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 powers and exponents scientific notation use of exponents power of 10

If $ { 9 }^{ x-1 }={ 3 }^{ 2x-1 }-486 $,then the value of x is:

  1. $\dfrac{7}{2}$

  2. 4

  3. 1

  4. 0
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
$9^{x-1}=3^{2x-1}-486$
$3^{2x-1}=9^{x-1}t486$
$\dfrac{3^{2x}}{3}=3^{2x-2}+486$
$3^{2x}=\dfrac{3(3^{2x})}{3^{z}}t(486)3$
$3^{2q}=3^{2x-1}+1458$
$2177=729+1458$
$3^{7}=3^{6}+1458$
$2x=7$
$x=\dfrac{7}{2}$
Multiple choice maths powers and exponents scientific notation use of exponents power of 10

If $8^x = 16^{x-1}$, find $x $.

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

Given, $8^x=(16)^{x-1}$

Left hand side, $8^x=(2^3)^x=2^{3x}$
Right hand side $16^{x-1}=(2^4)^{x-1}=2^{4(x-1)}=2^{(4x-4)}$
Equating both sides, we get
$2^{3x}=2^{{4x-4}}$
As bases are equal, then powers must be equal, so
$\Rightarrow 3x=4x-4$
$ \Rightarrow x=4$

Multiple choice maths powers and exponents scientific notation use of exponents power of 10

If $4^{2x + 2} = 64$, then calculate the value of $x $.

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

Given is $4^{2x+2}=64$

LHS: 
$\Rightarrow 4^{ 2x+2 }=(2^{ 2 })^{ 2x+2 }\ \Rightarrow { 2 }^{ 4x+4 }$
RHS: 
$\Rightarrow 64=2^6$

Now, LHS $=$ RHS
$\Rightarrow { 2 }^{ 4x+4 }={ 2 }^{ 6 }$
As bases are equal, so powers must be equal,
$\Rightarrow 6=4x+4$
$\Rightarrow 4x=2\ \Rightarrow x=\dfrac { 1 }{ 2 } $

Multiple choice maths powers and exponents scientific notation use of exponents power of 10

If $64^{x} = 4^{x^{2} - 4}$, then find the value of $x$.

  1. $x = 4$ or $x = -1$
  2. $x = -4$ or $x = 1$
  3. $x = 10$
  4. $x = \sqrt {20}$
  5. $x = 3$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
  • ${ 64 }^{ x }={ 4 }^{ 3x }={ 4 }^{ { x }^{ 2 }-4 }$ , by equating powers , we get,
  • ${ x }^{ 2 }-4 = 3x$ , which implies ${ x }^{ 2 }-3x-4 = 0$
  • $\Rightarrow x^2-4x+x-4=0$
  • $\Rightarrow x(x-4)+1(x-4)= 0 $
  • $\Rightarrow (x-4)(x+1)=0$
  • The roots are $x=4,-1$
Multiple choice maths powers and exponents scientific notation use of exponents power of 10

If $2^{3x - 2} = 16$, then calculate the value of $x $.

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

Given, ${ 2 }^{ 3x-2 }=16$
$\Rightarrow { 2 }^{ 3x-2 }=16={ 2 }^{ 4 }$

As bases are equal, there powers must be equal.
$\therefore 3x-2=4$
$\therefore 3x=6$
$\therefore x=2$

Multiple choice reciprocal equations theory of equations maths

Find $x$,  $2^{x^2}:2^{2x}=8:1$

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

Given, $2^{x^2}:2^{2x}=8:1$

$\Rightarrow \dfrac {2{x^2}}{2^{2x}}=\dfrac {8}{1}$
$\Rightarrow 2^{x^2}=8.2^{2x}$
$\Rightarrow 2^{x^2}=2^3.2^{2x}$
$\Rightarrow 2^{x^2}=2^{2x+3}$
$\Rightarrow x^2=2x+3$ ....As bases are equal, powers must be equal
$\Rightarrow x^2-2x-3=0$
$\Rightarrow (x-3)(x+1)=0$
$\therefore x=3,-1$

