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 business maths linear programming problems structure of linear programming model linear programming problem operations research

Let $a _1,a _2....,a _n$ be a non negative real number such that $a _1+a _2....+a _n=m$ and let $S=\underset{i<j}\sum a _ia _j$, then

  1. $S\leq \dfrac {m^2}2$
  2. $S> \dfrac {m^2}4$
  3. $S< \dfrac {m}2$
  4. $S> \dfrac {m^2}2$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For non-negative real numbers, the sum of products S = sum(a_i * a_j) for i < j is related to the square of the sum. Since (sum a_i)^2 = sum(a_i^2) + 2*S, and sum(a_i^2) >= 0, then 2*S <= m^2, so S <= m^2/2.

Multiple choice business maths inverse of a matrix and linear equations non singular matrix singular & non-singular matrix determinants

If $\left[ {\begin{array}{*{20}{c}}1&{ - 1}&x\1&x&1\x&{ - 1}&1\end{array}} \right]$ has no inverse, then the real value of $x$ is 

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

$A=\begin{vmatrix} 1 & -1 & x \\ 1 & x & 1 \\ x & -1 & 1 \end{vmatrix}has\quad no\quad inverse\quad x$

$if\,\left| A \right| = 0$

$\begin{vmatrix} 1 & -1 & x \\ 1 & x & 1 \\ x & -1 & 1 \end{vmatrix}=0$

$1\left( {x + 1} \right) - 1\left( { - 1 + x} \right) + x\left( { - 1 - {x^2}} \right) = 0$

$x + 1 + 1 - x - x - {x^3} = 0$

$ - {x^3} - x + 2 = 0$

${x^3} + x - 2 = 0$

$\left( {x - 1} \right)\left( {{x^2} + x + 2} \right) = 0$

$x = 1\,is\,real\,value\,$

Multiple choice maths ratio, proportion and unitary method more on proportion terms related to proportion proportion

Solve it:-
$a:b$=$5:8$ +$b:c$=$16:25$
Find  $a:c.$

  1. $6:625$
  2. $6:825$
  3. $9:625$
  4. $6:25$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\dfrac{a}{b}=\dfrac58+\dfrac bc=\dfrac{16}{25}$


Hence, $a=\dfrac{16}{25}b\\dfrac bc=\dfrac{16}{25}-\dfrac58\Rightarrow c=\dfrac{b*25*8}{3}$

Now, $a:c=\dfrac ac=\dfrac{\dfrac{16}{25}b}{\dfrac{b\times25\times8}{3}}=\dfrac{6}{625}$

Multiple choice maths congruency of triangles triangle inequality inequalities in triangle inequalities in triangles

If $Re(z)$ is a positive integer, then value of the $|1+z+...+z^n|$ cannot be less than

  1. $|z^n| - \displaystyle\frac{1}{|z|}$
  2. $|z^n| + \displaystyle\frac{1}{|z|}$
  3. $n|z|^n$
  4. $n|z|^n + 1$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$|1+z+z^{2}+..z^{n}|\leq 1+|z|+|z|^{2}+..|z|^{n}$
$\leq \dfrac{|z|^{n+1}-1}{|z|-1}$

$\leq\dfrac{(|z|.|z|^{n}-1)}{|z|-1}$

$\leq\dfrac{|z|}{|z|-1}.[|z|^{n}-\dfrac{1}{|z|}]$
Hence 
It cannot be less than $[|z|^{n}-\dfrac{1}{|z|}]$.

Multiple choice maths application of derivatives - iii second derivative test maxima and minima application of derivatives

Let $g(x) =||x + 2| - 3|$. If a denotes the number of relative minima, $b$ denotes the number of relative maxima and $c$ denotes the product of the zeros. Then the value of $(a + 2b - c)$ is

  1. $-1$
  2. $-2$
  3. $8$
  4. $9$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

This function has $a = 2$ relative minima at the x-intercepts, $(-5, 0)$ and $(1, 0), b = 1$ relative minima at $(-2, 3)$ and the product of the zeros is $c = (-5)(1) = - 5$. Thus $a + 2b - c = 9$.

Multiple choice maths squares and square roots finding the square of a number finding square of a number patterns in square numbers

If $x^{2}+2(a-1)x+a+5=0$ has real roots to the interval $(1,3)$, then complete set of value of $'a'$ is

  1. $\left(-\infty,-\dfrac {8}{7}\right)$
  2. $(4,\infty)$
  3. $\left(-\infty,-\dfrac {48}{3}\right)$
  4. $a\ \epsilon \left(-\dfrac {8}{7},-1 \right]$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Given $f(x)=x^{2}+2(a-1)x+a+5=0$ has real roots $\implies b^{2}-4{a}{c}\geq 0 $ in $a{x^{2}}+b{x}+c=0$

                     $\implies (a-1)^{2}-(a+5)\geq 0\implies a\in (-\infty,-1]\cup[4,\infty)\cdots\cdots(1)$
 the roots lies in $(1,3)\implies f(1).f(3)> 0\implies (3{a}+4)(7{a}+8)>0$
                              $\implies a\in (-\infty,-\dfrac{4}{3})\cup(-\dfrac{8}{7},\infty)\cdots\cdots(2)$
from $(1),(2)$    $a\in\bigg(-\dfrac{8}{7},-1\bigg]$

Multiple choice maths squares and square roots finding the square of a number finding square of a number patterns in square numbers

If ${x}^{3}+\cfrac{1}{{x}^{3}}=110$, then $x+\cfrac{1}{x}$ is equal to

  1. $5$
  2. $10$
  3. $15$

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

$\left(x+\dfrac{1}{x}\right)^3=\left(x^3+\dfrac{1}{x^3}\right)+3\left(x+\dfrac{1}{x}\right)$


$\Rightarrow$  $\left(x+\dfrac{1}{x}\right)^3=110+3\left(x+\dfrac{1}{x}\right)$

Let $y=x+\dfrac{1}{x}$

$\Rightarrow$  $y^3=110+3y$

$\Rightarrow$  $y^3-3y-110=0$

$\Rightarrow$  $y^3-5y^2+5y^2-3y-110=0$

$\Rightarrow$  $y^2(y-5)+5y^2-25y+22y-110=0$

$\Rightarrow$  $y^2(y-5)+5y(y-5)+22(y-5)=0$

$\Rightarrow$  $(y-5)(y^2+5y+22)=0$

$\Rightarrow$  $y=5$ or $y=\dfrac{-5\pm3\sqrt{7}i}{2}$

$\therefore$  $x+\dfrac{1}{x}=5$

Multiple choice maths equivalent fractions comparing and ordering fractions comparing fractions fractions and its related operations

If $ \dfrac {1}{a} < \dfrac {1}{b} ,$ then :

  1. $|a| > |b| $
  2. $ a < b$
  3. $ a > b $

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

$\begin{array}{l} We\, have \ \frac { 1 }{ a } <\frac { 1 }{ b }  \ by\, reciprocal\, both\, side\, we\; get \ \frac { a }{ 1 } >\frac { b }{ 1 }  \ \therefore a>b \ Hence,\, the\, option\, C\, is\, the\, correct\, answer. \end{array}$

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.