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 powers and exponents scientific notation use of exponents power of 10

If $9^n = 27^{n+1}$, then calculate the value of $2^n $.

  1. $-\dfrac{10}{3}$
  2. $-\dfrac{8}{3}$
  3. $-\dfrac{3}{8}$
  4. $\dfrac{1}{8}$
  5. $\dfrac{3}{8}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Given ${ 9 }^{ n }={ 27 }^{ n+1 }$, which implies
$ { 3 }^{ 2n }={ 3 }^{ 3n+3 }$
Now compare powers, we get
$2n = 3n+3$ , which implies $n = -3$
Therefore ${ 2 }^{ n }={ 2 }^{ -3 }=\dfrac { 1 }{ 8 } $

Multiple choice maths powers and exponents scientific notation use of exponents power of 10

Which of the following has the greatest value?

  1. $(6^{2} \times 6)^{4}$
  2. $(36)^{5}$
  3. $(36^{2} \times 6^{3})^{2}$
  4. $(216)^{4}$
  5. $(6^{4})^{4}$
Reveal answer Fill a bubble to check yourself
E Correct answer
Explanation

  1. ${ ({ 6 }^{ 2 }\times 6) }^{ 4 }={ { (6 }^{ 3 }) }^{ 4 }={ 6 }^{ 12 }$
  2. ${ 36 }^{ 5 }={ ({ 6 }^{ 2 }) }^{ 5 }={ 6 }^{ 10 }$
  3. ${ ({ 36 }^{ 2 }\times { 6 }^{ 3 }) }^{ 2 }={ ({ 6 }^{ 4 }\times { 6 }^{ 3 }) }^{ 2 }={ ({ 6 }^{ 7 }) }^{ 2 }={ 6 }^{ 14 }$
  4. ${ 216 }^{ 4 }={ ({ 6 }^{ 3 }) }^{ 4 }={ 6 }^{ 12 }$
  5. ${ ({ 6 }^{ 4 }) }^{ 4 }={ 6 }^{ 16 }$
  • Therefore option $E$ has maximum value

Multiple choice maths powers and exponents scientific notation use of exponents power of 10

If $3^{n} = n^{6}$, find the value of $ n^{18} $

  1. $3^{n} n^{3}$
  2. $3^{n} n^{12}$
  3. $9^{n}$
  4. $3^{12n}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Given, $3^n=n^6$
We need to find the value of $n^{18}$
$\therefore {n}^{18}= n^6. n^{12}=3^n.n^{12}$
$\therefore n^{6+12}=3^n. n^{12}$
$\therefore n^{18}= {3}^{n}{n}^{12}$
Multiple choice maths powers and exponents scientific notation use of exponents power of 10

If $5^{k^2}(25^{2k})(625) = 25\sqrt{5}$ and $k < -1$, find the value of $k$.

  1. $-3.581$
  2. $-3.162$
  3. $-2.613$
  4. $-1.581$
  5. $-0.419$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given, ${ 5 }^{ { k }^{ 2 } }({ 25 }^{ 2k })(625)=25\sqrt { 5 } $
${ 5 }^{ { k }^{ 2 }+4k+4 }={ 5 }^{ \tfrac 52 }$ 

By comparing powers, we get
${ k }^{ 2 }+4k+4=\cfrac 52$ which implies ${ 2k }^{ 2 }+8k+3=0$
$\Rightarrow k = -3.581$           ...(given $k<-1$)

Multiple choice maths place value, ordering and rounding order operations and algebra using brackets in algebraic expressions order of operations

$9+\cfrac { 3 }{ 4 } +7+\cfrac { 2 }{ 17 } -\left( 9+\cfrac { 1 }{ 15 }  \right) =$?

  1. $7+\cfrac { 719 }{ 1020 } $
  2. $9+\cfrac { 817 }{ 1020 } $
  3. $9+\cfrac { 719 }{ 1020 } $
  4. $7+\cfrac { 817 }{ 1020 } $

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

Given the sum $=9+\cfrac { 3 }{ 4 } +7+\cfrac { 2 }{ 17 } -\left( 9+\cfrac { 1 }{ 15 }  \right) $
$=(9+7-9)+\left( \cfrac { 3 }{ 4 } +\cfrac { 2 }{ 17 } -\cfrac { 1 }{ 15 }  \right) $
$=7+\cfrac { 765+120-68 }{ 1020 } $
$=7+\cfrac { 817 }{ 1020 } $

Multiple choice maths calculations and mental strategies 4 mental additions and subtractions of decimals multiplication and division of decimals mental multiplication and division

If k is an integer, and if $0.02468 \times 10^k$ is greater than 10,000, what is the least possible value of k?

