Quantitative Aptitude

Algebra

3,715 Questions
All Polynomial Roots and Coefficients 2451 Zero-Product Root Finding 667 Linear System Manipulation 162 Polynomial Theorems and Reduction 85 Algebraic Identity Simplification 82 Surds and Indices 70 Modulus and Absolute Value 54 Algebraic Expression Substitution 46 Algebraic Fraction Simplification 33 Logarithmic Equations 25 Repeated Series Sum 12 Linear Equation Word Problems 9

Algebra Questions

Multiple choice maths equation reducing simple equations to simpler form solving linear equations solution of a linear equation in one variable

Solve the following linear equations. If $\cfrac{x-5}{3} = \cfrac{x-3}{5}$, then $x  $is equal to

  1. $8$
  2. $6$
  3. $2$
  4. $3$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Given, $\cfrac { x - 5 }{ 3 }  =\cfrac { x-3 }{ 5 }$
Taking L.C.M., we get
$ \cfrac { 5x-25 }{ 15 }  = \cfrac { 3x - 9 }{ 15 } $
$5x - 25 = 3x -9$
Take $x$ terms on one sides and constants on another side, we get
$2x = 16$
$x = 8$
Multiple choice maths equation reducing simple equations to simpler form solving linear equations solution of a linear equation in one variable

Solve the following equations: $\cfrac{9x}{7-6x}=15$

  1. $x = \cfrac{25}{72}$
  2. $x = \cfrac{35}{33}$
  3. $x = \cfrac{45}{39}$
  4. $x = \cfrac{22}{45}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given, $\dfrac{9x}{7-6x}=15$
$9x=105-90x$

Add $90x$ on both the sides, we get
$9x+90x=105-90x+90x $
$99x=105$
$\therefore x=\dfrac{105}{99}=\dfrac{35}{33}$

Multiple choice maths calculations and mental strategies 1 equations from statements forming equations from statements writing mathematical statements

If a $\in { 1,2,3,4 } ,$ then number of equations of the form $x ^ { 2 } + a x + 1 = 0$ having real roots is

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

For real roots, the discriminant D = a^2 - 4 >= 0. This implies a^2 >= 4. Given a in {1, 2, 3, 4}, the values satisfying this are a=2, 3, 4. There are 3 such values.

Multiple choice maths factorization-1 solution of a quadratic equation by completing the square solving a quadratic equation solving a quadratic equation: completing the square

Solve the following quadratic equation by completing the square: $ x^2+(\sqrt{3}+1)x+\sqrt{3}=0$

  1. $\left \{ \sqrt{3}, 1\right \}$
  2. $\left \{ \sqrt{3}, 2\right \}$
  3. $\left \{ \sqrt{3}, 3\right \}$

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

Given equation is $x^2-(\sqrt{3}+1)x+\sqrt{3}=0$


$\Rightarrow$  $x^2-(\sqrt{3}+1)x=-\sqrt{3}$


Now, adding $\left(\dfrac{\sqrt{3}+1}{2}\right)^2$ on both sides,

$\Rightarrow$  $x^2-2\times x\times \left(\dfrac{\sqrt{3}+1}{2}\right)+\left(\dfrac{\sqrt{3}+1}{2}\right)^2=-\sqrt{3}+\left(\dfrac{\sqrt{3}+1}{2}\right)^2$

$\Rightarrow$  $\left[x-\dfrac{(\sqrt{3}+1)}{2}\right]^2=\left[\dfrac{(\sqrt{3}+1)^2}{4}-\sqrt{3}\right]$

$\Rightarrow$  $\left[x-\dfrac{(\sqrt{3}+1)}{2}\right]^2=\left[\dfrac{3+2\sqrt{3}+1-4\sqrt{3}}{4}\right]$

$\Rightarrow$  $\left[x-\dfrac{(\sqrt{3}+1)}{2}\right]^2=\left[\dfrac{3-2\sqrt{3}+1}{4}\right]$

