Pair of Linear Equations Test Class 10 Maths — 40 Marks
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PERIODIC TEST — Pair of Linear Equations in Two Variables (2026-27) | CLASS 10 | MATHEMATICS
Max Marks: 40 · Time: 1 hour 30 minutes (Time left: 90:00)
| Section | Marks | Questions | Time | Strategy |
|---|---|---|---|---|
| A | 12 | 12 MCQ | 15 min | Compare the ratios a1/a2, b1/b2, c1/c2 |
| B | 8 | 4 Short | 20 min | State the method used (substitution or elimination) |
| C | 12 | 4 Long | 30 min | Form the equations first, then solve |
| D | 8 | 2 Case | 20 min | Read the paragraph, label the two unknowns |
| Review | — | — | 5 min | Substitute your answer back to check |
General Instructions: All questions are compulsory. Section A has 12 MCQs of 1 mark each. Internal choice is given in Sections B, C and D — attempt any ONE of the two options. For graphical questions, a rough sketch is enough. Substitute your final answer back into both equations to check it.
What is this paper? A printable 40-mark Class 10 Maths chapter test on a Pair of Linear Equations in Two Variables with a full per-question marking scheme, internal choice and practice/test modes. It is built to the current CBSE 2026-27 syllabus and is board-pattern practice, not a prediction of any examination paper.
a1x + b1y + c1 = 0
a1 / a2 ≠ b1 / b2 (unique solution)
a1 / a2 = b1 / b2 = c1 / c2 (infinitely many)
a1 / a2 = b1 / b2 ≠ c1 / c2 (no solution)
These three ratio conditions decide whether the two lines meet at one point, are parallel, or lie on top of each other.
Section A — Multiple Choice (12 × 1 = 12 marks)
Choose the single best option.
The pair of equations x + y = 5 and x − y = 1 has
Marking Scheme: Correct option: 1 (no partial marks).
Total: 1 Mark
Total: 1 Mark
Compare the ratios
a1 = 1, b1 = 1; a2 = 1, b2 = −1
a1/a2 = 1, b1/b2 = −1
They are unequal → the lines intersect at one point → unique solution.
a1/a2 = 1, b1/b2 = −1
They are unequal → the lines intersect at one point → unique solution.
Answer: (b) a unique solution
If a1/a2 = b1/b2 ≠ c1/c2, then the two lines are
Marking Scheme: Correct option: 1 (no partial marks).
Total: 1 Mark
Total: 1 Mark
Meaning of the condition
Equal a-ratio and b-ratio means the same slope.
The c-ratio differing means they are not the same line.
Same slope, different line → parallel → no solution.
The c-ratio differing means they are not the same line.
Same slope, different line → parallel → no solution.
Answer: (b) parallel
The pair 2x + 3y = 7 and 4x + 6y = 14 has
Marking Scheme: Correct option: 1 (no partial marks).
Total: 1 Mark
Total: 1 Mark
Compare the ratios
a1/a2 = 2/4 = 1/2
b1/b2 = 3/6 = 1/2
c1/c2 = 7/14 = 1/2
All three equal → coincident lines → infinitely many solutions.
b1/b2 = 3/6 = 1/2
c1/c2 = 7/14 = 1/2
All three equal → coincident lines → infinitely many solutions.
Answer: (c) infinitely many solutions
The pair x + 2y = 5 and 3x + 6y = 20 has
Marking Scheme: Correct option: 1 (no partial marks).
Total: 1 Mark
Total: 1 Mark
Compare the ratios
a1/a2 = 1/3
b1/b2 = 2/6 = 1/3
Equal, so check c: c1/c2 = 5/20 = 1/4 ≠ 1/3
Parallel lines → no solution.
b1/b2 = 2/6 = 1/3
Equal, so check c: c1/c2 = 5/20 = 1/4 ≠ 1/3
Parallel lines → no solution.
