Quantitative Aptitude · Mathematics

Algebraic Simplification

146 Questions

Algebraic simplification involves reducing mathematical expressions to their simplest forms using exponent rules and identities. Test takers must manipulate equations to find rapid solutions. This skill is frequently assessed in SSC, banking, and railway quantitative aptitude sections.

Exponent rulesSurd operationsAlgebraic identitiesTrigonometric simplificationLogical expressions

Algebraic Simplification Questions

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

Simplify: $3x[x^2+1]-[2x(x^2+x-1)+1]-x^2$

  1. $x^3-3x^2+x+1$
  2. $x^3-3x^2+5x-1$
  3. $x^3+x^2-5x+1$
  4. $x^3+3x^2+5x-1$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$3x[x^2+1]-[2x(x^2+x-1)+1]-x^2$
We need to follow BODMAS rule.
=> Brackets (parts of a calculation inside brackets always come first).
=> Orders (numbers involving powers or square roots).
=> Division.
=> Multiplication.
=> Addition.
=> Subtraction.
$3x[x^2+1]-[2x(x^2+x-1)+1]-x^2$
$=$ $3x.x^2+3x.1-[2x.x^2+2x.x-2x.1+1]-x^2$
$=$ $3x^3+3x-2x^3-2x^2+2x-1-x^2$
$=$ $x^3-3x^2+5x-1$

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

Simplify using BODMAS rule: $[((100+x)x^4)\div x^2]\times 2 - (x+x^2-1)$.

  1. $x^3+199x^2-x+1$
  2. $2x^3+199x^2-x+1$
  3. $2x^3-199x^2-x+1$
  4. $2x^3+199x^2-x-1$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$[((100+x)x^4)\div x^2]\times 2 - (x+x^2-1)$
We need to follow BODMAS rule.
=> Brackets (parts of a calculation inside brackets always come first).
=> Orders (numbers involving powers or square roots).
=> Division.
=> Multiplication.
=> Addition.
=> Subtraction.
$=$ $[((100+x)x^4)\div x^2]\times 2 - (x+x^2-1)$
$=$ $\dfrac{(100+x)(x^4)}{x^2}\times 2-x-x^2+1$
$=$ $200x^2+2x^3-x-x^2+1$
$=$ $2x^3+199x^2-x+1$

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

Simplify the expression: $4x^3[(3x-x^2)-1]+(x^2)[x+1]$.

  1. $-4x^5-12x^4-3x^3+x^2$
  2. $-4x^5+12x^4+3x^3+x^2$
  3. $-4x^5+12x^4-3x^3-x^2$
  4. $-4x^5+12x^4-3x^3+x^2$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$4x^3[(3x-x^2)-1]+(x^2)[x+1]$
We need to follow BODMAS rule.
=> Brackets (parts of a calculation inside brackets always come first).
=> Orders (numbers involving powers or square roots).
=> Division.
=> Multiplication.
=> Addition.
=> Subtraction.
$=$ $4x^3[(3x-x^2)-1]+(x^2)[x+1]$
$=$ $4x^3[3x-x^2-1]+x^3+x^2$
$=$ $12x^4-4x^5-4x^3+x^3+x^2$
$=$ $-4x^5+12x^4-3x^3+x^2$

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

Use the BODMAS rule to simplify the expression: 

  1. $-x^4-4x^3-x^2+xy^2$
  2. $-x^4+4x^2-x^2+xy^2$
  3. $-x^4+4x^3+x^2+xy^2$
  4. $-x^4+4x^3-x^2+xy^2$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$xy^2+x^3-x[x^2-x][2x]+(x-1)x^2$
We need to follow BODMAS rule.
=> Brackets (parts of a calculation inside brackets always come first).
=> Orders (numbers involving powers or square roots).
=> Division.
=> Multiplication.
=> Addition.
=> Subtraction.
$=$ $xy^2+x^3+[-x^3+x^2]2x+x^3-x^2$
$=$ $xy^2+x^3-x^4+2x^3+x^3-x^2$
$=$ $xy^2+4x^3-x^4-x^2$
$=$ $-x^4+4x^3-x^2+xy^2$

