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 negative numbers and integers even and odd numbers sum of numbers odd and even numbers

If $|a|$ denotes the absolute value of an integer, then which of the following are correct?
1.$|ab| = |a| |b|$
2. $|a+b| \le |a|+|b|$
3. $|a-b| \ge| |a| -|b||$
Select the correct answer using the code given below.

  1. 1 and 2 only

  2. 2 and 3 only

  3. 1 and 3 only

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

Given $\left| a \right| $ is the absolute value of an integer,

From the definition,
$\left| a \right| =a$ if $ a\ge 0$,
$=-a\quad $ is $a\le0$
$\therefore$ $\left| ab \right| =\left| a \right| \left| b \right| $ where $a,b$ are real numbers.
We know that from the triangle inequality sum of any two sides is always greater than the third side,
i.e.,$\left| a+b \right| \le \left| a \right| +\left| b \right| $,
We can also prove by considering 
Absolute part of the difference between any two sides is always less than the third side,
$\Longrightarrow \left| a-b \right| \ge \left| \left| a \right| -\left| b \right|  \right| $

Multiple choice maths brackets order operations and algebra using brackets in algebraic expressions order of operations

Evaluate:
$( b - c + d + a ) ( d + a - b + c ) + ( c - d + a + b ) ( b + c + d - a )$

  1. $4 ( a d + b c )$
  2. $2 ( a d + b c )$
  3. $3 ( a d + b c )$
  4. $ ( a d + b c )$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
$(b-c+d+a)(d+a-b+c)+(c-d+a+b)(b+c+d-a)$

$=[(d+a)+(b-c)][(d+a)-(b-c)]+[(b+c)+(a-d)][(b+c)-(a-d)]$

$[\because (a-d)^2=(d-a)^2]$

$=(d+a)^2-(b-c)^2+(b+c)^2-(a-d)^2$

$=(d+a)^2-(d-a)^2+(b+c)^2-(b-c)^2$

$=4ad+4bc$

$=4(ad+bc)$.
Multiple choice maths brackets order operations and algebra using brackets in algebraic expressions order of operations

If ${a}^{2}+{b}^{2}+{c}^{2}-ab-bc-ca=0$, then

  1. $a+b=c$
  2. $b+c=a$
  3. $c+a=b$
  4. $a=b=c$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Given,

$a^2+b^2+c^2-ab-bc-ca=0$

$\Rightarrow 2(a^2+b^2+c^2-ab-bc-ca)=2(0)$

$\Rightarrow 2a^2+2b^2+2c^2-2ab-2bc-2ca=0$

$\Rightarrow (a^2-2ab+b^2)+(b^2-2bc+c^2)+(c^2-2ca+a^2)=0$

$(a-b)^2+(b-c)^2+(c-a)^2=0$

$\Rightarrow a-b=b-c=c-a=0$

$\therefore a=b=c$
Multiple choice maths brackets order operations and algebra using brackets in algebraic expressions order of operations

If $1\le a\le 2$, then $\sqrt { a-2\sqrt { a-1 }  } -\sqrt { a+2\sqrt { a-1 }  } =$.......

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

$\sqrt{a - 2 \sqrt{a - 1}} - \sqrt{a + 2 \sqrt{a - 1}}$

$\Rightarrow \sqrt{(\sqrt{a - 1})^2 - 2 (1) \sqrt{a - 1} + 1^2} - \sqrt{(\sqrt{(a - 1)})^2 + 2 \sqrt{a - 1} + 1^2}$
$\Rightarrow \sqrt{(\sqrt{a - 1} - 1)^2} - \sqrt{(\sqrt{a - 1} + 1)^2}$
$\Rightarrow (\sqrt{a - 1} - 1) - ( \sqrt{a - 1} + 1)$
$= -2$

Multiple choice properties of proportion ratio and proportions ratio and proportion maths

If $\cfrac{{x}^{3}+{x}^{2}+x+1}{{x}^{3}-{x}^{2}+x-1}=\cfrac{{x}^{2}+x+1}{{x}^{2}-x+1}$, then the number of real-value of $x$ satisfying are

  1. 0

  2. 1

  3. 2

  4. 3
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given that


$\dfrac{x^3+x^2+x+1}{x^3-x^2+x+1}=\dfrac{x^2+x+1}{x^2-x+1}$

Performing Componendo Dividendo, we get

$\Rightarrow \dfrac{x^3+x^2+x+1+x^3-x^2+x+1}{x^3+x^2+x+1-x^3+x^2-x+1}=\dfrac{x^2+x+1+x^2-x+1}{x^2+x+1-x^2+x-1}$

$\Rightarrow \dfrac{2x^3+2x}{2x^2+2}=\dfrac{2x^2+2}{2x} $

$\Rightarrow \dfrac{x^3+x}{x^2+1}=\dfrac{x^2+1}{x} $

$\Rightarrow x(x^3+x)=(x^2+1)^2$

$\Rightarrow x^4+x^2=x^4+1+2x^2$

$\Rightarrow x^2=-1$

Hence no real values of $x$ satisfy this equation.

