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 applications of matrices and determinants elementary transformations of a matrix multiplicative inverse of a matrix inverse of a matrix

If $
A=\left[ \begin{array}{ll}{x} & {1} \ {1} & {0}\end{array}\right]
 $ and $
A^{2}=I
 $, $
A^{-1}
 $ is equal to ...............

  1. $

    \left[ \begin{array}{ll}{0} & {1} \\ {1} & {0}\end{array}\right]

    $
  2. $

    \left[ \begin{array}{ll}{1} & {0} \\ {0} & {1}\end{array}\right]

    $
  3. $

    \left[ \begin{array}{ll}{1} & {1} \\ {1} & {1}\end{array}\right]

    $
  4. $

    \left[ \begin{array}{ll}{0} & {0} \\ {0} & {0}\end{array}\right]

    $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
$A=\left[\begin{matrix} x & 1 \\ 1 & 0  \end{matrix}\right]$
Given: ${A}^{2}=I$ where $I$ is $2\times 2$ identity matrix
Let us find ${A}^{2}$
$=\left[\begin{matrix} x & 1 \\ 1 & 0  \end{matrix}\right]\left[\begin{matrix} x & 1 \\ 1 & 0  \end{matrix}\right]$
$=\left[\begin{matrix} {x}^{2}+x & x+0 \\ x+0 & 1+0  \end{matrix}\right]$
Given ${A}^{2}=I$
$\Rightarrow \left[\begin{matrix} {x}^{2}+x & x+0 \\ x+0 & 1+0  \end{matrix}\right]=\left[\begin{matrix} 1 & 0 \\ 0 & 1  \end{matrix}\right]$
Equating,we get
${x}^{2}+x=1,x=0$
Put $x=0$ in $A$
$A=\left[\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix}\right]$
We have ${A}^{2}=I$
Pre-multiply ${A}^{-1}$ both sides,we get
${A}^{-1}{A}^{2}={A}^{-1}I$
$\Rightarrow A={A}^{-1}$
Hence,${A}^{-1}=\left[\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix}\right]$
Multiple choice maths fun with numbers some special sequences triangular numbers properties and patterns of perfect squares

$\cfrac { { \left( 963+476 \right)  }^{ 2 }+{ \left( 963-476 \right)  }^{ 2 } }{ \left( 973\times 963+476\times 476 \right)  } =$?

  1. $1449$
  2. $497$
  3. $2$
  4. $4$

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

Given Exp.$=\cfrac { { \left( a+b \right)  }^{ 2 }+{ \left( a-b \right)  }^{ 2 } }{ \left( { a }^{ 2 }+{ b }^{ 2 } \right)  } =\cfrac { 2\left( { a }^{ 2 }+{ b }^{ 2 } \right)  }{ \left( { a }^{ 2 }+{ b }^{ 2 } \right)  } =2$

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 $\left[\begin{array}{ll}
\mathrm{x} & \mathrm{y}^{3}\
2 & 0
\end{array}\right]=\left[\begin{array}{ll}
1 & 8\
2 & 0
\end{array}\right]$, 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 mathematics and statistics ellipse standard equation of ellipse introduction to ellipse standard equation of an ellipse

Find the  Lactus Rectum of  $\displaystyle 9y^{2}-4x^{2}=36$ 

  1. $ 9$
  2. $6$
  3. $11$
  4. $15$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\displaystyle \frac{y^{2}}{4}-\frac{x^{2}}{9}= 1.$ 
Here the coefficient of $\displaystyle y^{2}$ is + ive and that of $\displaystyle x^{2}$ is -ive and hence it represents a hyperbola whose transerse axis is vertical, i.e.
$\displaystyle a^{2}=4, b^{2}=9.$
$\displaystyle b^{2}= a^{2}\left ( e^{2}-1 \right )$
or $\displaystyle \frac{9}{4}+1=e^{2}\therefore e= \frac{\sqrt{13}}{2}$ 
Foci lie on y-axis $\displaystyle \left ( 0, \pm ae \right )$ i.e $\displaystyle \left ( 0, \pm ae \sqrt{13} \right )$ 
$\displaystyle L.R.= \frac{2b^{2}}{a}= 2.\frac{9}{2}= 9$

Multiple choice business mathematics and statistics insurance and annuity amount of an annuity annuities financial mathematics

The Future amount of annuity, $(M)$, can be found by

  1. $ M=\dfrac{A}{r} \times \left[(1+r)^{n}+1\right] $
  2. $ M=\dfrac{r}{A} \times \left[(1+r)^{n}-1\right] $
  3. $ M=\dfrac{A}{r} \times \left[(1+r)^{n}-1\right] $
  4. $ M=\dfrac{r}{A} \times \left[(1+r)^{n}+1\right] $
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The future amount of annuity (M) can be founded by the formula

$M=\dfrac { A }{ r } \times \left[ { \left( 1+r \right)  }^{ n }-1 \right] $
Where $A$ is amount and $r$ is rate and $n$ is duration

Multiple choice business mathematics and statistics insurance and annuity amount of an annuity annuities financial mathematics

Present value of annuity, $(V)$, can be found by

  1. $ V=\dfrac{r}{A} \times \left[1-(1+r)^{(-n)}\right] $
  2. $ V=\dfrac{A}{r} \times \left[1-(1+r)^{(-n)}\right] $
  3. $ V=\dfrac{A}{r} \times \left[1-(1+r)^{(n)}\right] $
  4. $ V=\dfrac{r}{A} \times \left[1-(1+r)^{(n)}\right] $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The present value annuity factor is used for simplifying the process of calculating the present value of an annuity. A table is used to find the present value per dollar of cash flows based on the number of periods and rate per period. Once the value per dollar of cash flows is found, the actual periodic cash flows can be multiplied by the per dollar amount to find the present value of the annuity.
$v= \frac{A}{r} \times [1-(1+r)^{(-n)}]$
where , A =annuity , r =rate per period , n= number of periods