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

Choose the correct option:
$\left[\dfrac{{100}}{{101}}\right]^3$

  1. $\dfrac{{100}^3}{{101}^3}$
  2. $\dfrac{{100}^4}{{101}^4}$
  3. $\dfrac{{1000}^2}{{101}^2}$
  4. $\dfrac{{100}}{{101}}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$Now\quad \left[ \dfrac { 100 }{ 101 }  \right] ^{ 3 }\quad \ \quad \quad =\quad \dfrac { { 10 }0^{ 3 } }{ 101^{ 3 } } \quad \left( \because \left( \dfrac { { a }^{ m } }{ { b }^{ m } }  \right) =\left( \dfrac { a }{ b }  \right) ^{ m } \right) \ $

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

Choose the correct options:$\dfrac{{10}^2}{{11}^2}$

  1. $\left[\dfrac{{10}}{{11}}\right]^2$
  2. $\left[\dfrac{{100}}{{11}}\right]^2$
  3. $\left[\dfrac{{10}}{{11}}\right]^4$
  4. $\left[\dfrac{{5}}{{11}}\right]^2$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$Now\quad \dfrac { { 10 }^{ 2 } }{ 11^{ 2 } } \ =\quad \left( \dfrac { 10 }{ 11 }  \right) ^{ 2 }\left( \because \left( \dfrac { { a }^{ m } }{ { b }^{ m } }  \right) =\left( \dfrac { a }{ b }  \right) ^{ m } \right) $

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

Choose the correct option:
$\left(\dfrac{5^5\times6^5}{3^5}\right)$

  1. $\left(\dfrac{5\times6}{3}\right)^5$
  2. $\left(\dfrac{5\times6}{3}\right)^6$
  3. $\left(\dfrac{5\times6}{5}\right)^3$
  4. $\left(\dfrac{5\times6}{5}\right)^5$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$Now\quad \left( \dfrac { { 5 }^{ 5 }\times { 6 }^{ 5 } }{ { 3 }^{ 5 } }  \right) \quad \ \quad \quad =\quad \left( \dfrac { 5\times 6 }{ 3 }  \right) ^{ 5 }\quad \left( \because \left( \dfrac { { a }^{ m }\times { c }^{ m } }{ { b }^{ m } }  \right) =\left( \dfrac { a\times c }{ b }  \right) ^{ m } \right) $

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

$\left(\dfrac{1}{x^{a-b}}\right)^{\tfrac{1}{(a-c)}}. \left(\dfrac{1}{x^{b-c}}\right)^{\tfrac{1}{(b-a)}}. \left(\dfrac{1}{x^{c-a}}\right)^{\tfrac{1}{(c-b)}}=$

  1. $0$
  2. $1$
  3. $a+b+c$
  4. $(a-b+c)^2$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

We can write the given equation as, 

$(x^{b-a})^{\frac{1}{a-c}}\cdot (x^{c-b})^{\frac{1}{b-a}}\cdot (x^{a-c})^{\frac{1}{c-b}}$

$=x^{\cfrac{b-a}{a-c}}\cdot x^{\cfrac{c-b}{b-a}}\cdot x^{\cfrac{a-c}{c-b}}$
On adding all the powers of $x$, We get
$x^{\Bigl(\cfrac{(b-a)^2(c-b)+(c-b)^2(a-c)+(a-c)^2(b-a)}{(a-c)(b-c)(c-b)}\Bigr)}\ =x^0=1$