  1. 7

  2. 4

  3. 6

  4. 5
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Multiplying 0.02468 by a positive power of 10 will shift the decimal point to the right. Simply shift the decimal point to the right until the result is greater than 10,000. Keep track of how many times you shift the decimal point. Shifting the decimal point 5 times results in 2,468. This is still less than 10,000. Shifting one more place yields 24,680, which is greater than 10,000.

Multiple choice maths calculating and mental strategies 3 finding percentage of a number how many in all? problems on percentage

Estimated value of square root of $650$ is

  1. $25.495$
  2. $24.495$
  3. $2.5495$

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

$\Rightarrow$   Here we have to find square root of $650$

$\Rightarrow$  $\sqrt{650}=\sqrt{25\times 26}$
$\Rightarrow$  $\sqrt{650}=5\sqrt{26}$
$\Rightarrow$  $\sqrt{650}=5\times  5.09$
$\Rightarrow$  $\sqrt{650}=25.495$

Multiple choice maths binomial theorem, sequence and series series introduction to series introduction to sequences and series

$ \left{ a _ { n } \right} $ and $ \left{ b _ { n } \right} $ are two sequences given by $ a _ { n } = ( x ) ^ { 1 / 2 ^ { \circ } } + ( y ) ^ { 1 / 2 ^ { \circ } } $ and $ b _ { n } = ( x ) ^ { 1 / 2 ^ { 2 } } - ( y ) ^ { 1 / 2 ^ { \circ } } $ for all $ \mathrm { n } \in \mathrm { N } . $ The value of $ \mathrm { a } _ { 1 } \mathrm { a } _ { 2 } \mathrm { a } _ { 3 } \dots \ldots \ldots \mathrm { a } _ { \mathrm { n } } $ is equal to

  1. x-y
  2. $

    \frac { x + y } { b _ { n } }

    $
  3. $

    \frac { x - y } { b _ { n } }

    $
  4. $

    \frac { x y } { b _ { n } }

    $
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice reciprocal equations theory of equations maths

$\cfrac { \left( 2x-1 \right) { \left( x-1 \right)  }^{ 4 }{ \left( x-2 \right)  }^{ 4 } }{ (x-2){ \left( x-4 \right)  }^{ 4 } } \le 0$

  1. $(\dfrac{1}{2},2)$
  2. $R$
  3. $\phi$
  4. $(1/3,2)$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\dfrac{(2{x}-1)(x-1)^{4}(x-2)^{4}}{(x-2)(x-4)^{4}} \le 0\implies x\neq 2$


$(2{x}-1)(x-1)^{4}(x-2)^{3}\le 0$


$(x-\dfrac{1}{2})(x-2)^{3}\le 0$

$x\in \bigg(\dfrac{1}{2},2\bigg)$

Multiple choice maths part number dividing fractions division of a fractions division of a fraction

Solve $\left[\dfrac{170}{3} +\dfrac{6}{7}\right] \div \left[\dfrac{2}{7} \times \dfrac{11}{2}\right]$

  1. $\dfrac{1208}{3\times 11}$
  2. $\dfrac{1208}{11}$
  3. $\dfrac{1208}{3}$
  4. $\dfrac{1208}{9\times 11}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\left[\dfrac{170}{3} +\dfrac{6}{7}\right] \div \left[\dfrac{2}{7} \times \dfrac{11}{2}\right]$


$=\left[ \dfrac{1190+18}{21}\right] \div \left[ \dfrac{11}{7}\right]$


$=\left[ \dfrac{1190+18}{21}\right] \times \left[ \dfrac{7}{11}\right]$

$=\left[ \dfrac{1208}{11\times 3}\right]$

Multiple choice maths part number dividing fractions division of a fractions division of a fraction

$\left(\large{\frac{-5}{3}}\right)^5$ $\div$ $\left(\large{\frac{-5}{3}}\right)^{7}$

  1. $\large{\frac{25}{9}}$
  2. $\large{\frac{9}{25}}$
  3. $\large{\frac{16}{25}}$
  4. $\large{\frac{25}{16}}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\left(\large{\frac{-5}{3}}\right)^5$ $\div$ $\left(\large{\frac{-5}{3}}\right)^{7}$


$=-\left(\large{\frac{5}{3}}\right)^5$ $\times$ $-\left(\large{\frac{3}{5}}\right)^{7}$


$=\left(\dfrac{3}{5}\right)^2$

$=\dfrac{9}{25}$.