Mathematics · Quantitative Aptitude

Algebra and Arithmetic

333 Questions

Algebra and arithmetic questions cover fundamental mathematical operations, inequalities, and binomial products. They assess core quantitative reasoning skills required for various aptitude tests. Solving these problems strengthens the understanding of number systems and algebraic identities.

Binomial productsLinear inequalitiesLeast common multipleQuadratic equationsInteger properties

Algebra and Arithmetic Questions

Multiple choice maths squares and square roots approximation of square roots estimating square roots square root of non perfect squares

If $x\, \ast\, y\, =\, \sqrt{x^2\, +\, y^2}$, then the value of $(1^{\ast}\, 2\, \sqrt{2})(1^{\ast}\, - 2\, \sqrt{2})$ is:  

    • 7

  2. 0

  3. 2

  4. 9
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$1^{\ast} 2 \sqrt{2}\, =\, \sqrt{(1)^2\, +\, (2 \sqrt{2}^2}\, =\, \sqrt{1\, +\, 8}\, =\, 3$

$1^{\ast} -2 \sqrt{2}\, =\, \sqrt{(1)^2\, +\, (-2 \sqrt)^2}\, =\, \sqrt{1\, +\, 8}\, =\, 3$
$(1\, \ast\, 2 \sqrt{2})(1\, \ast\, -2 \sqrt{2})\, =\, (3)(3)\, =\, 9$

Multiple choice maths algebraic functions, equations and inequalities descartes rule sign of quadratic expression sum and product of the roots of a polynomial equation

Given $P(x) = {x^4} + a{x^3} + b{x^2} + cx + d$ such that $x=0$ is the only real root of $P(x) = 0$. If $P(-1) < P(1) $,then in the interval $[-1,1]$

  1. $P(-1)$ is the minimum and $P(1)$ is the maximum of P
  2. $P(-1)$ is not the minimum but $P(1)$ is the maximum of P
  3. $P(-1)$ is the minimum and $P(1)$ is not the maximum of P
  4. neither $P(-1)$ is the minimum nor $P(1)$ is the maximum of P
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Given:$P\left(x\right)={x}^{4}+a{x}^{3}+b{x}^{2}+cx+d$

${P}^{\prime}\left(x\right)=4{x}^{3}+3a{x}^{2}+2bx+c$

Since, $x=0$ is a solution for ${P}^{\prime}\left(x\right)=0$

$\Rightarrow\,c=0$

So, $P\left(x\right)={x}^{4}+a{x}^{3}+b{x}^{2}+d$   

Also we have $P\left(−1\right)<P\left(1\right)$

$\Rightarrow\,1-a+b+d<1+a+b+d$

$\Rightarrow\,A>0$

Since ${P}^{\prime}\left(x\right)=0,$ only when $x=0$

and $P\left(x\right)$ is differentiable in $\left(−1,1\right)$, we should have the maximum and minimum at the points

$x=−1,0$ and $1$ only.

Also, we have $P\left(−1\right)<P\left(1\right)$

So,Maximum of $P\left(x\right)=Max\left\{P\left(0\right),P\left(1\right)\right\}$ and
Minimum of $P\left(x\right)=Min\left\{P\left(−1\right),P\left(0\right)\right\}$

In the interval $\left[0,1\right]$

${P}^{\prime}{\left(x\right)}=4{x}^{3}+3a{x}^{2}+2bx=x\left(4{x}^{2}+3ax+2b\right)$

Since ${P}^{\prime}{\left(x\right)}$ has only one root $x=0$, then $4{x}^{2}+3ax+2b=0$ has no real roots.

So,${\left(3a\right)}^{2}-32b<0$

$\Rightarrow\,\dfrac{3{a}^{2}}{32}>b$

So,$b>0$

Thus, we have $a>0$ and $b>0$

So,${P}^{\prime}{\left(x\right)}=4{x}^{3}+3a{x}^{2}+2bx>0,$ for $x\in\left(0,1\right)$

Hence, $P\left(x\right)$ is increasing in $\left[0,1\right]$ and $P\left(x\right)$ is decreasing in $\left[−1,0\right]$

Therefore, Maximum of $P\left(x\right)=P\left(1\right)$ and Minimum $P\left(x\right)$ does not occur at $x=−1$ respectively.
Multiple choice maths rational numbers proof of irrationality of numbers proofs of irrationality introduction to irrational numbers

If $a=\sqrt{11}+\sqrt{3}, b =\sqrt{12}+\sqrt{2}, c=\sqrt{6}+\sqrt{4}$, then which of the following holds true ?