$\Rightarrow$  $\left[x-\dfrac{(\sqrt{3}+1)}{2}\right]^2=\left[\dfrac{\sqrt{3}-1}{2}\right]^2$

Taking square root on both sides, 
  
$\Rightarrow$  $\left[x-\dfrac{(\sqrt{3}+1)}{2}\right]=\pm\dfrac{\sqrt{3}-1}{2}$

$\Rightarrow$  $\left[x-\dfrac{(\sqrt{3}+1)}{2}\right]=\dfrac{\sqrt{3}-1}{2}$ 
and $\left[x-\dfrac{(\sqrt{3}+1)}{2}\right]=-\dfrac{\sqrt{3}-1}{2}$

$\therefore$  $x=\dfrac{2\sqrt{3}}{2}$  and $ x=\dfrac{2}{2}$

$\therefore$  $x=\sqrt{3}$ and $x=1$

Multiple choice maths remainder and factor theorems linear and synthetic method of division factorising expressions using factor theorem division algorithm for polynomials

What is the degree of the remainder atmost, when a fourth degree polynomial is divided by a quadratic polynomial?

  1. $2$
  2. $0$
  3. $4$
  4. $1$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Here $f(x)$ represent dividend and $g(x)$ represent division

$g(x)=$ quadratic polynomial $=ax^2+bx+c$

$\therefore deg(g(x))=2$, $deg(f(x))=4$

quotient $q(x)$ is of degree $=2$ $(=4-2)$

Remainder $R(x)=$ degree $1$ or less than $1$.
Multiple choice maths power and exponent power of powers laws of exponents and powers law of indices

The sum of roots of the equation $(1.25)^{1-x^2} = (0.4096)^{1+x}$

  1. Infinite
  2. $1$
  3. $2$
  4. $4$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
$\left ( 1.25 \right )^{1-x^{2}}=\left ( 0.4096 \right )^{1+x}$

$\left ( \dfrac{125}{100} \right )^{1-x^{2}}=\left ( \dfrac{4096}{10000} \right )^{1+x}$

$\left ( \dfrac{5}{4} \right )^{1-x^{2}}=\left ( \left ( \dfrac{8}{10} \right )^{4} \right )^{1+x}$

$\left ( \dfrac{5}{4} \right )^{1-x^{2}}=\left ( \dfrac{4}{5} \right )^{4+4x}$

$\left ( \dfrac{5}{4} \right )^{1-x^{2}}=\left ( \dfrac{5}{4} \right )^{-4-4x}$

$\Rightarrow 1-x^{2}=-4-4x$

$x^{2}-4x-5=0$

$x^{2}-5x+x-5=0$
$x(x-5)+1(x-5)=0$
$(x+1)(x-5)=0$
$x=-1,5$

Therefore, Sum of the roots of equation  is $4$
Multiple choice de moivre’s theorem and its applications demoivre's theorem complex numbers maths

Construct an equation whose roots are $n^{th}$ powers of the roots of the equation $\displaystyle x^{2}-2x\cos \theta +1= 0.$

  1. $\displaystyle x^{2}-2n\cos n\theta x+1= 0$
  2. $\displaystyle x^{2}-2n\cos \theta x+1= 0$
  3. $\displaystyle x^{2}-2\cos n\theta x+1= 0$
  4. $\displaystyle x^{2}-2\cos ^{n}\theta x+1= 0$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

We know, $\displaystyle \alpha = \cos \theta +i\sin \theta , \beta = \cos \theta -i\sin \theta $
$\displaystyle \alpha ^{n}= \cos n\theta +i\sin n\theta ,$
$\displaystyle \beta ^{n}= \cos n\theta -i\sin n\theta $
$\displaystyle S= 2\cos n\theta , P= 1 \therefore x^{2}-Sx+P= 0$
or $\displaystyle x^{2}-2\cos n\theta x+1= 0$ is the required equation.

Ans: C