Mathematics · Quantitative Aptitude

Algebraic Equations and Expressions

152 Questions

Algebraic equations and expressions test mathematical skills in solving for unknown variables. Questions involve linear equations, ratios, and evaluating complex expressions. This topic forms a core component of quantitative aptitude sections in various competitive exams.

Linear equationsEvaluating expressionsRatio proportionsExponential equationsHypergeometric functions

Algebraic Equations and Expressions Questions

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

If $\sqrt{\left(12+\sqrt{12+\sqrt{12+....}}\right)}=x$, then the value of x is ____________.

  1. $3$
  2. $4$
  3. $6$
  4. Greater than $6$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\sqrt{12+\sqrt{12+\sqrt{12+....}}} = x$.........................(1)

$\sqrt{12+x}=x$
$(12+x)=x^2$
$x^2-x-12=0$
$(x-4)(x+3)=0$
$ x=4$ or $x=-3$
since x is a positive number (eq 1)
$x=4$

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

If $x+\cfrac{1}{x}=4$, then ${x}^{4}+\cfrac{1}{{x}^{4}}$ is equal to

  1. $196$
  2. $194$
  3. $192$
  4. $190$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Given,

$x+\dfrac{1}{x}=4$

squaring on both sides, we get,

$x^2+\dfrac{1}{x^2}+2=16$

$x^2+\dfrac{1}{x^2}=16-2=14$

squaring on both sides, we get,

$x^4+\dfrac{1}{x^4}+2=196$

$x^4+\dfrac{1}{x^4}=196-2=194$
Multiple choice maths algebraic functions, equations and inequalities descartes rule sign of quadratic expression sum and product of the roots of a polynomial equation

The equation $x-\dfrac{2}{x-1}=1-\dfrac{2}{x-1}$ has

  1. no root

  2. one root

  3. two equal roots

  4. infinitely many roots
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For equation $x-\dfrac { 2 }{ x-1 } =1-\dfrac { 2 }{ x-1 } $, the term $'x-1'$ is in the denominator. Hence the solution isn't defined. For $x=1$ $\Rightarrow $ $x\neq 1$

We have our equation as $x-\dfrac { 2 }{ x-1 } =1-\dfrac { 2 }{ x-1 } $
cancelling the common term on both sides,we get $x=1$. 
But for well defined solution $x\neq 1$. Hence,this equation has no solution.

Multiple choice maths algebraic functions, equations and inequalities descartes rule sign of quadratic expression sum and product of the roots of a polynomial equation

The equation $\displaystyle x - \frac{5}{x - 2} = 2 - \frac{5}{x - 2}$ has

  1. No real roots

  2. Only one real root

  3. Two real roots

  4. Infinitely many roots
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$x-\dfrac { 5 }{ x-2 } =2-\dfrac { 5 }{ x-2 } $       ...(1)
Equation (1) is valid when $x\neq 2$
Rewriting eq. (1), we get $x=2$
But $x\neq 2$
Therefore, number of roots satisfying eq. (1) are zero.

Multiple choice maths rational numbers proof of irrationality of numbers proofs of irrationality introduction to irrational numbers

Find x if $\dfrac{\sqrt{3x+1}+\sqrt{3x-6}}{\sqrt{3x+1}-\sqrt{3x-6}}=7$.

  1. $2$
  2. $5$
  3. $3$
  4. $7$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
$\dfrac { \sqrt { 3x+1 } +\sqrt { 3x-6 }  }{ \sqrt { 3x+1 } -\sqrt { 3x-6 }  } =7$

Rotational give :-

$\dfrac { \left( 3x+1 \right) +\left( 3x-6 \right) +2\sqrt { \left( 3x+1 \right) \left( 3x-6 \right)  }  }{ \left( 3x+1 \right) -\left( 3x-6 \right)  } =7$

$\Rightarrow 6x-5+2\sqrt { \left( 3x+1 \right) \left( 3x-6 \right)  } =49$

$\Rightarrow 2\sqrt { \left( 3x+1 \right) \left( 3x-6 \right)  } =-6x+54$

$\Rightarrow \sqrt { \left( 3x+1 \right) \left( 3x-6 \right)  } =-3x+27$
which gives, $x=5$
Multiple choice sum to infinite terms of a gp sequence, progression and series maths

The value of x that satisfies the relation 
$x=1-x+{ x }^{ 2 }-{ x }^{ 3 }+{ x }^{ 4 }-{ x }^{ 5 }+........\infty $ 

  1. $2cos{ 3 }6^{ \circ }$
  2. $2cos144^{ \circ }$
  3. $2sin18^{ \circ }$

  4. none
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The series $1-x+x^2-....$ form $GP$ with $a=1 ,r=-x$

Sum of infinte GP is $x=\dfrac{a}{1-r}\x=\dfrac{1}{1+x}\x+x^2=1\x^2+x-1=0$
By quadratic formulae 
$x=\dfrac{-1\pm\sqrt{1+4}}2\2\dfrac{-1\pm\sqrt5}{4}\2\cos 36^{\circ}$

Multiple choice maths geometric sequences sum of terms of g.p sum of n terms of an gp summing geometric series

The value of $x$ that satisfies the relation $x=1-x+x^{2}-x^{3}+x^{4}-x^{5}+.\infty$ if $|x|<1$ 

  1. $\dfrac{-1\pm\sqrt5}{2}$
  2. $\dfrac{-1\pm3i}{2}$
  3. $0$
  4. $none$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Here the first term is $a$ 
Common ratio is given by $-x$
The sum of infinite series is of an GP is given by $\dfrac{a}{1-r}\\x=dfrac{1}{1+x}\\x^2+x=1\\x^2+x-1=0$
Using quadratic formulae $\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$
Here $a=1b=1c=-1$
$\implies x=\dfrac{-1\pm\sqrt{1+4}}{2}\\x=\dfrac{-1\pm\sqrt{5}}{2}$
Multiple choice maths parts and whole multiplication of a fraction multiplication of a fractions multiplication of fraction finding the whole when a fraction is given

Find $x$ if $\left (\cfrac {1}{2}\times \cfrac {1}{3}\right )\times \cfrac {1}{4}= x \times \left (\cfrac {1}{3}\times \cfrac {1}{4}\right )$.

  1. 1
  2. $\dfrac {1}{5}$
  3. $\dfrac {1}{2}$
  4. $\dfrac {1}{3}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The given question shows the associative property of multiplication i.e. 

$\left( a\times b \right) \times c=a\times \left( b\times c \right) $
$\left (\cfrac {1}{2}\times \cfrac {1}{3}\right )\times \cfrac {1}{4}= x \times \left (\cfrac {1}{3}\times \cfrac {1}{4}\right )$
Therefore, $x=\dfrac{1}{2}$
Hence, the correct answer is option C.