Pair of Linear Equations in Two Variables: CBSE Class 10 Previous Year Questions

176 different questions from Pair of Linear Equations in Two Variables (NCERT Chapter 3) asked in CBSE Class 10 Maths board exams 2022–2026. Pick a mark group to practise, each with answers and step-by-step solutions.

1 mark questions68 questions 2 marks questions30 questions 3 marks questions40 questions 4 marks questions9 questions 5 marks questions29 questions

Most asked Pair of Linear Equations in Two Variables questions

Q35 (OR) (OR)5 marksLong AnswerPair of Linear Equations in Two Variables CBSE 2024 · Basic 430/3/1

The sum of the digits of a 2-digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number.

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Answer: 18

Let the tens digit be $x$ and the units digit be $y$ ; the number is $10 x + y$ .
$x + y = 9$ ... (1)
$9 (10 x + y) = 2 (10 y + x)$ gives $88 x = 11 y$ , i.e. $y = 8 x$ ... (2)
From (1) and (2), $9 x = 9$ , so $x = 1$ and $y = 8$ .
The number is 18. (Check: $9 \times 18 = 162 = 2 \times 81$ .)

Also asked in: 2024 Basic 430/3/2, 2024 Basic 430/3/3, 2024 Basic 430/5/3

Q51 markMCQPair of Linear Equations in Two Variables CBSE 2026 · Standard 30/1/1

If a pair of linear equations in two variables is represented by two coincident lines, then the pair of equations has :

(A)a unique solution
(B)two solutions
(C)no solution
(D)an infinite number of solutions

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Answer: (D) an infinite number of solutions

Coincident lines have every point in common.
Every common point is a solution, so the pair has infinitely many solutions.

Also asked in: 2026 Standard 30/1/2, 2026 Standard 30/1/3

Q191 markAssertion–ReasonPair of Linear Equations in Two Variables CBSE 2026 · Standard 30/5/1

Assertion (A): The system of linear equations $3 x - 5 y + 7 = 0$ and $- 6 x + 10 y + 14 = 0$ is inconsistent.
Reason (R): When two linear equations don’t have unique solution, they always represent parallel lines.

(A)Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
(B)Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
(C)Assertion (A) is true, but Reason (R) is false.
(D)Assertion (A) is false, but Reason (R) is true.

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Answer: (C) Assertion (A) is true, but Reason (R) is false.

$\frac{a _{1}}{a _{2}} = \frac{3}{- 6} = - \frac{1}{2}$ , $\frac{b _{1}}{b _{2}} = \frac{- 5}{10} = - \frac{1}{2}$ , $\frac{c _{1}}{c _{2}} = \frac{7}{14} = \frac{1}{2}$
$\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} \neq = \frac{c _{1}}{c _{2}}$ , so the lines are parallel and the system is inconsistent. A is true.
If there is no unique solution, the lines may be parallel or coincident (infinitely many solutions). So R is false.

Also asked in: 2026 Standard 30/5/2, 2026 Standard 30/5/3

Q263 marksShort AnswerPair of Linear Equations in Two Variables CBSE 2026 · Standard 30/4/1

In a class test, Veer scored 6 more than twice as many marks as Kevin scored. If one of them had scored 4 more marks, their total score would have been 40. Find the marks obtained by Veer and Kevin.

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Answer: Veer: 26 marks, Kevin: 10 marks

Let Veer score x marks and Kevin score y marks.
$x = 2 y + 6$ ... (1)
If one of them scored 4 more, total = 40: $x + y + 4 = 40 \Rightarrow x + y = 36$ ... (2)
Substituting (1) in (2): $2 y + 6 + y = 36 \Rightarrow 3 y = 30 \Rightarrow y = 10$
$x = 2 \times 10 + 6 = 26$
Veer scored 26 and Kevin scored 10.

Also asked in: 2026 Standard 30/4/2, 2026 Standard 30/4/3

Q26 (OR) (OR)3 marksShort AnswerPair of Linear Equations in Two Variables CBSE 2026 · Standard 30/4/1

Solve the linear equations $3 x + y = 14$ and $y = 2$ graphically.

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Answer:

x = 4

y = 2

For $3 x + y = 14$ : points (4, 2), (3, 5), (5, $- 1$ ).
For $y = 2$ : points (0, 2), (2, 2), (4, 2); a line parallel to the x-axis.
Plot both lines on the same axes.
They intersect at (4, 2).
Solution: $x = 4$ , $y = 2$ .

Also asked in: 2026 Standard 30/4/2, 2026 Standard 30/4/3

Q364 marksCase StudyPair of Linear Equations in Two Variables CBSE 2026 · Basic 430/5/1

Seema daily goes to a park to exercise on machines available there. When Seema spent 15 minutes on exercise bicycle and 30 minutes on double cross walker, she received a message of burning 435 calories on her fitness watch. When she spent 30 minutes on exercise bicycle and 40 minutes on double cross walker, she received a message of burning 690 calories.
To find the number of calories burned per minute on each machine, answer the following :
(i) Represent the above situation in terms of a pair of linear equations in two variables.
(ii) Show that the equations have unique solution.
(iii) (a) Solve both equations to find the values of the variables using elimination method.
OR (b) Solve both equations to find the values of the variables using substitution method.

