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

Find the missing term in the following problem.
$\left (\dfrac {3x}{4} - \dfrac {4y}{3}\right )^{2} = \dfrac {9x^{2}}{16} + \dfrac {16y^{2}}{9} + ?$.

  1. $2xy$
  2. $-2xy$
  3. $12xy$
  4. $-12xy$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
We know that $(a-b)^2=a^2+b^2-2ab$
$\left (\dfrac{3x}{4}-\dfrac{4y}{3}\right)^2=\left (\dfrac{3x}{4}\right)^2+\left (\dfrac{4y}{3}\right)^2-2\left(\dfrac{3x}{4}\right)\left (\dfrac{4y}{3}\right)$
$=\dfrac{9x^2}{16}+\dfrac{16y^2}{9}-2xy$
Therefore, the missing term is $-2xy$.

Option (B) is correct
Multiple choice maths squares and square roots finding the square of a number finding square of a number patterns in square numbers

$\sqrt { 3+2\sqrt { 2 }  } +\sqrt { 3-2\sqrt { 2 }  } =...$ ?

  1. $2+2\sqrt {2}$
  2. $2\sqrt {2}$
  3. $1$
  4. $0$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\sqrt { 3+2\sqrt { 2 }  } +\sqrt { 3-2\sqrt { 2 }  }$ 


$\Rightarrow$  $\left(\sqrt { 3+2\sqrt { 2 }  } +\sqrt { 3-2\sqrt { 2 }  }\right)^2$          [ Squaring both  sides ]

$\Rightarrow$  $(\sqrt{3+2\sqrt{2}})^2+(\sqrt{3-2\sqrt{2}})^2+2(\sqrt{3+2\sqrt{2}})(\sqrt{3-2\sqrt{2}})$

$\Rightarrow$  $3+2\sqrt{2}+3-2\sqrt{2}+2\sqrt{(3)^2-(2\sqrt{2})^2}$

$\Rightarrow$  $6+2\sqrt{9-8}$

$\Rightarrow$  $8$
Taking square root
$\Rightarrow$  $2\sqrt{2}$

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

Evaluate each of the following using identities :
i) $(399)^2$
ii) $(0.98)^2$
iii) $991 \times 1009$

  1. i) 159876ii) 0.91
    iii) 876590

  2. i) 135879ii) 0.87
    iii) 896750

  3. i) 159201ii) 0.9604
    iii) 999919

  4. i) 138760ii) 0.9
    iii) 999999
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$(i)$

$(399)^2=(400-1)^2$
             $=(400)^2+(1)^2-2\times 400\times 1$                [ $(a-b)=a^2+b^2-2ab$ ]
             $=160000+1-800$
             $=159201$
$\therefore$  $(399)^2=159201$

$(ii)$
$(0.98)^2=(1-0.02)^2$
              $=(1)^2+(0.02)^2-2\times 1\times 0.02$                 [ $(a-b)=a^2+b^2-2ab$ ]
              $=1+0.0004-0.04$
              $=0.9604$

$(iii)$ 
$991\times 1009=(1000-9)(1000+9)$
                       $=(1000)^2-(9)^2$                               [ $(a+b)(a-b)=a^2-b^2$ ]     
                       $=1000000-81$ 
                       $=999919$

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

Evaluating the following :
$(3+\sqrt{2})^{5}-(3-\sqrt{2})^{5}$

  1. $1718\sqrt 3$
  2. $1718\sqrt 2$
  3. $1178\sqrt 3$
  4. $1178\sqrt 2$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Given term is $(3+\sqrt{2})^{5}-(3-\sqrt{2})^{5}$


$\Rightarrow 2\left[\ ^{5}C _{1}\times 3^{4}\times (\sqrt{2})^{1}+\ ^{5}C _{3}\times 3^{2}\times (\sqrt{2})^{3}+\ ^{5}C _{5}\times 3^{0}\times (\sqrt{2})^{5}\right]$

$\Rightarrow 2\left[5\times 81\times\sqrt{2}+10\times 9\times 2\sqrt{2}+4\sqrt{2}\right]$

$\Rightarrow 2\sqrt{2}(405+180+4)$

$\Rightarrow 1178\sqrt{2}$

Multiple choice maths solving equations numerically finding roots by iteration fundamental theorem of algebra complex numbers and linear inequations

The product of $\left( { 23 x }^{ 2 }{ y }^{ 2 }z \right)$ and $\left( -15{ x }^{ 3 }{ yz }^{ 2 } \right) $ is .........................  .

