Consistency of Pair of Equations PYQs

Class 10 Previous Year Questions (2014 – 2026)

Topic Overview

Master the conditions for consistency and inconsistency based on the ratios of coefficients $a_1/a_2, b_1/b_2$ and $c_1/c_2$. These are frequent 1-mark and 2-mark questions in board exams.

Q1 2026

00:00

Assertion (A): The pair of equations $x + 2y = 5$ and $3x + 6y = 15$ has infinitely many solutions.
Reason (R): If $a_1/a_2 = b_1/b_2 = c_1/c_2$, the pair of linear equations is consistent with infinitely many solutions.

For $x + 2y = 5$ and $3x + 6y = 15$:

$a_1/a_2 = 1/3, b_1/b_2 = 2/6 = 1/3, c_1/c_2 = 5/15 = 1/3$.

Since $a_1/a_2 = b_1/b_2 = c_1/c_2$, the system has infinitely many solutions.

Ans: (A) Both A and R are true; R is the correct explanation

Q2 2025

00:00

The pair $2x + 3y = 5$ and $4x + 6y = 10$ represents:

(a)(A) intersecting lines

(b)(B) parallel lines

(c)(C) coincident lines

(d)(D) perpendicular lines

$a_1/a_2 = 2/4 = 1/2$

$b_1/b_2 = 3/6 = 1/2$

$c_1/c_2 = 5/10 = 1/2$

Since $a_1/a_2 = b_1/b_2 = c_1/c_2$, the lines are coincident.

Ans: (C) coincident lines

Q3 2024

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The pair of lines $3x + 2y = 7$ and $4x + 8y - 11 = 0$ are:

(a)(A) perpendicular

(b)(B) parallel

(c)(C) intersecting

(d)(D) coincident

$a_1/a_2 = 3/4$

$b_1/b_2 = 2/8 = 1/4$

Since $a_1/a_2 \neq b_1/b_2$, the lines are intersecting.

Ans: (C) intersecting

Q4 2023

00:00

A system of two linear equations is inconsistent if the lines are:

(a)(A) coincident

(b)(B) parallel

(c)(C) intersecting at one point

(d)(D) at right angles

A system is inconsistent if it has no solution.

This happens when the lines are parallel.

Ans: (B) parallel

Q5 2022

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For what value of $k$ do $3x - y + 8 = 0$ and $6x - ky = -16$ represent coincident lines?

(a)(A) 1/2

(b)(B) -1/2

(c)(C) 2

(d)(D) -2

For coincident lines: $a_1/a_2 = b_1/b_2 = c_1/c_2$

$3/6 = -1/(-k) = 8/16 \Rightarrow 1/2 = 1/k \Rightarrow k = 2$.

Ans: (C) 2

Q6 2022

00:00

Assertion (A): $x - 2y = 3$ and $3x + ky = 1$ have unique solution for $k = -6$.
Reason (R): $a_1/a_2 \neq b_1/b_2$ gives a unique solution.

For $k = -6$, $a_1/a_2 = 1/3$ and $b_1/b_2 = -2/(-6) = 1/3$.

Since $a_1/a_2 = b_1/b_2$, it does NOT have a unique solution (it has no solution).

Thus Assertion (A) is false, but Reason (R) is a true general statement.

Ans: (D) A is false, R is true

Q7 2021

00:00

Find the value of $k$ for which $kx + 2y = 5$ and $3x + y = 1$ has a unique solution.

For a unique solution: $a_1/a_2 \neq b_1/b_2$

$k/3 \neq 2/1 \Rightarrow k \neq 6$.

Ans: $k \neq 6$

Q8 2019

00:00

Find the value of $k$ for which $kx - y = 2$ and $6x - 2y = 3$ has no solution.

For no solution: $a_1/a_2 = b_1/b_2 \neq c_1/c_2$

$k/6 = -1/(-2) \neq 2/3 \Rightarrow k/6 = 1/2 \Rightarrow k = 3$.

Ans: $k = 3$

Q9 2018

00:00

Find the value of $p$ for which $2x + 3y = 7$ and $(p+1)x + (2p-1)y = 21-4p$ has infinitely many solutions.