Answer: (b) no solution
The solution of x + y = 10 and x − y = 4 is
Marking Scheme: Correct option: 1 (no partial marks).
Total: 1 Mark
Total: 1 Mark
Add the equations
(x + y) + (x − y) = 10 + 4
2x = 14 → x = 7
Then y = 10 − 7 = 3
2x = 14 → x = 7
Then y = 10 − 7 = 3
Answer: (a) (7, 3)
The graph of the equation x = 3 is a line
Marking Scheme: Correct option: 1 (no partial marks).
Total: 1 Mark
Total: 1 Mark
Every point has x = 3
Points like (3, 0), (3, 1), (3, 2) … all have x = 3.
Joining them gives a vertical line, parallel to the y-axis.
Joining them gives a vertical line, parallel to the y-axis.
Answer: (b) parallel to the y-axis
If 3x + 4y = 11 and 3x − 4y = −5, then x equals
Marking Scheme: Correct option: 1 (no partial marks).
Total: 1 Mark
Total: 1 Mark
Add (elimination)
(3x + 4y) + (3x − 4y) = 11 + (−5)
6x = 6 → x = 1
6x = 6 → x = 1
Answer: (a) 1
The pair of equations y = 0 and y = −5 has
Marking Scheme: Correct option: 1 (no partial marks).
Total: 1 Mark
Total: 1 Mark
Two horizontal lines
y = 0 is the x-axis; y = −5 is parallel to it.
Parallel lines never meet → no solution.
Parallel lines never meet → no solution.
Answer: (b) no solution
For which value of k does the pair x + y = 3, kx + 2y = 6 have infinitely many solutions?
Marking Scheme: Correct option: 1 (no partial marks).
Total: 1 Mark
Total: 1 Mark
All three ratios equal
a1/a2 = 1/k, b1/b2 = 1/2
Set 1/k = 1/2 → k = 2
Check c: 3/6 = 1/2 ✓
Set 1/k = 1/2 → k = 2
Check c: 3/6 = 1/2 ✓
Answer: (b) 2
A pair of linear equations which has the unique solution x = 2, y = 3 is
Marking Scheme: Correct option: 1 (no partial marks).
Total: 1 Mark
Total: 1 Mark
Substitute x = 2, y = 3
(a) 2 + 3 = 5 ✓; 4 − 3 = 1 ✓
(b) 2 + 6 = 8 ✓; 2 − 3 = −1 ≠ 1 ✗
(c) 4 + 3 = 7 ✓; 2 + 3 = 5 ≠ 6 ✗
(d) 6 + 3 = 9 ✓; 2 + 6 = 8 ≠ 7 ✗
(b) 2 + 6 = 8 ✓; 2 − 3 = −1 ≠ 1 ✗
(c) 4 + 3 = 7 ✓; 2 + 3 = 5 ≠ 6 ✗
(d) 6 + 3 = 9 ✓; 2 + 6 = 8 ≠ 7 ✗
Answer: (a) x + y = 5, 2x − y = 1
The lines representing 3x + 2y = 6 and 6x + 4y = 12 are
Marking Scheme: Correct option: 1 (no partial marks).
Total: 1 Mark
Total: 1 Mark
Compare the ratios
3/6 = 2/4 = 6/12 = 1/2
All three equal → the same line → coincident.
All three equal → the same line → coincident.
Answer: (b) coincident
The sum of two numbers is 15 and their difference is 3. The numbers are
Marking Scheme: Correct option: 1 (no partial marks).
Total: 1 Mark
Total: 1 Mark
Form and solve
x + y = 15, x − y = 3
Adding: 2x = 18 → x = 9
y = 15 − 9 = 6
Adding: 2x = 18 → x = 9
y = 15 − 9 = 6
Answer: (a) 9 and 6
Section B — Short Answer (4 × 2 = 8 marks)
Solve by substitution: x + y = 7 and 2x − 3y = 9.
