Mathematics · Quantitative Aptitude

Linear Equations

153 Questions

Linear equations involve solving for unknown variables in single or multi variable systems. These questions test algebraic manipulation and logical consistency skills. They are a core component of quantitative aptitude and advanced mathematics tests.

Solving simultaneous equationsSingle variable equationsSystem consistency checksIndeterminate equationsMatrix form solutions

Linear Equations 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 linear equations. If $\cfrac{3t-2}{4}-\cfrac{2t+3}{3} = \cfrac{2}{3}-t$, then $t  $ is equal to

  1. $1$
  2. $2$
  3. $3$
  4. $4$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Given $\cfrac { 3t-2 }{ 4 } -\cfrac { (2t+3) }{ 3 }  = \cfrac { 2 }{ 3 }  - t$
L.C.M. of $4,3,3$ is $ 12$
$\Rightarrow  3(3t-2) - 4(2t+3) = 8 - 12t$
$\Rightarrow 9t - 6 -(8t +12) = 8 -12t$
$\Rightarrow t -18  = 8 -12t$
$\Rightarrow 13 t = 26$
$\Rightarrow t = 2$
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: $m-\cfrac{m-1}{2} = 1-\cfrac{m-2}{3}$

  1. $m = \cfrac{7}{5}$
  2. $m = \cfrac{2}{3}$
  3. $m = \cfrac{10}{3}$
  4. $m = \cfrac{3}{8}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Given, $m - \cfrac { (m-1) }{ 2 }  = 1-\cfrac { (m-2) }{ 3 }$ 
L.C.M. of $2$ and $3$ is $6$
$6m -3(m-1) = 6 - 2(m-2)$
$ 6m - 3m +3 = 6 - 2m +4$
$5m = 7$
$ m = \cfrac { 7 }{ 5 }$
Multiple choice maths equation reducing simple equations to simpler form solving linear equations solution of a linear equation in one variable

Find the Solution  : $x - cy - bz = 0 $

                                 $ cx - y +az = 0 $
                                 $bx+ ay -z = 0 $

  1. $a^3 + b^3 + c^3 + 3abc = 1$
  2. $a^2 + b^2 + c^2 + 2abc = 1$
  3. $a^4 + b^4 + c^4 + 4abc = 1$
  4. $a^5 + b^5 + c^5 + 5abc = 1$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The given equations are 

$x-cy-bz = 0$
$cx - y + az = 0$
$bx + ay - z = 0$
Since x, y, z are not all zero, the system will have non-trivial solution if
$\triangle = \begin{vmatrix} 1 & -c & -b \ c & -1 & a \ b & a & -1\end{vmatrix} = 0$
or $1(1 - a^2) + c (-c -ab) -b (ac + b) = 0$
or $1 - a^2 - c^2 -abc - abc - b^2 = 0$
or $a^2 + b^2 + c^2 + 2abc = 1$

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

Which equation is  non- linear 

  1. $x + 1 = 2x - 3$
  2. $9 - 5{x^2} = 4$
  3. $- 2x - 5 - \left( {x + 4} \right) = - 9$
  4. ${x} - \frac{1}{{15}} = 2$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Linear equations are those which have the degree of equation as $1$


Here every equation has the degree as $1$

Except $9-5x^2=4$

So its is not a linear equation 

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

Solve:
$x\, +\, y\, =\, 7xy$
$2x\, -\, 3y\, =\, -xy$

  1. $x\, =\, \displaystyle \frac{1}{8}$ and $y\, =\, \displaystyle \frac{1}{4}$
  2. $x\, =\, \displaystyle \frac{1}{7}$ and $y\, =\, \displaystyle \frac{1}{4}$
  3. $x\, =\, \displaystyle \frac{1}{2}$ and $y\, =\, \displaystyle \frac{1}{4}$
  4. $x\, =\, \displaystyle \frac{1}{3}$ and $y\, =\, \displaystyle \frac{1}{4}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Given equations are 

$x+y=7xy$ ....(1)
and $2x-3y=-xy$ ....(2)
Multiply equation (1) by $3$, we get
$3x+3y=21xy$ ....(3)
Add equations (2) and (3),
$5x=20xy$
$\Rightarrow y=\dfrac {1}{4}$
Put this value in equation (1), we get
$x+\dfrac {1}{4}=7x\times \dfrac {1}{4}$
$\Rightarrow x-\dfrac {7x}{4}=-\dfrac {1}{4}$
$\Rightarrow \dfrac {3x}{4}=\dfrac {1}{4}$
$\Rightarrow x=\dfrac {1}{3}$
Therefore, solution is $x=\dfrac {1}{3}, y=\dfrac {1}{4}$.

Multiple choice maths ratio, proportion and unitary method converting to ratios finding ratios other quantities

If $\dfrac{1}{x} + y = 3$ and $x + \dfrac{1}{y} = 2$ then $x:y$ is 

  1. $3:2$
  2. $2:3$
  3. $1:2$
  4. $2:1$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
The question states '...then x: is' 
It should state '..then x:y is'

Given
$\dfrac { 1 }{ x } +y=3$ --- Eqn (1)
$x+\dfrac { 1 }{ y } =2$ ---Eqn (2)

Multiplying Eqn (1) by x and Eqn (2) by y, we get:

$x\left( \dfrac { 1 }{ x } +y \right) =3x\quad \Rightarrow 1+yx=3x$ --- Eqn (3)
$y\left( x+\dfrac { 1 }{ y }  \right) =2y\quad \Rightarrow xy+1=2y$ --- Eqn (4)

Subtracting Eqn (4) and Eqn (5), we get:

$1+yx-1-yx=3x-2y$
$\Rightarrow 0=3x-2y$
$\Rightarrow 3x=2y$
$\Rightarrow \dfrac { x }{ y } =\dfrac { 2 }{ 3 } $
$\therefore x:y=2:3$

Hence the answer is B
Multiple choice

Solve the following system of equations using the substitution method: 2x + 3y = 11 x - y = 1

  1. (x, y) = (2, 3)

  2. (x, y) = (3, 2)

  3. (x, y) = (4, 1)

  4. (x, y) = (1, 4)
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

To solve using substitution, solve one equation for one variable and substitute it into the other equation. Solving x - y = 1 for x, we get x = y + 1. Substituting this into 2x + 3y = 11, we get 2(y + 1) + 3y = 11. Solving for y, we get y = 2. Substituting y = 2 back into x = y + 1, we get x = 3. Therefore, the solution is (x, y) = (2, 3).

Multiple choice

Solve the following system of equations using the elimination method: 3x + 2y = 7 2x - 3y = 1

  1. (x, y) = (2, 1)

  2. (x, y) = (1, 2)

  3. (x, y) = (3, -1)

  4. (x, y) = (-1, 3)
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

To solve using elimination, multiply one or both equations by a constant to make the coefficients of one variable the same. Multiplying the first equation by 2 and the second equation by 3, we get: 6x + 4y = 14 6x - 9y = 3 Subtracting the second equation from the first, we get: 13y = 11 Solving for y, we get y = 1. Substituting y = 1 back into either of the original equations, we can solve for x. Using the first equation, we get 3x + 2(1) = 7, which gives x = 2. Therefore, the solution is (x, y) = (2, 1).