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 fun with numbers some special sequences triangular numbers properties and patterns of perfect squares

Fourth roots of $193-4\sqrt{2178}$ is

  1. $(7-\sqrt{2})$
  2. $(5-\sqrt{2})$
  3. $(3-\sqrt{2})$
  4. $(10-\sqrt{7})$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
According to Question

$(193-4\sqrt{2178} )^{1/4}$

$=(193-4\sqrt{11\times11\times3\times3\times2} )^{1/4}$

$=(193-4\times11\times3\sqrt2 )^{1/4}$

$=(121+72-132\sqrt{2} )^{1/4}$

$=(11^2+(6\sqrt2)^2-2\times11\times6\sqrt2)^{1/4}$                          $Using\ a^2+b^2-2ab=(a-b)^2$

$=(11-6\sqrt2)^{2\times0.25}$

$=(9+2-6\sqrt{2})^{0.5}$

$=(3^2+\sqrt{2}\ ^2-2\times3\times\sqrt{2})^{0.5}$                          $Using\ a^2+b^2-2ab=(a-b)^2$

$=(3-\sqrt{2})^{0.5\times2}$

$=3-\sqrt{2}$

$C$ is the right answer

Multiple choice sum to infinite terms of a gp sequence, progression and series maths

If $x=\sqrt{4}.\sqrt[4]{4}. \sqrt[8]{4}.\sqrt[16]{4}........ \infty$, then 

  1. $x^2-8x+16=0$
  2. $x^2-3x+2=0$
  3. $x^2-5x+4=0$
  4. $x^2+5x+4=0$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\begin{array}{l} x={ 4^{ 1/2 } }{ 4^{ 1/4 } }{ 4^{ 1/8 } }\cdots \infty  \ ={ 4^{ \frac { 1 }{ 2 } +\frac { 1 }{ 4 } +\cdots +\infty  } } \ ={ 4^{ \frac { { 1/2 } }{ { 1-\frac { 1 }{ 2 }  } }  } } \ =4 \ { \left( { x-4 } \right) ^{ 2 } }=0 \ { x^{ 2 } }+16-8x=0 \end{array}$

Multiple choice business maths matrix properties of matrix multiplication properties of multiplication of matrix multiplication of matrices

If $M = \left[ \begin{array}{l}0\,\,\,\,2\5\,\,\,\,\,0\end{array} \right]\,\,\,and\,\,N = \left[ \begin{array}{l}0\,\,\,\,5\2\,\,\,\,\,0\end{array} \right]$,then ${M^{2011}}$ is-

  1. ${10^{1005}}M$
  2. ${10^{1005}}N$
  3. ${10^{2010}}M$
  4. ${10^{2011}}M$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Computing M^2 gives matrix multiplication resulting in [10 0; 0 10], which is 10I. By observing patterns for higher powers, M^(2k) = 10^k I and M^(2k+1) = 10^k M. For M^2011, since 2011 is odd, it simplifies to 10^1005 M.

Multiple choice business maths matrix properties of matrix multiplication properties of multiplication of matrix multiplication of matrices

If $\left[\begin{array}{ll}
\mathrm{x} & \mathrm{y}^{3}\
2 & 0
\end{array}\right]=\left[\begin{array}{ll}
1 & 8\
2 & 0
\end{array}\right]$, then  $\left[\begin{array}{ll}
\mathrm{x} & \mathrm{y}\
2 & 0
\end{array}\right]^{-1}$ is equal to

  1. $-\dfrac{1}{4}$$\left[\begin{array}{ll}

    0 &-2\\

    -2 & 1

    \end{array}\right]$
  2. $\dfrac{2}{4}$$\left[\begin{array}{ll}

    1 & 0\\

    0 & 1

    \end{array}\right]$
  3. $\dfrac{1}{4}$$\left[\begin{array}{ll}

    0 & -8\\

    -2 & 1

    \end{array}\right]$
  4. $\dfrac{1}{4}\left[\begin{array} \ 1&4 \\7 &2 \end{array}\right]$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given the matrix equality, x=1 and y^3=8, which implies y=2. The matrix to invert is [[1, 2], [2, 0]]. The determinant is (1*0) - (2*2) = -4. The inverse is (1/det) * [[0, -2], [-2, 1]], which simplifies to -1/4 * [[0, -2], [-2, 1]].

Multiple choice maths geometric sequences sum of terms of g.p sum of n terms of an gp summing geometric series

The value of the sum $\sum _{ n=1 }^{ 13 }{ \left( { i }^{ n }+{ i }^{ n+1 } \right)  } $ where $i=\sqrt { -1 } $ is:

  1. $i$
  2. $-i$
  3. $0$
  4. $i-1$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$\displaystyle \sum _{ n=1 }^{ 13 }{ ({ i }^{ n }+{ i }^{ n+1 }) }$ 

$=\displaystyle (i+1)\sum _{ n=1 }^{ 13 }{ { i }^{ n } } $
$=(i+1)\dfrac { (i({ i }^{ 13 }-1)) }{ i-1 } $
$=(i+1)\dfrac { (i(i-1)) }{ i-1 } $
$=(i+1)(i)$
$=i-1$
Hence, the correct answer is D.

Multiple choice maths parts and whole multiplication of a fraction multiplication of a fractions multiplication of fraction finding the whole when a fraction is given

The value of $\left (-\dfrac {7}{2}\right )^{-1}$ is _________.