For infinitely many solutions: $a_1/a_2 = b_1/b_2 = c_1/c_2$

$2/(p+1) = 3/(2p-1) = 7/(21-4p)$

From first two: $2(2p-1) = 3(p+1) \Rightarrow 4p-2 = 3p+3 \Rightarrow p = 5$.

Verify with third: $2/6 = 3/9 = 7/1 = 1/3$. Verified.

Ans: $p = 5$

Q10 2017

00:00

For what value of $k$ will $kx + 3y = k - 3$ and $12x + ky = k$ have infinitely many solutions?

$a_1/a_2 = b_1/b_2 = c_1/c_2 \Rightarrow k/12 = 3/k = (k-3)/k$

From $k/12 = 3/k \Rightarrow k^2 = 36 \Rightarrow k = \pm 6$.

If $k=6$: $3/6 = 3/6$. If $k=-6$: $3/-6 = -9/-6 = 3/2$ (Not equal).

Thus $k = 6$.

Ans: $k = 6$

Q11 2015

00:00

For what value of $k$ will $2x + ky = 1$ and $3x - 5y = 7$ have no solution?

$a_1/a_2 = b_1/b_2 \neq c_1/c_2 \Rightarrow 2/3 = k/(-5) \Rightarrow k = -10/3$.

Ans: $k = -10/3$

Q12 2016

00:00

For what value of $k$ will $kx + 3y = k-3$ and $12x + ky = k$ have infinitely many solutions?

Identical to Q13. $k/12 = 3/k = (k-3)/k$.

Solving gives $k = 6$.

Ans: $k = 6$

Q13 2023

00:00

For what value of $k$ does $2x + 3y = 7$ and $(k-1)x + (k+2)y = 3k$ have infinitely many solutions?

$2/(k-1) = 3/(k+2) = 7/3k$

From $2/(k-1) = 3/(k+2) \Rightarrow 2k+4 = 3k-3 \Rightarrow k = 7$.

Verify: $2/6 = 3/9 = 7/21 = 1/3$. Correct.

Ans: $k = 7$

Q14 2021

00:00

Solve: $x + 2y - 8 = 0$ and $2x + 4y = 16$. Find any two solutions.

Equations are dependent ($a_1/a_2 = b_1/b_2 = c_1/c_2$).

Coincident lines, infinitely many solutions.

Solutions: $(0,4), (8,0), (2,3)$.

Ans: (0,4) and (8,0)

00:00

For $x + 2y = 5$ and $3x + 6y = 15$:

$a_1/a_2 = 1/3, b_1/b_2 = 2/6 = 1/3, c_1/c_2 = 5/15 = 1/3$.

Since $a_1/a_2 = b_1/b_2 = c_1/c_2$, the system has infinitely many solutions.

Ans: (A) Both A and R are true; R is the correct explanation

Q2 2025

00:00

The pair $2x + 3y = 5$ and $4x + 6y = 10$ represents:

(a)(A) intersecting lines

(b)(B) parallel lines

(c)(C) coincident lines

(d)(D) perpendicular lines

$a_1/a_2 = 2/4 = 1/2$

$b_1/b_2 = 3/6 = 1/2$

$c_1/c_2 = 5/10 = 1/2$

Since $a_1/a_2 = b_1/b_2 = c_1/c_2$, the lines are coincident.

Ans: (C) coincident lines

Q3 2024

00:00

The pair of lines $3x + 2y = 7$ and $4x + 8y - 11 = 0$ are:

(a)(A) perpendicular

(b)(B) parallel

(c)(C) intersecting

(d)(D) coincident

$a_1/a_2 = 3/4$

$b_1/b_2 = 2/8 = 1/4$

Since $a_1/a_2 \neq b_1/b_2$, the lines are intersecting.

Ans: (C) intersecting

Q4 2023

00:00

A system of two linear equations is inconsistent if the lines are:

(a)(A) coincident

(b)(B) parallel

(c)(C) intersecting at one point

(d)(D) at right angles

A system is inconsistent if it has no solution.

This happens when the lines are parallel.

Ans: (B) parallel

Q5 2022

00:00

For what value of $k$ do $3x - y + 8 = 0$ and $6x - ky = -16$ represent coincident lines?