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

$a=\sqrt{11}+\sqrt{3}$

$a^{2}=11+3+2\sqrt{33}=14+2\sqrt{33}$

$b=\sqrt{12}+\sqrt{2}$

$b^{2}=14+2\sqrt{24}$

As $\sqrt{33} > \sqrt{24}, a^{2} > b^{2}, a>b$

$c=\sqrt{6}+\sqrt{4}$

$c^{2}=10+2\sqrt{24}$

As $14>10, b^{2} > c^{2}, b>c$

Hence, $a>b>c$.
'

Multiple choice maths applications of matrices and determinants elementary transformations of a matrix multiplicative inverse of a matrix inverse of a matrix

Let A=$\left( {\begin{array}{{20}{c}}{ - 5}&{ - 8}&{ - 7}\3&5&4\2&3&3\end{array}} \right),B = \left( {\begin{array}{{20}{c}}x\y\z\end{array}} \right)$. If AB is scalar $\left( { \ne 0} \right)$ multiple of B, then x+y=

  1. $z$
  2. $-z$
  3. $0$
  4. $2z$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
$A=\begin{pmatrix} -5 & -8 & -7 \\ 3 & 5 & 4 \\ 2 & 3 & 3 \end{pmatrix}\quad B=\begin{pmatrix} x \\ y \\ z \end{pmatrix}$
Given $AB=k\ B$
$AB\Rightarrow \begin{pmatrix} -5 & -8 & -7 \\ 3 & 5 & 4 \\ 2 & 3 & 3 \end{pmatrix}\begin{pmatrix} x \\ y \\ z \end{pmatrix}\quad k=\begin{pmatrix} x \\ y \\ z \end{pmatrix}$
$\Rightarrow \begin{pmatrix} -5x & -8y & -7z \\ 3x & +5y & +4z \\ x2 & +3y & +3z \end{pmatrix}=\quad k\begin{pmatrix} x \\ y \\ z \end{pmatrix}$
On adding all the elements on left and right side 
$\Rightarrow \ (-5x-8y-7z)+(3x+5y+47)+(2x+3y+3z)$
$=k(x+y+z)$
$\Rightarrow \ D=k(x+y+z)$
$k\neq 0$
$\Rightarrow \ x+y+z=0$
$\Rightarrow \ x+y=-z$

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

For real number $a,b,c$ and $d$ , if $a^2+b^2=4$ and $c^2+d^2=1$, then possible value of $ac+bd$ is / are 

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

According to Cauchy-Schwarz inequality:


$(ac+bd)^2\le(a^2+b^2)(c^2+d^2)$      $(\forall a, b, c, d \in \mathbb{R})$

Substituting the values here gives:

$(ac+bd)^2\le4$

$(ac+bd)^\le2$

Therefore, it can take all values given in the options except 3.

Multiple choice geography weather maps atmosphere and its structures what is around us? the atmosphere

In a place the average rainfall was 4 centimeter during Monday to Saturday. Due to heavy rainfall on Sunday the average of the week was increased to 6 cm. The rainfall on Sunday will be?

  1. 17 cm

  2. 18 cm

  3. 19cm

  4. 20cm
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The total rainfall from Monday to Saturday (6 days) at an average of 4 cm is 6 * 4 = 24 cm. The total rainfall for the entire week (7 days) at an average of 6 cm is 7 * 6 = 42 cm. Therefore, the rainfall on Sunday is the difference: 42 - 24 = 18 cm.