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

Simplify the expression: $x^2\times(x-1)+[(2x+2)\times 4x]-1$

  1. $x^3+7x^2+8x+1$
  2. $x^3-7x^2+8x-1$
  3. $x^3+7x^2+8x-1$
  4. $x^3+7x^2-8x-1$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$x^2\times(x-1)+[(2x+2)\times 4x]-1$
We need to follow BODMAS rule.
=> Brackets (parts of a calculation inside brackets always come first).
=> Orders (numbers involving powers or square roots).
=> Division.
=> Multiplication.
=> Addition.
=> Subtraction.
$=$ $x^2\times(x-1)+[(2x+2)\times 4x]-1$
$=$ $x^3-x^2+8x^2+8x-1$
$=$ $x^3+7x^2+8x-1$

Multiple choice the nth roots of unity complex numbers maths

Simplify the expressions of the sums

$\displaystyle cot^2 \frac{\pi}{2n + 1} + cot^2 \frac{2\pi}{2n + 1} + cot^2 \frac{3\pi}{2n + 1} + ...... + cot^2 \frac{n\pi}{2n + 1}=$

  1. $\displaystyle \frac{n (2n +1)}{3}$
  2. $\displaystyle \frac{n (2n + 1)}{6}$
  3. $\displaystyle \frac{n (2n - 1)}{3}$
  4. $\displaystyle \frac{n (2n - 1)}{6}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

${ (cosx+i sinx) }^{ n }=cos(nx)+i sin(nx)$
Comparing imaginary parts: $sin(nx)= _{ 1 }^{ n }{ C }sinx{ .cos }^{ n-1 }x- _{ 3 }^{ n }{ C }sin^{ 3 }x{ .cos }^{ n-3 }x+.....$
Divide both sides by $sin^{ n }x$
$\dfrac { sin(nx) }{ sin^{ n }x } = _{ 1 }^{ n }{ C }cot^{ n-1 }x- _{ 3 }^{ n }{ C }cot^{ n-3 }x+...$
Replace $n$ with $2n+1:$ $\dfrac { sin((2n+1)x) }{ sin^{ 2n+1 }x } = _{ 1 }^{ 2n+1 }{ C }cot^{ 2n }x- _{ 3 }^{ 2n+1 }{ C }cot^{ 2n-2 }x+...$
$ = _{ 1 }^{ 2n+1 }{ C }(cot^{ 2 }x)^{ n }- _{ 3 }^{ 2n+1 }{ C }(cot^{ 2 }x)^{ n-1 }+.... ----1)$
So $cot^{ 2 }x _{ i }$ are the roots of the polynomial for LHS=0 where $cot^{ 2 }x _{ i }=cot^{ 2 }(\dfrac { \pi i }{ 2n+1 } )$ where i=1,2,3,...n.
Sum of the roots is the ratio of the first two coefficients on the RHS.
Thus $\sum _{ i=1 }^{ n }{ cot^{ 2 }(\dfrac { \pi i }{ 2n+1 } )= } -\dfrac { - _{ 3 }^{ 2n+1 }{ C } }{ _{ 1 }^{ 2n+1 }{ C } } =\dfrac { (2n+1)2n(2n-1) }{ 6 } .\dfrac { 1 }{ 2n+1 } =\dfrac { n(2n-1) }{ 3 } $
Hence, (c) is correct.

Multiple choice reciprocal equations theory of equations maths

Simplify the reciprocal equation $\dfrac{3}{12}=\dfrac{3}{2x}$

  1. $0$
  2. $3$
  3. $6$
  4. $1$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given equation is $\dfrac {3}{12}=\dfrac {3}{2x}$

$\Rightarrow 6x=36$
$\Rightarrow x=\dfrac {36}{6}$
$\Rightarrow x=6$
Therefore, the reciprocal of the given function is $6$.

