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 power and exponent power of powers laws of exponents and powers law of indices

The value of $\left (\dfrac {a^{-2} \times b^{-3}}{a^{-3}\times b^{-4}}\right )$ is _________.

  1. $a^{-1}\times b$
  2. $a \times b^{-1}$
  3. $(ab)^{-1}$
  4. $ab$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
We need to find value of $\left (\dfrac {a^{-2} \times b^{-3}}{a^{-3}\times b^{-4}}\right )$
By using $\dfrac {a^m}{a^n}=a^{m-n}$
Then it can be written as,
$a^{-2-(-3)}\times b^{-3-(-4)}$ $=$ $ab$   
Hence, option D is correct.
Multiple choice maths power and exponent power of powers laws of exponents and powers law of indices

$\left(\dfrac{5^a}{5^b}\right)^{a+b}.\left(\dfrac{5^b}{5^c}\right)^{b+c}.\left(\dfrac{5^c}{5^a}\right)^{c+a} =$ 

  1. $1$
  2. $4$
  3. $5$
  4. $0$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

We have, $\Bigr(\dfrac{5^a}{5^b}\Bigl)^{a+b}\cdot\Bigl(\dfrac{5^b}{5^c} \Bigr)^{b+c}\cdot \Bigl(\dfrac{5^c}{5^a} \Bigr)^{c+a}$


$=(5^{a-b})^{a+b}\cdot(5^{b-c})^{b+c}\cdot(5^{c-a})^{c+a}$

$=5^{a^2-b^2}\cdot 5^{b^2-c^2}\cdot 5^{c^2-a^2}\ $

$=\dfrac{5^{a^2}}{5^{b^2}}\cdot \dfrac{5^{b^2}}{5^{c^2}}\cdot \dfrac{5^{c^2}}{5^{a^2}}\\=1$

Multiple choice maths power and exponent power of powers laws of exponents and powers law of indices

Simplicity
$\left[ \left{ \left( 625 \right) ^{ -\dfrac { 1 }{ 2 } } \right} ^{ -\dfrac { 1 }{ 4 } } \right] $

  1. $\dfrac{1}{\sqrt5}$
  2. $\sqrt5$

  3. 5

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

$\begin{array}{l}\left[ {{{\left( {{{\left( {625} \right)}^{\frac{{ - 1}}{2}}}} \right)}^{\frac{{ - 1}}{4}}}} \right] = \left[ {{{\left( {{{\left( {{{25}^2}} \right)}^{\frac{{ - 1}}{2}}}} \right)}^{\frac{{ - 1}}{4}}}} \right]\ = \left[ {{{\left( {{{25}^{ - 1}}} \right)}^{\frac{{ - 1}}{4}}}} \right]\ = {25^{\frac{1}{4}}}\ = {5^{2 \times \frac{1}{4}}}\ = {5^{\frac{1}{2}}}\ = \sqrt 5 \end{array}$

Multiple choice maths power and exponent power of powers laws of exponents and powers law of indices

The value of $\left(\dfrac{x^q}{x^r}\right)^{\dfrac{1}{qr}} \times \left(\dfrac{x^r}{x^p}\right)^{\dfrac{1}{rp}}\times \left(\dfrac{x^p}{x^q}\right)^{\dfrac{1}{pq}}$ is equal to ___.

  1. $x^{\frac{1}{p}+\frac{1}{q}+\frac{1}{2}}$
  2. $0$
  3. $x^{pq+qr+rp}$
  4. $1$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$(x^{q-r})^{\cfrac{1}{qr}}\times (x^{r-p})^{\cfrac{1}{rp}}\times (x^{p-q})^{\cfrac{1}{pq}} $


$=x^{\cfrac{q-r}{qr}}\times x^{\cfrac{r-p}{rp}}\times x^{\cfrac{p-q}{pq}}$
On adding all the powers of $x$, we get
$\Rightarrow x^{\bigl(\cfrac{q-r}{qr}+\cfrac{r-p}{rp}+\cfrac{p-q}{pq}\bigr)}$

$=x^{\cfrac{p(q-r)+q(r-p)+r(p-q)}{pqr}}$

$=x^{\cfrac{0}{pqr}}=1$

Multiple choice de moivre’s theorem and its applications demoivre's theorem complex numbers maths

If ${ x }^{ 6 }={ \left( 4-3i \right)  }^{ 5 }$, then the product of all of its roots is (where $\displaystyle \theta =-\tan ^{ -1 }{ \frac { 3 }{ 4 }  } $)

