Mathematics · Quantitative Aptitude

Algebra and Arithmetic

406 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 inverse of a matrix and linear equations properties of inverse matrix properties of inverses of matrices applications of matrices and determinants

If $A$ satisfies the equation $x^3-5x^2+4x+kI=0,$ then $A^{-1}$ exists if

  1. $k\neq -1$
  2. $k\neq 0$
  3. $k\neq 1$

  4. none of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Since A satisfies the given equation, therefore ${ A }^{ 3 }-5{ A }^{ 2 }+4A+kI=0$

${ A }^{ -1 }$ exits if $k\neq 0$ since if $k=0$ then the above equation gives $A=0$ and in that case ${ A }^{ -1 }$ wont exist.

Multiple choice maths inverse of a matrix and linear equations properties of inverse matrix properties of inverses of matrices applications of matrices and determinants

The value of $(\mathrm{A}$dj $\mathrm{A})^{-1}$ is equal to 

  1. $\mathrm{A}$dj $(\mathrm{A}^{-1})$
  2. $\mathrm{A}$dj $[-\mathrm{A}]$
  3. $(\mathrm{A}$dj$\mathrm{A})^{\mathrm{T}}$
  4. $\mathrm{A}$dj $(\mathrm{A}^{\mathrm{T}})$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The property (Adj A)^-1 = Adj(A^-1) is a standard identity in matrix algebra.

Multiple choice maths inverse of a matrix and linear equations properties of inverse matrix properties of inverses of matrices applications of matrices and determinants

$\mathrm{A}\mathrm{B}\mathrm{A^{-1}}$ $=\mathrm{X}$ then $\mathrm{B}^{2}=$

  1. $\mathrm{x}^{2}$
  2. $\mathrm{A}\mathrm{x}\mathrm{A}^{-1}$
  3. $\mathrm{A}\mathrm{x}^{2}\mathrm{A}^{-1}$
  4. $\mathrm{A}^{-1}\mathrm{x}^{2}\mathrm{A}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$ABA^{-1}=X$
$\Rightarrow (ABA^{-1})(ABA^{-1})=XX=X^2$
$\Rightarrow ABAA^{-1}BA^{-1} =X^2$
$\Rightarrow ABBA^{-1} =X^2[\because AA^{-1}=I]$
$\Rightarrow AB^2A^{-1} =X^2$
Pre and post multiplying both sides with $A^{-1}$ and $A$ respectively we get,
$\Rightarrow B^2 =A^{-1}X^2A$

Multiple choice business maths applications of matrices and determinants non-homogeneous linear equations system of simultaneous equations matrices

Let $a,\ b,\ c$ be any real numbers. Suppose that there are real numbers $x, y, z$ not all zero such that $x=cy+bz,\ y=az+cx$ and $z=bx+ay$. Then $a^{2}+b^{2}+c^{2}+2abc$ is equal to 

  1. 0

  2. 1

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

For $x,y,z$ not to be simultaneously zero

determinant of the coefficients should be zero.
$\left| \begin{matrix}-1 & c & b\ c& -1 & a \ b & a & -1 \end{matrix}\right|=0$
$\Rightarrow a^2+b^2+c^2+2abc = 1$

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

$\sqrt{(a - b)^2} + \sqrt{(b - a)^2}$ is

  1. Always zero

  2. Never zero

  3. Positive if and only if a > b
  4. Positive only if a $\ne$ b
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$\sqrt{(a - b)^2} + \sqrt{(b - a)^2}$
$= |a - b| + |b - a|$

Now, If $a > b$
$= a - b + a - b$
$= 2a - 2b$...+ ve

If $b > a$
$= b - a + b - a$
$= 2b - 2a$...+ ve
Therefore, if $a \ne b$ then the given equation is always positive.
Hence, option 'D' is correct.

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

$\sqrt{1\, +\, \sqrt{1\, +\, \sqrt{1\, +\, ..........}}}\, =\, ..........$   

  1. Equals 1

  2. Lies between 0 and 1

  3. Lies between 1 and 2

  4. Is greater than 2
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let 


$x=\sqrt{1+\sqrt{1-----}}$

Squaring both sides

$x^2=1+\sqrt{1+\sqrt{1+\sqrt{1------}}}$

$x^2=1+x$                $(\because x=\sqrt{1+\sqrt{1------}})$

$x^2-x-1=0$

finding roots, we get

$\dfrac{1\pm\sqrt{1+4}}{2}$

$=\dfrac{1\pm\sqrt{5}}{2}$

$=-0.615$ and $1.615$

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 squares and square roots approximation of square roots estimating square roots square root of non perfect squares

$\sqrt{1\, +\, \sqrt{1\, +\, \sqrt{1\, +\, .....}}}$ = ........

  1. Equals $1$
  2. Lies between $0$ and $1$
  3. Lies between $1$ and $2$
  4. Is greater than $2$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let $x=\sqrt { 1+\sqrt { 1+\sqrt { 1+....... }  }  }$

$\therefore { x }^{ 2 }=1+\sqrt { 1+\sqrt { 1+\sqrt { 1+....... }  }  }$
$\Longrightarrow { x }^{ 2 }=1+x$
$\Longrightarrow { x }^{ 2 }-x-1=0$
$\Longrightarrow x=\cfrac { 1\pm \sqrt { 1+4 }  }{ 2 }$
$ \Longrightarrow x=\cfrac { 1\pm \sqrt { 5 }  }{ 2 }$
$ \Longrightarrow \cfrac { 1\pm 2.236 }{ 2 } =\left( -0.618, 1.618 \right) $
We reject the negative value because, from the given expression,

$x$ is positive.
So, $x=1.618$ approximately.
$\therefore 1<x<2$

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, $a>b>c$.
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