  1. $-345 { x }^{ 5 } { y }^{ 3 } { z }^{ 3 }$
  2. $345 { x }^{ 2 } { y }^{ 2 } { z }^{ 3 }$
  3. $145 { x }^{ 2 } { y }^{ 2 } { z }^{ 3 }$
  4. $170 { x }^{ 2 } { y }^{ 2 } { z }^{ 3 }$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
$\left(23{x}^{2}{y}^{2}z\right)\times\left(-15{x}^{3}y{z}^{2}\right)$
$=-345‬{x}^{2+3}{y}^{2+1}{z}^{1+2}$
$=-345{x}^{5}{y}^{3}{z}^{3}$
Multiple choice maths solving equations numerically finding roots by iteration fundamental theorem of algebra complex numbers and linear inequations

$x$ and $y$ are real numbers such that ${7^x} - 16y = 0\;{\text{and}}\;{4^x} - 49y = 0,$ then the value of $\left( {y - x} \right)$ is

  1. $\dfrac{5}{2}$
  2. $\dfrac{{19}}{5}$
  3. $\dfrac{{4115}}{{2013}}$
  4. $\dfrac{{1569}}{{784}}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
$7^{x}=16y$

$4^{x}=49 y$

$\Rightarrow \dfrac{7^{x}}{4^{x}} = \dfrac{16}{49}$

$\Rightarrow \left( \dfrac{7}{4} \right)^{x} = \left( \dfrac{4}{7} \right)^{2} = \left( \dfrac{7}{4} \right)^{-2}$

$\Rightarrow x=-2$

$y= \dfrac{7^{x}}{16}$

$\Rightarrow y= \dfrac{1}{49 \times 16}$

So, $y-x = \dfrac{1}{49 \times 16}+2$

$=\dfrac{1}{784}+2$

$=\dfrac{1569}{784}$
Multiple choice maths the trapezium rule approximation errors and approximations the need for approximation

By Newton - Raphson's method the formula for finding the square root of any number $y$ is:

  1. $x _{n + 1} = \dfrac {1}{2}\left [x _{n} + \dfrac {y}{x _{n}}\right ]$
  2. $x _{n + 1} = \dfrac {1}{2}\left [x _{0} + \dfrac {y}{x _{0}}\right ]$
  3. $x _{n + 1} = \dfrac {1}{3}\left [2x _{n} + \dfrac {y}{x _{n}^{2}}\right ]$
  4. $x _{n + 1} = \dfrac {1}{3}\left [2x _{0} + \dfrac {y}{x _{0}^{2}}\right ]$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Let $x = \sqrt { y } $
$\implies { x }^{ 2 } = y$
$\implies { x }^{ 2 }-y = 0$
Iterative eqn. for Newton Raphson method is

${ x } _{ n+1 } = { { x } _{ n } }-\dfrac { f({ { x } _{ n } }) }{ f\prime ({ { x } _{ n } }) } $ 

Substitute $f(x) = { x }^{ 2 }-y$
$\implies f'(x) = 2x$ ........... $[\because\ y$ is any number $\therefore  f'(y) =0]$

     ${ x } _{ n+1 } = { { x } _{ n } }-\dfrac { { x } _{ n }^{ 2 }-y }{ 2{ x } _{ n } } $ 
              $= { x } _{ n }-\dfrac { { x } _{ n }^{ 2 } }{ 2{ x } _{ n } } +\dfrac { y }{ 2{ x } _{ n } } $ 

               $= { x } _{ n }-\dfrac { { x } _{ n } }{ 2 } +\dfrac { y }{ 2{ x } _{ n } } $
 
    ${ x } _{ n+1 } = \dfrac { { x } _{ n } }{ 2 } +\dfrac { y }{ 2{ x } _{ n } } $ 

    $\boxed { { x } _{ n+1 } = \dfrac { 1 }{ 2 } \left[ { x } _{ n }+\dfrac { y }{ { x } _{ n } }  \right]  } $ 
Multiple choice maths the trapezium rule approximation errors and approximations the need for approximation

Using Newton-Raphson method, the cube root of $24$ is?