Multiple choice business maths functions and graphs some functions and their graphs -i graphs of the form y=ax^2+bx+c introduction to sets

If $ P ( X ) = x ^ { 3 } - 3 x ^ { 2 } + 2 x + 5 $ and P ( a ) = P ( b ) = P ( c ) = 0 then the value of ( 2 - a ) ( 2 - b ) ( 2 - c ) is

  1. 3

  2. 5

  3. 7

  4. 9
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Since P(a) = P(b) = P(c) = 0, a, b, and c are the roots of the cubic polynomial P(x) = x^3 - 3x^2 + 2x + 5. This means P(x) can be factored as (x - a)(x - b)(x - c). We need the value of (2 - a)(2 - b)(2 - c), which is precisely P(2). Substituting x = 2 into P(x) gives 2^3 - 3(2^2) + 2(2) + 5 = 8 - 12 + 4 + 5 = 5.

Multiple choice maths complex numbers and linear inequations identities of complex numbers powers of imaginary unit i algebra of complex numbers

If $(i^{413})(i^x)=1$, then determine the one possible value of x.

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

${ i }^{ 413 }{ i }^{ x }=1$

$\Rightarrow \quad { i }^{ 413 }=1$
now, $\left( 413+x \right) $ must be a multiple of 4 becouse ${ i }^{ 4 }=1$
$\therefore \quad \left( 413+3 \right) $ is divisible by $4$
                                     hence $x=3$

Multiple choice physics logarithms introduction to logarithm logarithmic notation exponential and logarithms

If $2\log y -\log x-3=0$ express $x$ in terms of $y.$

  1. $x^2=1000y$
  2. $x^2= \dfrac{y^2}{e^3}$
  3. $y^2= \dfrac{x}{1000}$
  4. $y^2= 1000x$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Given: $2\log y -\log x-3=0$

$\log { { y }^{ 2 } } -\log { x } -3\log { { e }=0 } $.......$(\log e=1)$

$\log { { y }^{ 2 } } -\log { x } -\log { { e }^{ 3 }=0 } $

$ \log { x } =\log { { y }^{ 2 } } -\log { { e }^{ 3 } } =\log { \left (\cfrac { { y }^{ 2 } }{ { e }^{ 3 } } \right ) } $

$ x=\cfrac { { y }^{ 2 } }{ { e }^{ 3 } } $
Multiple choice physics logarithms introduction to logarithm logarithmic notation exponential and logarithms

The equation  ${ \left( \log _{ 10 }{ x+2 }  \right)  }^{ 3 }+{ \left( \log _{ 10 }{ x-1 }  \right)  }^{ 3 }={ \left( 2\log _{ 10 }{ x+1 }  \right)  }^{ 3 }$ has

  1. no natural solution

  2. two rational solutions

  3. no prime solution

  4. one irrational solution
Reveal answer Fill a bubble to check yourself
B,C,D Correct answer
Explanation

Let $ \log _{ 10 }{ x+2 } =a$ and $ \log _{ 10 }{ x-1 } =b$
$\therefore a+b=2\log _{ 10 }{ x+1 } $ (from the question)
Thus, the given equation(in the question) reduces to ${a}^{3}+{b}^{3}={(a+b)}^{3}$
$\Rightarrow 3ab(a+b)=0$
$\Rightarrow a=0$ or $b=0$ or $a+b=0$
$\Rightarrow \log _{ 10 }{ x+2 }=0$  or $\log _{ 10 }{ x-1 }=0$ or $2\log _{ 10 }{ x } +1=0$
$\Rightarrow x={10}^{-2}$  or  $x=10$ or  $x={ 10 }^{ -\frac { 1 }{ 2 }  }$
Hence  $x=\left{ \dfrac { 1 }{ 100 },10 ,\dfrac { 1 }{ \sqrt { 10 }  }  \right} $