  1. $-1$
  2. $\dfrac {7}{2}$
  3. $-\dfrac {2}{7}$
  4. $\dfrac {-7}{2}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

To find $\left(-\dfrac{7}{2}\right)^{-1}$


Any fraction raised to negative power yields same result as of its reciprocal with modulus of the power.

$\therefore \left(-\dfrac{7}{2}\right)^{-1} = -\dfrac{2}{7}$

Multiple choice maths parts and whole multiplication of a fraction multiplication of a fractions multiplication of fraction finding the whole when a fraction is given

The value of the expression $\sqrt {34-24\sqrt 2}\times (4+3\sqrt 2)$ is

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

$\sqrt {34-24\sqrt 2}\times (4+3\sqrt 2)$


$=\sqrt {34-24\sqrt 2}\sqrt {(4+3\sqrt 2)^2}$

$=\sqrt {(34-24\sqrt 2)(16+18+24\sqrt 2)}$

$=\sqrt {(34-24\sqrt 2)(34+24\sqrt 2)}$

$=\sqrt {(34)^2(24\sqrt 2)^2}$

$=\sqrt {1156-1152}=\sqrt 4=2$

Multiple choice maths parts and whole multiplication of a fraction multiplication of a fractions multiplication of fraction finding the whole when a fraction is given

When simplified, the product $\left( 1-\cfrac { 1 }{ 3 }  \right) \left( 1-\cfrac { 1 }{ 4 }  \right) \left( 1-\cfrac { 1 }{ 5 }  \right) ...\left( 1-\dfrac 1n \right) $ becomes

  1. $\dfrac { 1 }{ n } $
  2. $\dfrac { 2 }{ n } $
  3. $\dfrac { 2(n-1) }{ n } $
  4. $\dfrac { 2 }{ n(n+1) } $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\left( 1-\cfrac { 1 }{ 3 }  \right) \left( 1-\cfrac { 1 }{ 4 }  \right) \left( 1-\cfrac { 1 }{ 5 }  \right) ...\left( 1- \cfrac 1n \right) =\cfrac { 2 }{ 3 } .\cfrac { 3 }{ 4 } .\cfrac { 4 }{ 5 } ....\cfrac { n-2 }{ n-1 } .\cfrac { n-1 }{ n } =\cfrac { 2 }{ n } $

Multiple choice maths rational and irrational numbers existence of irrational numbers irrational numbers properties of irrational numbers

Read out each of the following numbers carefully and specify the natural numbers in it.
$87, 54, 0, -13, -4.7, \sqrt{7}, 2{1}{7}, \sqrt{15}, -{8}{7}, 3\sqrt{7}, 4.807, 0.002, \sqrt{16}$ and $2+\sqrt{3}.$

  1. $0,87,54,\sqrt{16}$
  2. $87, 54,$ $\sqrt{16}$, $217$
  3. $0, -13, -4,7, 217, 54, 87$
  4. $\sqrt{7}$, $\sqrt{15}$, $3 \sqrt{7}$, $\sqrt{16}$, $2 + \sqrt{3}$,
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Natural numbers from the given list are 87, 54,  $\sqrt { 16 } =4$ and 217

Multiple choice maths rational and irrational numbers existence of irrational numbers irrational numbers properties of irrational numbers

Which of the following number is different from others?

  1. $\sqrt 7$
  2. $\sqrt 6$
  3. $\sqrt {25}$
  4. $\sqrt{10}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$\sqrt{7}$ is an irrational number

$\sqrt{6}$ is an irrational number
$\sqrt{10}$ is an irrational number
$\sqrt{25}=5$ is different from others because others are irrational number but $\sqrt{25}$ is a rational number
Hence, option C is correct.

Multiple choice maths rational and irrational numbers existence of irrational numbers irrational numbers properties of irrational numbers

$\left ( 2+\sqrt{5} \right )\left ( 2+\sqrt{5} \right )$ expression is :

  1. A rational number

  2. A whole number

  3. An irrational number

  4. A natural number
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

${ (2+\sqrt { 5 } ) }^{ 2 }\ =4+5+4\sqrt { 5 } \ =9+4\sqrt { 5 } $

In the above equation $4\sqrt { 5 } $ is irrational number so $9+4\sqrt { 5 } $ will also be irrational number 
So correct answer is option C.

Multiple choice maths rational and irrational numbers existence of irrational numbers irrational numbers properties of irrational numbers

$\sqrt{21-4\sqrt{5}+8\sqrt{3}-4\sqrt{15}}=$...........

  1. $\sqrt{5}-2+2\sqrt{3}$
  2. $\sqrt{5}-\sqrt{4}-\sqrt{12}$
  3. $-\sqrt{5}+\sqrt{4}+\sqrt{12}$
  4. $-\sqrt{5}-\sqrt{4}+\sqrt{12}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The expression is sqrt(21 - 4*sqrt(5) + 8*sqrt(3) - 4*sqrt(15)). This is of the form sqrt((a+b+c)^2) = |a+b+c|. Expanding (sqrt(5) - 2 - 2*sqrt(3))^2 gives 5 + 4 + 12 - 4*sqrt(5) - 4*sqrt(15) + 8*sqrt(3) = 21 - 4*sqrt(5) + 8*sqrt(3) - 4*sqrt(15). Thus the square root is |sqrt(5) - 2 - 2*sqrt(3)|, which equals -sqrt(5) + 2 + 2*sqrt(3).