  1. $2.884$
  2. $3.256$
  3. $5.231$
  4. $4.526$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

To find the cube root of $24$ using Newton - Raphson method,
we need to solve $f(x)=x^3-24$.

$ \Rightarrow f'(x)=3x^2$

Notice $3^3=27$

Therefore the cube root of $24$ is slightly less than $3$.

We have $f(x)=x^3-24, f'(x)=3x^2$

Let us start estimating the root $x$

Let the first estimation be $a=2.9$ (slightly less than 3)

Hence the subsequent estimates will be $b=a-\dfrac{f(a)}{f'(a)},c=b-\dfrac{f(b)}{f'(b)}$.

$f(a)=f(2.9)=(2.9)^3-24=0.389$ and $f'(a)=f'(2.9)=3(2.9)^2=25.23$

Therefore $b=2.9-\dfrac{0.389}{25.23}\approx 2.88458$

Now $c=2.88458-\dfrac{f(2.88458)}{f'(2.88458)}=2.88449$

Hence the cube root of $24$ is $2.884$

Multiple choice maths the trapezium rule approximation errors and approximations the need for approximation

The third approximation of roots of $x^3-x^2-1=0$ in the interval $(1,2)$ by the method of false position is?

  1. $2.430$
  2. $1.340$
  3. $1.430$
  4. $1.230$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Here, $x^3-x^2-1=0$

Let $f(x)=x^3-x^2-1$
First Iteration:
Here, $f(1)=-1<0$ and $f(2)=3>0$
Now, Root lies between $x _0=1$ and $x _1=2$
$x _2=x _0-f(x _0)\times \dfrac{x _1-x _0}{f(x _1)-f(x _0)}=1-(-1)\times \dfrac{2-1}{3-(-1)}=$
Second Iteration:
Here, $f(1.25)=-0.60938<0$ and $f(2)=3>0$

Now, Root lies between $x _0=1.25$ and $x _1=2$
$x _3=x _0-f(x _0)\times \dfrac{x _1-x _0}{f(x _1)-f(x _0)}=1.25-(-0.61)\times \dfrac{2-1.25}{3-(-0.61)}=1.37662$

Third Iteration:

Here, $f(1.37662)=-0.28626<0$ and $f(2)=3>0$
Now, Root lies between $x _0=1.37662$ and $x _1=2$
$x _4=x _0-f(x _0)\times \dfrac{x _1-x _0}{f(x _1)-f(x _0)}=1.38-(-0.29)\times \dfrac{2-1.38}{3-(-0.29)}=1.43093$

Multiple choice maths the trapezium rule approximation errors and approximations the need for approximation

The second approximation of roots of $x^3-5x-7=0$ in the interval $(2,3)$ by the method of false position is?

  1. $1.735$
  2. $2.375$
  3. $3.735$
  4. $2.735$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Here, $x^3-5x-7=0$

Let $f(x)=x^3-5x-7$
First Iteration:
Here, $f(2)=-9<0$ and $f(3)=5>0$
Now, Root lies between $x _0=2$ and $x _1=3$
$x _2=x _0-f(x _0)\times \dfrac{x _1-x _0}{f(x _1)-f(x _0)}=2-(-9)\times \dfrac{3-2}{5-(-9)}=2.64286$
Second Iteration:
Here, $f(2.64286)=-1.75474$ and $f(3)=5>0$

Now, Root lies between $x _0=2.64286$ and $x _1=3$
$x _3=x _0-f(x _0)\times \dfrac{x _1-x _0}{f(x _1)-f(x _0)}=2.64-(-1.75)\times \dfrac{3-2.64}{5-(-1.75)}=2.73564$

Multiple choice maths the trapezium rule approximation errors and approximations the need for approximation

The third approximation of root of $x^3-x^2-1=0$ in the interval $(1,2)$ using successive bisection method is?