  1. ${ 5 }^{ 5 }\left( \cos { 5\theta } +i\sin { 5\theta } \right) $
  2. $-{ 5 }^{ 5 }\left( \cos { 5\theta } +i\sin { 5\theta } \right) $
  3. ${ 5 }^{ 5 }\left( \cos { 5\theta } -i\sin { 5\theta } \right) $
  4. $-{ 5 }^{ 5 }\left( \cos { 5\theta } -i\sin { 5\theta } \right) $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\displaystyle { x }^{ 6 }={ \left( 4-3i \right)  }^{ 5 }\Rightarrow { x }^{ 6 }={ 5 }^{ 6 }\left( \frac { 4 }{ 5 } -\frac { 3i }{ 5 }  \right) ={ 5 }^{ 5 }{ \left( \cos { \theta  } +i\sin { \theta  }  \right)  }^{ 5 }$

where $\displaystyle \theta =-\tan ^{ -1 }{ \frac { 3 }{ 4 }  } ={ 5 }^{ 5 }\left( \cos { 5\theta  } +i\sin { 5\theta  }  \right) $
$\displaystyle x={ 5 }^{ 5/6 }{ \left( \cos { 5\theta  } +i\sin { 5\theta  }  \right)  }^{ 1/6 }={ 5 }^{ 5/6 }\left[ \cos { \left( \frac { 2k\pi +5\theta  }{ 6 }  \right)  } +i\sin { \left( \frac { 2k\pi +5\theta  }{ 6 }  \right)  }  \right] $
${ x } _{ 1 }{ x } _{ 2 }{ x } _{ 3 }...{ x } _{ 6 }={ 5 }^{ 5 }\left( \cos { \left( 5\pi +5\theta  \right)  } +i\sin { \left( 5\pi +5\theta  \right)  }  \right) \ ={ 5 }^{ 5 }\left( -\cos { 5\theta  } -i\sin { 5\theta  }  \right) =-{ 5 }^{ 5 }\left( \cos { 5\theta  } +i\sin { 5\theta  }  \right) $

Multiple choice maths fraction lowest form of a fraction simplest ratio lowest form of fractions

The simplest rationalizing factor of $\sqrt{75}$ is.

  1. $(75)^{1/3}$
  2. $5\sqrt3$
  3. $3$
  4. $\sqrt{150}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let us first factorize $75$ as shown below:


$75=3\times 5\times 5=3\times { 5 }^{ 2 }$

Now consider $\sqrt {75}$ as follows:

$\sqrt { 75 } =\sqrt { 3\times 5\times 5 } =\sqrt { 3\times { 5 }^{ 2 } } =\sqrt { 3 } \times \sqrt { { 5 }^{ 2 } } =5\sqrt { 3 }$

Hence, the simplest rationalizing factor of $\sqrt {75}$ is $5\sqrt { 3 }$.

Multiple choice maths geometric sequences introduction to geometric progression understanding geometric progressions introduction to geometric progressions

$S = {3^{10}} + {3^9} + \frac{{{3^9}}}{4} + \frac{{{3^7}}}{2} + \frac{{{{5.3}^6}}}{{16}} + \frac{{{3^2}}}{{16}} + \frac{{{{7.3}^4}}}{{64}} + .........$ upto infinite terms, then $\left( {\frac{{25}}{{36}}} \right)S$ equal to 

  1. ${6^9}$
  2. ${3^{10}}$
  3. ${3^{11}}$
  4. ${2.3^{10}}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\begin{array}{l}S = {3^{10}} + {3^9} + \cfrac{{{3^9}}}{4} + \cfrac{{{3^7}}}{2} + \cfrac{{{{5.3}^6}}}{{16}} + \cfrac{{{3^2}}}{{16}} + \cfrac{{{{7.3}^4}}}{{64}} + .........\infty \S = \cfrac{{1 \times {3^{10}}}}{{{2^0}}} + \cfrac{{2 \times {3^9}}}{{{2^1}}} + \cfrac{{3 \times {3^8}}}{{{2^2}}} + \cfrac{{4 \times {3^7}}}{{{2^3}}} + \cfrac{{5 \times {3^6}}}{{{2^4}}} + \cfrac{{6 \times {3^5}}}{{{2^5}}} + \cfrac{{7 \times {3^4}}}{{{2^6}}} + .....\infty \\cfrac{S}{6} = \cfrac{{{3^9}}}{2} + \cfrac{{2 \times {3^8}}}{{{2^2}}} + .......\infty \S - \cfrac{S}{6} = \cfrac{{{3^{10}}}}{2^0} + \cfrac{{{3^9}}}{{{2^1}}} + \cfrac{{{3^8}}}{{{2^2}}} + ........\infty \\cfrac{{6S - S}}{6} = \cfrac{{{3^{10}}}}{{1 - \cfrac{1}{6}}} = \cfrac{{{3^{10}}}}{5}\left( 6 \right) \end{array}$

$\dfrac56S=\dfrac{3^{10}\times 6 }{5}$
$\therefore \dfrac {25}{36}S=3^{10}$

Multiple choice

What is the value of the square root of 2 according to Vagbhata?

  1. 1.414

  2. 1.4142

  3. 1.41421

  4. 1.414213
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Vagbhata calculated the square root of 2 to five decimal places, which is an impressive achievement for his time.