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Answer: (i)

15 x + 30 y = 435

and

30 x + 40 y = 690

, i.e.

x + 2 y = 29

and

3 x + 4 y = 69

, where x and y are calories burned per minute on the exercise bicycle and the double cross walker (ii)

\frac{a _{1}}{a _{2}} = \frac{1}{3} \neq = \frac{b _{1}}{b _{2}} = \frac{1}{2}

, so a unique solution (iii)

x = 11

y = 9

(by either method)

(i) Let x and y be the calories burned per minute on the exercise bicycle and the double cross walker.
$15 x + 30 y = 435 \Rightarrow x + 2 y = 29$ ; $30 x + 40 y = 690 \Rightarrow 3 x + 4 y = 69$
(ii) For $x + 2 y = 29$ and $3 x + 4 y = 69$ : $\frac{a _{1}}{a _{2}} = \frac{1}{3}$ , $\frac{b _{1}}{b _{2}} = \frac{2}{4} = \frac{1}{2}$ ; these are unequal, so the pair has a unique solution.
(iii)(a) Elimination: multiply $x + 2 y = 29$ by 2: $2 x + 4 y = 58$ . Subtract from $3 x + 4 y = 69$ : $x = 11$ . Then $2 y = 29 - 11 = 18$ , $y = 9$ .
(iii)(b) Substitution: $x = 29 - 2 y$ ; $3 (29 - 2 y) + 4 y = 69 \Rightarrow 87 - 2 y = 69 \Rightarrow y = 9$ , $x = 29 - 18 = 11$ .
So 11 calories per minute on the exercise bicycle and 9 calories per minute on the double cross walker.

Also asked in: 2026 Basic 430/5/2, 2026 Basic 430/5/3

Q355 marksLong AnswerPair of Linear Equations in Two Variables CBSE 2026 · Basic 430/4/1

The difference between two numbers is 12. The greater number is 6 less than twice the smaller one.
(i) Representing the above situation, frame two linear equations in two variables.
(ii) Show that the equations have unique solution.
(iii) Solve the equations and hence find the numbers.

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Answer: (i)

x - y = 12

x - 2 y = - 6

(ii) unique solution (iii) 30 and 18

(i) Let the greater number be x and the smaller be y: $x - y = 12$ and $x = 2 y - 6$ , i.e. $x - 2 y = - 6$ .
(ii) $\frac{a _{1}}{a _{2}} = \frac{1}{1} = 1$ , $\frac{b _{1}}{b _{2}} = \frac{- 1}{- 2} = \frac{1}{2}$ ; since $\frac{a _{1}}{a _{2}} \neq = \frac{b _{1}}{b _{2}}$ , the lines intersect and there is a unique solution.
(iii) Subtracting the second equation from the first: $y = 18$ .
Then $x = 12 + 18 = 30$ .
The numbers are 30 and 18.

Also asked in: 2026 Basic 430/4/2, 2026 Basic 430/4/3

Q35 (OR) (OR)5 marksLong AnswerPair of Linear Equations in Two Variables CBSE 2026 · Basic 430/4/1

Solve the following equations graphically :
$x + y = 7$ and $2 x - 5 y = 7$

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Answer:

x = 6

, y = 1

For $x + y = 7$ : points (0, 7), (7, 0), (6, 1).
For $2 x - 5 y = 7$ : points (1, – 1), (6, 1), (– 4, – 3).
Plot both lines on the same axes.
The lines intersect at (6, 1).
So $x = 6$ , $y = 1$ .

Also asked in: 2026 Basic 430/4/2, 2026 Basic 430/4/3

Q201 markAssertion–ReasonPair of Linear Equations in Two Variables CBSE 2025 · Basic 430/1/1

Assertion (A) : The value of p for which the system of equations $4 x + p y + 8 = 0$ and $2 x + 2 y + 2 = 0$ is consistent is 4.
Reason (R) : The system of equations $a_{1} x + b_{1} y = c_{1}$ and $a_{2} x + b_{2} y = c_{2}$ is consistent with infinitely many solutions, if $\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} = \frac{c _{1}}{c _{2}}$ .

(A)Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
(B)Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
(C)Assertion (A) is true, but Reason (R) is false.
(D)Assertion (A) is false, but Reason (R) is true.

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Answer: (D) Assertion (A) is false, but Reason (R) is true.

For $p = 4$ : $\frac{a _{1}}{a _{2}} = \frac{4}{2} = 2$ , $\frac{b _{1}}{b _{2}} = \frac{4}{2} = 2$ , $\frac{c _{1}}{c _{2}} = \frac{8}{2} = 4$ .
Since $\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} \neq = \frac{c _{1}}{c _{2}}$ , the lines are parallel and the system is inconsistent. So A is false.
R is the standard condition for infinitely many solutions, so R is true.

Also asked in: 2025 Basic 430/1/2, 2025 Basic 430/1/3

Q141 markMCQPair of Linear Equations in Two Variables CBSE 2025 · Basic 430/4/1

The value of $m$ for which lines $14 x + m y = 20$ and $- 3 x + 2 y = 16$ are parallel, is :

(A) $- \frac{3}{14}$
(B) $- \frac{7}{3}$
(C) $- \frac{28}{3}$
(D) $- \frac{3}{28}$

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Answer: (C)

- \frac{28}{3}

For parallel lines, $\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} \neq = \frac{c _{1}}{c _{2}}$ .
$\frac{14}{- 3} = \frac{m}{2}$ , so $m = - \frac{28}{3}$ .
Check: $\frac{c _{1}}{c _{2}} = \frac{20}{16} \neq = - \frac{14}{3}$ , so the lines are parallel.

Also asked in: 2025 Basic 430/4/2, 2025 Basic 430/4/3

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