  1. $1.475$
  2. $1.375$
  3. $2.213$
  4. $1.564$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

We have to find the third approximation of root of the equation $x^3-x^2-1=0$ in the interval $(1,2)$ using successive Bisection method.

$\textbf{Iteration 1: k=0}$

$c _0=\dfrac{a _0+b _0}{2}=\dfrac{1+2}{2}=1.5$

Since $f(c _0)f(a _0)=f(1.5)f(1)<0$

Therefore set $a _1=a _0,b _1=c _0$

$\textbf{Iteration 2: k=1}$

$c _1=\dfrac{a _1+b _1}{2}=\dfrac{1+1.5}{2}=1.25$

Since $f(c _1)f(a _1)=f(1.25)f(1)>0$

Therefore set $a _2=c _1,b _2=b _1$

$\textbf{Iteration 3: k=2}$

$c _2=\dfrac{a _2+b _2}{2}=\dfrac{1.25+1.5}{2}=1.375$

Thus the third approximation of the root is $1.375$ respectively.

Multiple choice maths the trapezium rule approximation errors and approximations the need for approximation

By successive bisection method, the cube root of $2$ between the interval (1,1.5)_is?

  1. $1.2813$
  2. $1.2121$
  3. $1.013$
  4. $1.475$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Function can be written as $f(x)=x^3-2$


First Iteration:
$f(1)=-1<0$ and $f(1.5)=1.375>0$
Now, root lies between $1$ and $1.5$
So,
$x _0=\dfrac{1+1.5}{2}=1.25$
$f(x _0)=-0.04688<0$

Second Iteration:

$f(1.25)=-0.04688<0$ and $f(1.5)=1.375>0$
Now, root lies between $1.25$ and $1.5$
So,
$x _1=\dfrac{1.25+1.5}{2}=1.375$

$f(x _1)=0.59961>0$

Third Iteration:

$f(1.25)=-0.04688<0$ and $f(1.375)=0.59961>0$
Now, root lies between $1.25$ and $1.375$
So,
$x _2=\dfrac{1.25+1.375}{2}=1.3125$
$f(x _2)=0.26099>0$

Fourth Iteration:
$f(1.25)=-0.04688<0$ and $f(1.3125)=0.26099>0$
Now, root lies between $1.25$ and $1.3125$
So,
$x _2=\dfrac{1.25+1.3125}{2}=1.28125\approx 1.2813$
Multiple choice maths the trapezium rule approximation errors and approximations the need for approximation

The third approximation of roots of $x^3-9x+1=0$ in the interval $(2,4)$ by the method of false position is?

  1. $8.23$
  2. $1.25$
  3. $2.85$
  4. $2.12$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Here, $x^3-9x+1=0$

Let $f(x)=x^3-9x+1$
First Iteration:
Here, $f(2)=-9<0$ and $f(4)=29>0$
Now, Root lies between $x _0=2$ and $x _1=4$
$x _2=x _0-f(x _0)\times \dfrac{x _1-x _0}{f(x _1)-f(x _0)}=2-(-9)\times \dfrac{4-2}{29-(-9)}=2.47368$
Second Iteration:
Here, $f(2.47368)=-6.1264$ and $f(2)=29>0$

Now, Root lies between $x _0=2.47368$ and $x _1=4$
$x _3=x _0-f(x _0)\times \dfrac{x _1-x _0}{f(x _1)-f(x _0)}=2.47-(-6.13)\times \dfrac{4-2.47}{29-(-6.13)}=2.73989$

Third Iteration:

Here, $f(2.73989)=-3.09067$ and $f(4)=29>0$
Now, Root lies between $x _0=2.73989$ and $x _1=4$
$x _4=x _0-f(x _0)\times \dfrac{x _1-x _0}{f(x _1)-f(x _0)}=2.74-(-3.09)\times \dfrac{4-2.74}{29-(-3.09)}=2.86125$