Multiple choice reciprocal equations theory of equations maths

Simplify $\sqrt { 1+{ \left( \cfrac { { x }^{ 4 } }{ -2{ x }^{ 2 } }  \right)  }^{ 2 } } $

  1. $\cfrac { { x }^{ 4 }+1 }{ 2{ x }^{ 2 } } $
  2. $\cfrac{\sqrt{{x}^{2}+1}}{2}$
  3. $\cfrac{{x}^{4}+2{x}^{2}-1}{2{x}^{2}}$
  4. $\cfrac { { x }^{ 4 }-1 }{ 2{ x }^{ 4 } } $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\sqrt{1+(\dfrac{x^4}{-2x^2})^2}$


$\Rightarrow \sqrt{1+\dfrac{x^8}{4x^4}}$

$\Rightarrow \sqrt{\dfrac{4x^4+x^8}{4x^4}}$

$\Rightarrow \dfrac{\sqrt{4x^4+x^8}}{2x^2}$

$\Rightarrow\cfrac{\sqrt{{x}^{2}+1}}{2}$

Multiple choice maths part number dividing fractions division of a fractions division of a fraction

Simplify $35\times 6\dfrac{1}{14}$(approximately)$=$

  1. $220.5$
  2. $220$
  3. $212$
  4. $231$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
$6 \cfrac{1}{14} = \cfrac{14 \times 6 + 1}{14} = \cfrac{85}{14}$
$\therefore \; 35 \times 6\cfrac{1}{14} = 35 \times \cfrac{85}{14} = \cfrac{5 \times 85}{2} = \cfrac{425}{2} = 212.5 \approx 212$
Hence, 212 is the correct answer.
Multiple choice maths part number dividing fractions division of a fractions division of a fraction

Simplify : $\displaystyle \frac{2+2\times 2}{2\div 2\times 2}\div \frac{\frac{1}{2}\div \frac{1}{2} \, \text{of} \, \frac{1}{2}}{\frac{1}{2}+\frac{1}{2} \, \text{of} \, \frac{1}{2}}$

  1. $1$
  2. $2$
  3. $\displaystyle 1\frac{1}{3}$
  4. $\displaystyle 1\frac{1}{8}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Given exp =$\displaystyle =\frac{2+4}{1\times 4}\div \frac{\frac{1}{2}\div \frac{1}{4}}{\frac{1}{2}+\frac{1}{4}}$
$\displaystyle =\frac{6}{2}\div\frac{\frac{1}{2}\times \frac{4}{1}}{\frac{3}{4}}=3\div \frac{2}{\frac{3}{4}} $
$\displaystyle =3\div \frac{8}{3}=3\times \frac{3}{8}=\frac{9}{8}=1\frac{1}{8}$

Multiple choice maths part number dividing fractions division of a fractions division of a fraction

Simplify : $\displaystyle \left [ 3\frac{1}{4}\div \left { 1\frac{1}{4}-\frac{1}{2}\left ( 2\frac{1}{2}-\frac{1}{4}-\frac{1}{6} \right ) \right } \right ]\div \left ( \frac{1}{2}of4\frac{1}{3} \right )$

  1. 18

  2. 36

  3. 39

  4. 78
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given exp.
$\displaystyle =\left [ \frac{13}{4}\div \left { \frac{5}{4}-\frac{1}{2}\left ( \frac{5}{2}-\frac{3\, \, \,2}{2} \right ) \right } \right ]\div \left ( \frac{1}{2}of\frac{13}{3} \right )$
$\displaystyle =\left [ \frac{13}{4}\div \left { \frac{5}{4}-\frac{1}{2}\left ( \frac{5}{2}-\frac{1}{12} \right ) \right } \right ]\div \frac{13}{6} $
$=\displaystyle \left [ \frac{13}{4}\div \left { \frac{5}{4}-\frac{1}{2}\times \frac{30-1}{12} \right } \right ]\div \frac{13}{6}$
$\displaystyle =\left [ \frac{13}{4}\div \left { \frac{5}{4}-\frac{29}{24} \right } \right ]\div \frac{13}{6}$
$\displaystyle =\left [ \frac{13}{4}\div \frac{30-29}{24} \right ]\div \frac{13}{6}$
$\displaystyle =\left ( \frac{13}{4} \div \frac{1}{24}\right )\div \frac{13}{4}=\frac{13}{4}\times 24\times \frac{6}{13}=36$