of a matrix multiplicative inverse of a matrix inverse of a matrix

If $
A=\left[ \begin{array}{ll}{x} & {1} \ {1} & {0}\end{array}\right]
 $ and $
A^{2}=I
 $, $
A^{-1}
 $ is equal to ...............

  1. $

    \left[ \begin{array}{ll}{0} & {1} \\ {1} & {0}\end{array}\right]

    $
  2. $

    \left[ \begin{array}{ll}{1} & {0} \\ {0} & {1}\end{array}\right]

    $
  3. $

    \left[ \begin{array}{ll}{1} & {1} \\ {1} & {1}\end{array}\right]

    $
  4. $

    \left[ \begin{array}{ll}{0} & {0} \\ {0} & {0}\end{array}\right]

    $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
$A=\left[\begin{matrix} x & 1 \\ 1 & 0  \end{matrix}\right]$
Given: ${A}^{2}=I$ where $I$ is $2\times 2$ identity matrix
Let us find ${A}^{2}$
$=\left[\begin{matrix} x & 1 \\ 1 & 0  \end{matrix}\right]\left[\begin{matrix} x & 1 \\ 1 & 0  \end{matrix}\right]$
$=\left[\begin{matrix} {x}^{2}+x & x+0 \\ x+0 & 1+0  \end{matrix}\right]$
Given ${A}^{2}=I$
$\Rightarrow \left[\begin{matrix} {x}^{2}+x & x+0 \\ x+0 & 1+0  \end{matrix}\right]=\left[\begin{matrix} 1 & 0 \\ 0 & 1  \end{matrix}\right]$
Equating,we get
${x}^{2}+x=1,x=0$
Put $x=0$ in $A$
$A=\left[\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix}\right]$
We have ${A}^{2}=I$
Pre-multiply ${A}^{-1}$ both sides,we get
${A}^{-1}{A}^{2}={A}^{-1}I$
$\Rightarrow A={A}^{-1}$
Hence,${A}^{-1}=\left[\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix}\right]$
Multiple choice maths fun with numbers some special sequences triangular numbers properties and patterns of perfect squares

$\cfrac { { \left( 963+476 \right)  }^{ 2 }+{ \left( 963-476 \right)  }^{ 2 } }{ \left( 973\times 963+476\times 476 \right)  } =$?

  1. $1449$
  2. $497$
  3. $2$
  4. $4$

  5. None of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given Exp.$=\cfrac { { \left( a+b \right)  }^{ 2 }+{ \left( a-b \right)  }^{ 2 } }{ \left( { a }^{ 2 }+{ b }^{ 2 } \right)  } =\cfrac { 2\left( { a }^{ 2 }+{ b }^{ 2 } \right)  }{ \left( { a }^{ 2 }+{ b }^{ 2 } \right)  } =2$