Marking Scheme: Expressing one variable and substituting 1 mark · Both values 1 mark.
Total: 2 Marks
Total: 2 Marks
Express x in terms of y
x = 7 − y
Substitute in the second equation
2(7 − y) − 3y = 9
14 − 2y − 3y = 9
−5y = −5 → y = 1
x = 7 − 1 = 6
14 − 2y − 3y = 9
−5y = −5 → y = 1
x = 7 − 1 = 6
Answer: x = 6, y = 1
Find the value of k for which 2x + 3y = 5 and 4x + ky = 10 have infinitely many solutions.
Marking Scheme: Setting the ratio condition 1 mark · Value of k 1 mark.
Total: 2 Marks
Total: 2 Marks
Condition for infinitely many solutions
a1/a2 = b1/b2 = c1/c2
2/4 = 3/k = 5/10
2/4 = 3/k = 5/10
Solve
3/k = 1/2 → k = 6
Check c: 5/10 = 1/2 ✓
Check c: 5/10 = 1/2 ✓
Answer: k = 6
Solve by elimination: 3x + 2y = 11 and 2x + 3y = 4.
Marking Scheme: Correct elimination step 1 mark · Both values 1 mark.
Total: 2 Marks
Total: 2 Marks
Add the equations
5x + 5y = 15
x + y = 3
x + y = 3
Subtract the equations
x − y = 7
Adding x + y = 3 and x − y = 7:
2x = 10 → x = 5, so y = 3 − 5 = −2
Adding x + y = 3 and x − y = 7:
2x = 10 → x = 5, so y = 3 − 5 = −2
Answer: x = 5, y = −2
Answer either Option 1 or Option 2.
Option 1 — No solution
Marking Scheme: Condition for no solution 1 mark · Value of k 1 mark.
Total: 2 Marks
Total: 2 Marks
Set up the condition
x + 2y = 5 and 3x + ky = 12 have no solution when
a1/a2 = b1/b2 ≠ c1/c2
1/3 = 2/k
a1/a2 = b1/b2 ≠ c1/c2
1/3 = 2/k
Solve and check c
k = 6
c ratio = 5/12 ≠ 1/3 ✓
c ratio = 5/12 ≠ 1/3 ✓
Answer: k = 6
— OR —
Option 2
Determine whether 4x + 5y = 20 and 8x + 10y = 45 have a unique solution, no solution, or infinitely many solutions.
Marking Scheme: Comparing the ratios 1 mark · Correct conclusion 1 mark.
Total: 2 Marks
Total: 2 Marks
Compare the ratios
a1/a2 = 4/8 = 1/2
b1/b2 = 5/10 = 1/2
c1/c2 = 20/45 = 4/9 ≠ 1/2
b1/b2 = 5/10 = 1/2
c1/c2 = 20/45 = 4/9 ≠ 1/2
Conclusion
a-ratio = b-ratio ≠ c-ratio → parallel lines → no solution.
Answer: No solution
[Attempt any ONE of the two options of Q16]
Section C — Long Answer (4 × 3 = 12 marks)
Solve: 2x + 3y = 13 and 5x − 4y = −2.
Marking Scheme: Choosing a valid method and one elimination step 1 mark · Finding one variable 1 mark · Finding the other variable and stating the pair 1 mark.
Total: 3 Marks
Total: 3 Marks
Multiply to match coefficients
(2x + 3y = 13) × 4 → 8x + 12y = 52
(5x − 4y = −2) × 3 → 15x − 12y = −6
(5x − 4y = −2) × 3 → 15x − 12y = −6
Add to eliminate y
23x = 46 → x = 2
Substitute to find y
2(2) + 3y = 13
3y = 9 → y = 3
Check: 5(2) − 4(3) = 10 − 12 = −2 ✓
3y = 9 → y = 3
Check: 5(2) − 4(3) = 10 − 12 = −2 ✓
Answer: x = 2, y = 3
Find the values of a and b for which 2x + 3y = 7 and (a + b)x + (2a − b)y = 21 have infinitely many solutions.