Multiple choice properties of proportion ratio and proportions ratio and proportion maths

If $\cfrac{a+3d}{a+9d}=\cfrac{a+d}{a+5d}=k$, then $k$ is equal to $(a,d> 0)$

  1. $\dfrac{1}{2}$
  2. $2$
  3. $6$
  4. $0$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\dfrac{a+3d}{a+9d}=\dfrac{a+d}{a+5d}$


Applying Componendo-Dividendo

$\Rightarrow \dfrac{a+3d+a+9d}{a+3d-a-9d}=\dfrac{a+d+a+5d}{a+d-a-5d}$ 

$\Rightarrow \dfrac{2a+12d}{-6d}=\dfrac{2a+6d}{-4d}$

$\Rightarrow \dfrac{a+6d}{-3}=\dfrac{a+3d}{-2}$

$\Rightarrow -2a-12d=-3a-9d$

$\Rightarrow a=3d$

Now, given that 

$\Rightarrow k=\dfrac{a+3d}{a+9d}$

$\Rightarrow k=\dfrac{3d+3d}{3d+9d}$

$\Rightarrow k=\dfrac{6d}{12d}$

$\Rightarrow k=\dfrac{1}{2}$

Multiple choice properties of proportion ratio and proportions ratio and proportion maths

If $x=\cfrac { 4ab }{ a+b } $ then value of $\cfrac { x+2a }{ x-2a } +\cfrac { x+2b }{ x-2b } $

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

Given, $x = 4ab/(a+b)$

=> $x = 2a\times 2b / (a+b)$
=> $x/2a = 2b / (a+b)$
Using componendo and dividendo 
=> $x+2a / x-2a = 2b+a+b / 2b-a-b$
=> $x+2a / x-2a = 3b+a / b-a$

Similarly, $x+2b / x-2b = 3a+b / a-b$

Now, adding both,
$[x+2a / x-2a] + [x+2b / x-2b ] = [3b+a / b-a ] + [ 3a +b / a-b ]$
RHS => $[3b+a / b-a ] - [ 3a+b / b-a ]$
=> $3b+a-3a-b / b-a$
=> $2b-2a / b-a$
=> $2(b-a)/b-a$
=> $2$

Hence, LHS $= 2$ 

Multiple choice properties of proportion ratio and proportions ratio and proportion maths

If $\cfrac { { a }^{ 3 }+3a{ b }^{ 2 } }{ 3{ a }^{ 2 }b+{ b }^{ 3 } } =\cfrac { { x }^{ 3 }+3x{ y }^{ 2 } }{ 3{ x }^{ 2 }y+{ y }^{ 3 } } $ then

  1. $bx=ay$
  2. $by=ax$
  3. ${ b }^{ 2 }y={ a }^{ 2 }x$
  4. ${ b }^{ 2 }x={ a }^{ 2 }y$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\dfrac{{{a^3} + 3a{b^2}}}{{3{a^2}b + {b^3}}} = \dfrac{{{x^3} + 3x{y^2}}}{{3{x^2}y + {y^3}}}$


Use of compodendo and Devidendo

$\dfrac{{{a^3} + 3a{b^2} + 3{a^2}b + {b^3}}}{{3{a^2}b + {b^3} - 3{a^2}b + {b^3}}} = \dfrac{{{x^3} + 3x{y^2} + 3{x^2}y + {y^3}}}{{3{x^2}y + {y^3} - 3{x^2}y + {y^3}}}$

$ \Rightarrow \dfrac{{{{\left( {a + b} \right)}^3}}}{{{{(a - b)}^3}}} = \dfrac{{{{\left( {x + y} \right)}^3}}}{{{{(x - y)}^3}}}$

$ \Rightarrow {\left( {\dfrac{{a + b}}{{a + b}}} \right)^3} = {\left( {\dfrac{{x + y}}{{x - y}}} \right)^3}$

$ \Rightarrow \dfrac{{a + b}}{{a - b}} = \dfrac{{x + y}}{{x - y}}$

$ \Rightarrow \left( {a + b} \right)\left( {x + y} \right) = \left( {x + y} \right)\left( {a - b} \right)$

$ \Rightarrow ax{\text{ }} - {\text{ }}ay{\text{ }} + {\text{ }}bx{\text{ }} - {\text{ }}by{\text{ }} = {\text{ }}xa{\text{ }} - {\text{ }}xb{\text{ }} + {\text{ }}ay{\text{ }} - {\text{ }}yb$

$ \Rightarrow bx{\text{ }} + {\text{ }}xb{\text{ }} = {\text{ }}ay{\text{ }} + {\text{ }}ay$