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

If $x=1+a+{ a }^{ 2 }+{ a }^{ 3 }+....$ to $\infty \left( \left| a \right| <1 \right) $ and 
$y=1+b+{ b }^{ 2 }+{ b }^{ 3 }+...$ to $\infty \left( \left| b \right| <1 \right) $ then
$1+ab+{ a }^{ 2 }{ b }^{ 2 }+{ a }^{ 3 }{ b }^{ 3 }+...$ to $\infty =\cfrac { xy }{ x+y-1 } $

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

${ x }=\dfrac { 1 }{ 1-a } $ ${ y }=\dfrac { 1 }{ 1-b } $ [summing infinite $G.P's$].
$\therefore a=\dfrac { x-1 }{ x } $, $b=\dfrac { y-1 }{ y } $
$\therefore 1+ab+{ a }^{ 2 }{ b }^{ 2 }+...\infty $
$=\dfrac { 1 }{ 1-ab } =\dfrac { 1 }{ 1-\dfrac { (x-1)(y-1) }{ xy }  } =\dfrac { xy }{ x+y-1 }. $

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

If $x=1+a+a^2+...\infty$ where $|a| <1 $ and $y=1+b+b^2+...\infty$, where $|b| < 1$, then $1+ab+a^2b^2+...\infty =\dfrac{xy}{x+y-1}$.

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

This is the same identity derived in question 466317. The sum of the series 1 + ab + a^2b^2 + ... is indeed xy / (x + y - 1) given the definitions of x and y.

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

${x}^{\cfrac{1}{2}}.{x}^{\cfrac{1}{4}}.{x}^{\cfrac{1}{8}}.{x}^{\cfrac{1}{16}}.....$ to $\infty$

  1. $0$
  2. $1$
  3. $x$
  4. $\infty$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$x^{1/2}.x^{1/4}......\infty =x^{\left ( \frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16} \right )}$ 


$=x\left ( 1/2+1/2^{2}+1/2^{3}+1/2^{4}+.......\infty  \right )$

$=x^{\frac{1}{2}\left ( \frac{1-(1/2)\infty }{1-1/2} \right )}$ 

$=x^{1}$ 

$=x$

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

If $p$ is positive, then the sum to infinity of the series, ${1 \over {1 + p}} - {{1 - p} \over {{{(1 + p)}^2}}} + {{{{(1 - p)}^2}} \over {{{(1 + p)}^3}}} - ......$ is

  1. $1/2$
  2. $3/4$
  3. $1$

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
$a=\dfrac { 1 }{ 1+P } \\ r=-\dfrac { 1-P }{ 1+P } $
Sum to infinity $=\dfrac { a }{ 1-r } \\ =\dfrac { \dfrac { 1 }{ 1+P }  }{ 1+\dfrac { 1-P }{ 1+P }  } \\ =\dfrac { \dfrac { 1 }{ 1+P }  }{ \dfrac { 1+P+1-P }{ 1+P }  } =\dfrac { 1 }{ 2 } $
Multiple choice sum to infinite terms of a gp sequence, progression and series maths

If $0<x,y,a,b<1$,then the sum of infinite terms of the series $\sqrt x (\sqrt a  + \sqrt x ) + \sqrt x (\sqrt {ab}  + \sqrt {xy} ) + \sqrt x (b\sqrt a  + y\sqrt x ) + .......$ is

  1. $\dfrac{{\sqrt {ax} }}{{1 + \sqrt b }} + \dfrac{x}{{1 + \sqrt y }}$
  2. $\dfrac{{\sqrt x }}{{1 + \sqrt b }} + \dfrac{{\sqrt x }}{{1 + \sqrt y }}$
  3. $\dfrac{{\sqrt x }}{{1 - \sqrt b }} + \dfrac{{\sqrt x }}{{1 - \sqrt y }}$
  4. $\dfrac{{\sqrt {ax} }}{{1 - \sqrt b }} + \dfrac{x}{{1 - \sqrt y }}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$0<x,y,a,b<1\ S=\sqrt { x } (\sqrt { a } +\sqrt { x } )+\sqrt { x } (\sqrt { ab } +\sqrt { xy } )+\sqrt { x } (b\sqrt { a } +y\sqrt { x } )\ =\sqrt { x } \left[ (\sqrt { a } +\sqrt { ab } +b\sqrt { a } ....) \right] +(\sqrt { x } +\sqrt { xy } +y\sqrt { x } )\ =\sqrt { x } \left[ (\cfrac { \sqrt { a }  }{ 1-\sqrt { b }  } )+\cfrac { \sqrt { x }  }{ 1-\sqrt { y }  }  \right] \ S=\cfrac { \sqrt { ax }  }{ 1-\sqrt { b }  } +\cfrac { x }{ 1-\sqrt { y }  } $