Mathematics · Quantitative Aptitude

Linear Equations

196 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 general knowledge math & puzzles

For what values of 'k' will the pair of equations 3x + 4y = 12 and kx + 12y = 30 not have a unique solution?

  1. 12

  2. 9

  3. 3

  4. 7.5

  5. 2.5
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Two linear equations have no unique solution when the lines are parallel or coincident, meaning the ratio of x and y coefficients is equal. For no unique solution: 3/4 = k/12, giving k = 9. Option B is correct. When k = 9, the second equation becomes 9x + 12y = 30, or 3x + 4y = 10, which is parallel to the first equation 3x + 4y = 12.

Multiple choice

Consider the following system of linear equations $$\left( \begin{array}{ccc} 2 & 1 & -4 \\ 4 & 3 & -12 \\ 1 & 2 & -8 \end{array} \right) \left( \begin{array}{ccc} x \\ y \\ z \end{array} \right) = \left( \begin{array}{ccc} \alpha \\ 5 \\ 7 \end{array} \right)$$ Notice that the second and the third columns of the coefficient matrix are linearly dependent. For how many values of $\alpha$, does this system of equations have infinitely many solutions?

  1. 0

  2. 1

  3. 2

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

$\left( \begin{array}{ccc} 2 & 1 & -4 \\ 4 & 3 & -12 \\ 1 & 2 & -8 \end{array} \right) \left( \begin{array}{ccc} x \\ y \\ z \end{array} \right) = \left( \begin{array}{ccc} \alpha \\ 5 \\ 7 \end{array} \right)$ We can write for this linear equation 2x + y - 4z = a             4x + 3y - 12z = 5 x + 2y - 8z = 7 For infinitely solutions, D = 0 $\left( \begin{array}{ccc} 2 & 1 & -4 \\ 4 & 3 & -12 \\ 1 & 2 & -8 \end{array} \right) = 0$ Because, 2nd and 3rd columns are linearly dependent For x D = $\left( \begin{array}{ccc} \alpha & 1 & -4 \\ 5 & 3 & -12 \\ 7 & 2 & -8 \end{array} \right) = 0$ Because, 2nd and 3rd columns are linearly dependent For y $\left( \begin{array}{ccc} 2 & \alpha & -4 \\ 4 & 5 & -12 \\ 1 & 7 & -8 \end{array} \right) = 0$ $\Rightarrow$2(- 40 + 84) -$\alpha$(- 32 + 12) - 4 ( 28 -5) = 0 $\Rightarrow$ 88 + 20$\alpha$-92 = 0 $\Rightarrow$$\alpha = \dfrac{4}{20} = \dfrac{1}{5}$ --- (i) For z D = $\left( \begin{array}{ccc} 2 & 1 & \alpha \\ 4 & 3 & 5 \\ 1 & 2 & 7 \end{array} \right) = 0$

$\Rightarrow$2 (21-10) - 1 ( 28 - 5) +$\alpha$ (8 - 3) = 0 $\Rightarrow$ 22 - 23 + 5 $\alpha$ = 0 $\Rightarrow$ $\alpha = \dfrac{1}{5}$--(ii) Hence from equation, we find that, $\alpha$ have only one value for infinitely solution.

Multiple choice

Consider the following system of linear equations $$\left( \begin{array}{ccc} 2 & 1 & -4 \\ 4 & 3 & -12 \\ 1 & 2 & -8 \end{array} \right) \left( \begin{array}{ccc} x \\ y \\ z \end{array} \right) = \left( \begin{array}{ccc} \alpha \\ 5 \\ 7 \end{array} \right)$$ Notice that the second and the third columns of the coefficient matrix are linearly dependent. For how many values of $\alpha$, does this system of equations have infinitely many solutions?

  1. 0

  2. 1

  3. 2

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

$\left( \begin{array}{ccc} 2 & 1 & -4 \\ 4 & 3 & -12 \\ 1 & 2 & -8 \end{array} \right) \left( \begin{array}{ccc} x \\ y \\ z \end{array} \right) = \left( \begin{array}{ccc} \alpha \\ 5 \\ 7 \end{array} \right)$ We can write for this linear equation 2x + y - 4z = a             4x + 3y - 12z = 5 x + 2y - 8z = 7 For infinitely solutions, D = 0 $\left( \begin{array}{ccc} 2 & 1 & -4 \\ 4 & 3 & -12 \\ 1 & 2 & -8 \end{array} \right) = 0$ Because, 2nd and 3rd columns are linearly dependent For x D = $\left( \begin{array}{ccc} \alpha & 1 & -4 \\ 5 & 3 & -12 \\ 7 & 2 & -8 \end{array} \right) = 0$ Because, 2nd and 3rd columns are linearly dependent For y $\left( \begin{array}{ccc} 2 & \alpha & -4 \\ 4 & 5 & -12 \\ 1 & 7 & -8 \end{array} \right) = 0$ $\Rightarrow$2(- 40 + 84) -$\alpha$(- 32 + 12) - 4 ( 28 -5) = 0 $\Rightarrow$ 88 + 20$\alpha$-92 = 0 $\Rightarrow$$\alpha = \dfrac{4}{20} = \dfrac{1}{5}$ --- (i) For z D = $\left( \begin{array}{ccc} 2 & 1 & \alpha \\ 4 & 3 & 5 \\ 1 & 2 & 7 \end{array} \right) = 0$

$\Rightarrow$2 (21-10) - 1 ( 28 - 5) +$\alpha$ (8 - 3) = 0 $\Rightarrow$ 22 - 23 + 5 $\alpha$ = 0 $\Rightarrow$ $\alpha = \dfrac{1}{5}$--(ii) Hence from equation, we find that, $\alpha$ have only one value for infinitely solution.

Multiple choice

The following system of equations $$x_1 + x_2 + 2x_3 = 1$$ $$x_1 + 2x_2 + 3x_3 = 2$$ $$x_1 + 4x_2 + αx_3 = 4$$ has a unique solution. The only possible value(s) for α is/are

  1. 0

  2. either 0 or 1

  3. one of 0, 1 or -1

  4. any real number
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

One of 0, 1 or -1, the system will have unique solution. If det A $\ne$0, where A = det A $\ne$ 0 $\Rightarrow$$\alpha$-5 $\ne$0 Since, $\alpha$-5 $\ne$5 Hence $\alpha$ could be any real number except 5.

Multiple choice

The system of linear equations 4x + 2y = 7 and 2x + y = 6 has

  1. a unique solution

  2. no solution

  3. an infinite number of solutions

  4. exactly two distinct solutions
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\text{The given system is} \\ \begin{bmatrix} \ 4 & 2 \ \ 2 & 1 \ \end{bmatrix} \begin{bmatrix} \ x \ \ y \ \end{bmatrix} = \begin{bmatrix} \ 7 \ \ 6 \ \end{bmatrix} \\ \text{We have} \hspace{1cm} A = \begin{bmatrix} \ 4 & 2 \ \ 2 & 1 \ \end{bmatrix} \\ \text{and} \hspace{1cm} |A| = \begin{vmatrix} \ 4 & 2 \ \ 2 & 1 \ \end{vmatrix} = 0 \hspace{1cm} \text{Rank of matrix $\rho(A) < 2$} \\ \text{Now} \hspace{1cm} C = \begin{vmatrix} \ 4 & 2 & | & 7\ \ 2 & 1 & | & 6\ \end{vmatrix} \hspace{1cm} \text{Rank of matrix $\rho(C) = 2$} \\ \text{Since $\rho{A} \neq \rho (C)$ there is no solution.}$