$ \Rightarrow 2bx{\text{ }} = {\text{ }}2ay$

$ \Rightarrow bx{\text{ }} = {\text{ }}ay$


Multiple choice applications of quadratic equations solving (simple) problems word problems based on quadratic equations quadratic equation maths

If the zeroes of the rational expression $ (ax+b)(3x+2)$ are $-\dfrac{2}{3}$ and $ \dfrac{1}{2}$, then $ a+b=$

  1. $4$
  2. $0$
  3. $-b$

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

$(ax+b)(3x+2)=0$

$\Rightarrow$  $3ax^2+2ax+3bx+2b$
$\Rightarrow$  $3ax^2+(2a+3b)x+2b=0$
It is given that $\dfrac{-2}{3}$ and $\dfrac{1}{2}$ are zeros of the equation $3ax^2+(2a+3b)x+2b=0$
Here, $A=3a,\,B=2a+3b,\,C=2b$
Let $\alpha=\dfrac{-2}{3}$ and $\beta=\dfrac{1}{2}$
We know,
$\Rightarrow$  $\alpha+\beta=\dfrac{-B}{A}$
$\Rightarrow$  $\dfrac{-2}{3}+\dfrac{1}{2}=\dfrac{-(2a+3b)}{3a}$

$\Rightarrow$  $\dfrac{-4+3}{6}\times 3a=-2a-3b$

$\Rightarrow$  $\dfrac{-a}{2}=-2a-3b$
$\Rightarrow$  $-a=-4a-6b$
$\Rightarrow$  $3a+6b=0$              ----- ( 1 )
Now,
$\Rightarrow$  $\alpha.\beta=\dfrac{CA}{A}$

$\Rightarrow$  $\dfrac{-2}{3}\times\dfrac{1}{2}=\dfrac{2b}{3a}$

$\Rightarrow$  $\dfrac{-1}{3}\times 3a=2b$

$\Rightarrow$  $-a+2b=0$          ----- ( 2 )
Adding equation ( 1 ) and ( 2 ) we get,
$b=0$
Put $b=0$ in ( 2 ) we get,
$a=0$
$\therefore$  $a+b=0+0=0$

Multiple choice the nth roots of unity complex numbers maths

For positive integers ${ n } _{ 1 },{ n } _{ 2 }$ the value of the expression; ${ (1+i) }^{ { n } _{ 1 } }+{ (1+i) }^{ { n } _{ 1 } }+{ (1+i) }^{ { n } _{ 2 } }+{ (1+i) }^{ { n } _{ 2 } }$, where $i=\sqrt { -1 } $, is a real number if :

  1. ${ n } _{ 1 }={ n } _{ 2 }+1$
  2. ${ n } _{ 1 }={ n } _{ 2 }-1$
  3. ${ n } _{ 1 }={ n } _{ 2 }$
  4. ${ n } _{ 1 }>0,{ n } _{ 2 }>0$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The expression (1+i)^n1 + (1+i)^n1 + (1+i)^n2 + (1+i)^n2 simplifies to 2 * ((1+i)^n1 + (1+i)^n2). For this to be real, the imaginary parts must cancel or be zero, which occurs when n1 = n2 + 1.

Multiple choice the nth roots of unity complex numbers maths

If $r$ is non-real and $r=\sqrt [ 5 ]{ 1 } $, then the value of $ { 2 }^{ \left| 1+r+{ r }^{ 2 }+{ r }^{ -2 }-{ r }^{ -1 } \right|  }$ is equal to

  1. $2$
  2. $4$
  3. $8$

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

$\left| 1+r+{ r }^{ 2 }+{ r }^{ -2 }-{ r }^{ -1 } \right| =\left| 1+r+{ r }^{ 2 }+{ r }^{ 3 }-{ r }^{ 4 } \right| $


$[\because { r }^{ 5 }=1\Rightarrow { r }^{ 3 }.{ r }^{ 2 }=1\Rightarrow { r }^{ -2 }={ r }^{ 3 }$ and ${ r }^{ 4 }.r=1\Rightarrow { r }^{ -1 }={ r }^{ 4 }]$


$\displaystyle=\left| 1+r+{ r }^{ 2 }+{ r }^{ 3 }+{ r }^{ 4 }-2{ r }^{ 4 } \right| =\left| \frac { 1-{ r }^{ 5 } }{ 1-r } -2{ r }^{ 4 } \right| =\left| 0-2{ r }^{ 4 } \right| \quad \quad \quad \left[ \because { r }^{ 5 }=1 \right] $

$=2{ \left| r \right|  }^{ 4 }=2\left( 1 \right) =2\quad \quad [\because \left| r \right| =1$ as ${ r }^{ 5 }=1]$

$\therefore { 2 }^{ \left| 1+r+{ r }^{ 2 }+{ r }^{ -2 }-{ r }^{ -1 } \right|  }={ 2 }^{ 2 }=4$