Marking Scheme: Writing the ratio condition 1 mark · Forming the two equations in a and b 1 mark · Solving for a and b 1 mark.
Total: 3 Marks
Total: 3 Marks
Condition
(a + b)/2 = (2a − b)/3 = 21/7 = 3
Form two equations
a + b = 6 … (1)
2a − b = 9 … (2)
2a − b = 9 … (2)
Solve
(1) + (2): 3a = 15 → a = 5
From (1): b = 6 − 5 = 1
From (1): b = 6 − 5 = 1
Answer: a = 5, b = 1
Answer either Option 1 or Option 2.
Option 1 — Two-digit number
Marking Scheme: Forming the equations 1 mark · Solving for the two digits 1 mark · Stating the number 1 mark.
Total: 3 Marks
Total: 3 Marks
Let the number be 10a + b
Reversed number = 10b + a
Sum: (10a + b) + (10b + a) = 99
11(a + b) = 99 → a + b = 9
Sum: (10a + b) + (10b + a) = 99
11(a + b) = 99 → a + b = 9
Difference of the digits
a − b = 3 (the tens digit is larger)
Adding: 2a = 12 → a = 6, b = 3
Adding: 2a = 12 → a = 6, b = 3
The number
10a + b = 63
Check: 63 + 36 = 99 ✓
Check: 63 + 36 = 99 ✓
Answer: 63
— OR —
Option 2
A father is 3 times as old as his son. After 12 years he will be twice as old as his son. Find their present ages.
Marking Scheme: Forming the two equations 1 mark · Solving for the son’s age 1 mark · Both ages 1 mark.
Total: 3 Marks
Total: 3 Marks
Let the son’s age be s
Father’s age = 3s
After 12 years: 3s + 12 = 2(s + 12)
After 12 years: 3s + 12 = 2(s + 12)
Solve
3s + 12 = 2s + 24
s = 12
s = 12
Both ages
Son = 12 years, Father = 3 × 12 = 36 years
Check: in 12 years, 48 = 2 × 24 ✓
Check: in 12 years, 48 = 2 × 24 ✓
Answer: Son 12 years, father 36 years
[Attempt any ONE of the two options of Q19]
Answer either Option 1 or Option 2.
Option 1 — Elimination
Marking Scheme: Simplifying one equation 1 mark · Eliminating a variable 1 mark · Both values 1 mark.
Total: 3 Marks
Total: 3 Marks
Simplify the second equation
2x − 2y = 2 → x − y = 1 → x = y + 1
Substitute in the first
3(y + 1) + 4y = 10
7y + 3 = 10 → y = 1
7y + 3 = 10 → y = 1
Find x
x = 1 + 1 = 2
Check: 3(2) + 4(1) = 10 ✓
Check: 3(2) + 4(1) = 10 ✓
Answer: x = 2, y = 1
— OR —
Option 2
The cost of 5 pens and 3 pencils is ₹55, and the cost of 3 pens and 5 pencils is ₹49. Find the cost of one pen and one pencil.
Marking Scheme: Forming the two equations 1 mark · Solving for one item 1 mark · Both costs 1 mark.
Total: 3 Marks
Total: 3 Marks
Let a pen cost p and a pencil cost q
5p + 3q = 55
3p + 5q = 49
3p + 5q = 49
Add and subtract
Adding: 8p + 8q = 104 → p + q = 13
Subtracting: 2p − 2q = 6 → p − q = 3
Subtracting: 2p − 2q = 6 → p − q = 3
Solve
2p = 16 → p = 8, so q = 5
Check: 5(8) + 3(5) = 55 ✓
Check: 5(8) + 3(5) = 55 ✓
Answer: Pen ₹8, pencil ₹5
[Attempt any ONE of the two options of Q20]
Section D — Case Study (2 × 4 = 8 marks)
A test paper has 20 questions in all. Each question is either Type A (3 marks) or Type B (5 marks), and the paper is worth 76 marks.