(a)(A) 1/2

(b)(B) -1/2

(c)(C) 2

(d)(D) -2

For coincident lines: $a_1/a_2 = b_1/b_2 = c_1/c_2$

$3/6 = -1/(-k) = 8/16 \Rightarrow 1/2 = 1/k \Rightarrow k = 2$.

Ans: (C) 2

Q6 2022

00:00

Assertion (A): $x - 2y = 3$ and $3x + ky = 1$ have unique solution for $k = -6$.
Reason (R): $a_1/a_2 \neq b_1/b_2$ gives a unique solution.

For $k = -6$, $a_1/a_2 = 1/3$ and $b_1/b_2 = -2/(-6) = 1/3$.

Since $a_1/a_2 = b_1/b_2$, it does NOT have a unique solution (it has no solution).

Thus Assertion (A) is false, but Reason (R) is a true general statement.

Ans: (D) A is false, R is true

Q7 2021

00:00

Find the value of $k$ for which $kx + 2y = 5$ and $3x + y = 1$ has a unique solution.

For a unique solution: $a_1/a_2 \neq b_1/b_2$

$k/3 \neq 2/1 \Rightarrow k \neq 6$.

Ans: $k \neq 6$

Q8 2019

00:00

Find the value of $k$ for which $kx - y = 2$ and $6x - 2y = 3$ has no solution.

For no solution: $a_1/a_2 = b_1/b_2 \neq c_1/c_2$

$k/6 = -1/(-2) \neq 2/3 \Rightarrow k/6 = 1/2 \Rightarrow k = 3$.

Ans: $k = 3$

Q9 2018

00:00

Find the value of $p$ for which $2x + 3y = 7$ and $(p+1)x + (2p-1)y = 21-4p$ has infinitely many solutions.

For infinitely many solutions: $a_1/a_2 = b_1/b_2 = c_1/c_2$

$2/(p+1) = 3/(2p-1) = 7/(21-4p)$

From first two: $2(2p-1) = 3(p+1) \Rightarrow 4p-2 = 3p+3 \Rightarrow p = 5$.

Verify with third: $2/6 = 3/9 = 7/1 = 1/3$. Verified.

Ans: $p = 5$

Q10 2017

00:00

For what value of $k$ will $kx + 3y = k - 3$ and $12x + ky = k$ have infinitely many solutions?

$a_1/a_2 = b_1/b_2 = c_1/c_2 \Rightarrow k/12 = 3/k = (k-3)/k$

From $k/12 = 3/k \Rightarrow k^2 = 36 \Rightarrow k = \pm 6$.

If $k=6$: $3/6 = 3/6$. If $k=-6$: $3/-6 = -9/-6 = 3/2$ (Not equal).

Thus $k = 6$.

Ans: $k = 6$

Q11 2015

00:00

For what value of $k$ will $2x + ky = 1$ and $3x - 5y = 7$ have no solution?

$a_1/a_2 = b_1/b_2 \neq c_1/c_2 \Rightarrow 2/3 = k/(-5) \Rightarrow k = -10/3$.

Ans: $k = -10/3$

Q12 2016

00:00

For what value of $k$ will $kx + 3y = k-3$ and $12x + ky = k$ have infinitely many solutions?

Identical to Q13. $k/12 = 3/k = (k-3)/k$.

Solving gives $k = 6$.

Ans: $k = 6$

Q13 2023

00:00

For what value of $k$ does $2x + 3y = 7$ and $(k-1)x + (k+2)y = 3k$ have infinitely many solutions?

$2/(k-1) = 3/(k+2) = 7/3k$

From $2/(k-1) = 3/(k+2) \Rightarrow 2k+4 = 3k-3 \Rightarrow k = 7$.

Verify: $2/6 = 3/9 = 7/21 = 1/3$. Correct.

Ans: $k = 7$

Q14 2021

00:00

Solve: $x + 2y - 8 = 0$ and $2x + 4y = 16$. Find any two solutions.

Equations are dependent ($a_1/a_2 = b_1/b_2 = c_1/c_2$).

Coincident lines, infinitely many solutions.

Solutions: $(0,4), (8,0), (2,3)$.

Ans: (0,4) and (8,0)