Marking Scheme: Part (a) 1 mark · Part (b) 1.5 marks · Part (c) 1.5 marks.
Total: 4 Marks
Total: 4 Marks
(a) Form the equations
Let Type A = a and Type B = b.
a + b = 20
3a + 5b = 76
a + b = 20
3a + 5b = 76
(b) Number of Type B questions
From a + b = 20: 3a + 3b = 60
Subtract from 3a + 5b = 76: 2b = 16 → b = 8
Subtract from 3a + 5b = 76: 2b = 16 → b = 8
(c) Number of Type A questions
a = 20 − 8 = 12
Check: 12(3) + 8(5) = 36 + 40 = 76 ✓
Check: 12(3) + 8(5) = 36 + 40 = 76 ✓
Answer: 12 Type A questions and 8 Type B questions
Answer either Option 1 or Option 2.
Option 1 — Taxi fare
A taxi charges a fixed amount plus a fixed rate for every kilometre. A 10 km ride costs ₹105 and a 15 km ride costs ₹155.
Marking Scheme: Part (a) 1 mark · Part (b) 1.5 marks · Part (c) 1.5 marks.
Total: 4 Marks
Total: 4 Marks
(a) Form the equations
Let the fixed charge be f and the per-km rate be r.
f + 10r = 105
f + 15r = 155
f + 10r = 105
f + 15r = 155
(b) Per-km rate
Subtracting: 5r = 50 → r = 10
(c) Fixed charge and 25 km fare
f = 105 − 10(10) = 5
For 25 km: 5 + 25(10) = 5 + 250 = 255
For 25 km: 5 + 25(10) = 5 + 250 = 255
Answer: (a) f + 10r = 105, f + 15r = 155 (b) ₹10 per km (c) fixed ₹5; 25 km = ₹255
— OR —
Option 2
A shopkeeper has small and large boxes. Three small boxes and two large boxes weigh 31 kg. Two small boxes and three large boxes weigh 34 kg.
Marking Scheme: Part (a) 1 mark · Part (b) 1.5 marks · Part (c) 1.5 marks.
Total: 4 Marks
Total: 4 Marks
(a) Form the equations
Let a small box weigh s kg and a large box weigh l kg.
3s + 2l = 31
2s + 3l = 34
3s + 2l = 31
2s + 3l = 34
(b) Weight of one small box
Adding: 5s + 5l = 65 → s + l = 13
Subtracting: s − l = −3 → l − s = 3
Solving: l = 8, s = 5
Subtracting: s − l = −3 → l − s = 3
Solving: l = 8, s = 5
(c) Weight of 4 small and 4 large boxes
4 × 5 + 4 × 8 = 20 + 32 = 52 kg
Check: 3(5) + 2(8) = 31 ✓
Check: 3(5) + 2(8) = 31 ✓
Answer: (a) 3s + 2l = 31, 2s + 3l = 34 (b) small 5 kg (c) large 8 kg, total 52 kg
[Attempt any ONE of the two options of Q22]
Self-assessment for Sections B, C and D (28 marks)
The MCQ section is scored for you. For the written questions, tick each one you got right out of the marks shown, and add them up honestly.
Written total: 0 / 28
Keep going with these
Written and checked by Pandey Sir · Pandey Sir’s BEST Classes, Dammaiguda, Hyderabad. Every answer on this paper is worked out and checked before publishing.
This is board-pattern practice for a Pair of Linear Equations in Two Variables, built to the current CBSE Class 10 Maths syllabus 2026-27. It is not a prediction of any examination paper.
Curriculum scope checked against cbseacademic.nic.in. Spotted a mistake? Tell us here.