Pair of Linear Equations in Two Variables: Important Questions and Answers for Class 10

Pair of Linear Equations in Two Variables is one of the most important chapters in Class 10 Mathematics, as it introduces students to methods of solving algebraic problems. The chapter covers graphical methods, substitution, elimination, and cross-multiplication techniques, which are essential for solving real-life mathematical situations. By solving a variety of important and exam-oriented questions, students can learn how to apply different methods effectively and identify the most suitable approach for each problem. In this guide, you will find carefully selected important questions covering all major concepts from the Class 10 chapter, Pair of Linear Equations in Two Variables, designed according to the latest syllabus and exam pattern to support effective preparation and revision.

Our subject experts have provided detailed solutions for these problems based on the latest CBSE syllabus and the NCERT textbook. This material helps students revise the chapter easily and perform well in the final examination.

Table of Contents

• Key Concepts: Pair of Linear Equations in Two Variables for Class 10
• Section A: 1-Mark Questions (MCQ & Very Short Answer)
• Section B: 2-Mark Questions (Short Answer)
• Section C: 3-Mark Questions
• Section D: 5-Mark Questions (Long Answer & Word Problems)
• Section E: Previous Year Questions (PYQs) with Solutions
• Section F: HOTS & Case-Study Based Questions
• Exam Tips & Common Mistakes to Avoid

Key Concepts: Pair of Linear Equations in Two Variables for Class 10

• Standard Form: 

A pair of linear equations in two variables is written as:

a₁x + b₁y + c₁ = 0

a₂x + b₂y + c₂ = 0

• Conditions for Nature of Solutions: 

By comparing the ratios of coefficients, you can immediately tell whether the system has one solution, no solution, or infinitely many without solving it. This is one of the most repeatedly tested concepts.

 

• Method 1: Substitution Method

Express one variable in terms of the other from one equation, then substitute into the second. Best used when one variable has a coefficient of 1 or −1.

• Method 2: Elimination Method

Multiply one or both equations by suitable numbers to make the coefficients of one variable equal, then add or subtract to eliminate it. This is generally the fastest method in exams.

• Method 3: Cross-Multiplication Method

Cross-Multiplication Formula: x / (b₁c₂ − b₂c₁) = y / (c₁a₂ − c₂a₁) = 1 / (a₁b₂ − a₂b₁)

Section A: 1-Mark Questions (MCQ & Very Short Answer)

Q1: How many solutions does the pair of equations y = 0 and y = −5 have?

Solution: y = 0 and y = −5 are two horizontal lines. They are parallel to each other (both parallel to the x-axis) and never meet.

The pair has NO solution. (Inconsistent system)

Q2: For what value of k do the equations 2x + 3y = 5 and 4x + ky = 10 have infinitely many solutions?

Solution: For infinitely many solutions: a₁/a₂ = b₁/b₂ = c₁/c₂

a₁/a₂ = 2/4 = 1/2

b₁/b₂ = 3/k

For infinitely many solutions:  3/k = 1/2

k = 6

Verify: c₁/c₂ = 5/10 = 1/2  

Q3: The pair of lines 6x − 3y + 10 = 0 and 2x − y + 9 = 0 are 

(a) intersecting at one point (b) parallel (c) coincident (d) intersecting at two points

Solution: 

a₁/a₂ = 6/2 = 3

b₁/b₂ = -3/-1 = 3

c₁/c₂ = 10/9

a₁/a₂ = b₁/b₂ ≠ c₁/c₂  

Therefore, the lines are parallel. 

Answer: (b)

Q4: The pair of lines 3x + 2y = 7 and 4x + 8y − 11 = 0 are — (a) perpendicular (b) parallel (c) intersecting (d) coincident

Solution:

a₁/a₂ = 3/4

b₁/b₂ = 2/8 = 1/4

3/4 ≠ 1/4  

Therefore, the lines intersect. 

Answer: (c) intersecting

Q5: Can a pair of linear equations in two variables have exactly two solutions? Justify.

Solution: No. Each linear equation represents a straight line. Two straight lines can either intersect at exactly one point (unique solution), be parallel (no solution), or coincide (infinitely many solutions). There is no geometric configuration where two distinct lines cross at two separate points.

Section B: 2-Mark Questions (Short Answer)

Q6: Solve the pair of equations by substitution method: x + y = 14 and x − y = 4

Solution:  From equation (1):

x + y = 14  

⇒  x = 14 − y   ...(1)

Substitute in equation (2):

(14 − y) − y = 4

⇒ 14 − 2y = 4

⇒ 2y = 10

y = 5

Back-substitute:

x = 14 − 5 = 9

Solution: x = 9, y = 5

Q7: Find the value of k for which the pair x + 2y − 5 = 0 and 3x + ky + 15 = 0 has a unique solution.

Solution: For a unique solution: a₁/a₂ ≠ b₁/b₂

a₁/a₂ = 1/3

b₁/b₂ = 2/k

Unique solution condition:  1/3 ≠ 2/k

⇒  k ≠ 6

The system has a unique solution for all values of k except k = 6.

Q8: Solve: 2x + y = 6 and 2x − y = 2 by elimination method.

Solution:

2x + y = 6   ...(1)

2x − y = 2   ...(2)

Adding (1) and (2):

4x = 8   

⇒  x = 2

Substituting in (1):

2(2) + y = 6

y = 6 − 4 = 2

Solution: x = 2, y = 2

Q9: The larger of two supplementary angles exceeds the smaller by 18°. Find both angles.

Solution: Let the larger angle = x, smaller angle = y

Supplementary angles:  x + y = 180°  ...(1)

Given condition: x − y = 18°   ...(2)

Adding (1) and (2):  2x = 198°  

⇒  x = 99°

From (1): y = 180 − 99 = 81°

The angles are 99° and 81°.

Section C: 3-Mark Questions

Q10: Solve by elimination method: 3x + 4y = 10 and 2x − 2y = 2

Solution: 

3x + 4y = 10   ...(1)

2x − 2y =  2   ...(2)

Multiply (2) by 2:

4x − 4y = 4    ...(3)

Add (1) and (3):

7x = 14   

⇒ x = 2

Substitute in (2):

2(2) − 2y = 2

4 − 2y    = 2

2y  = 2   

⇒ y = 1

Solution: x = 2, y = 1

Verification: 3(2) + 4(1) = 6 + 4 = 10 

Q11: A fraction becomes 1/3 when 2 is subtracted from its numerator, and becomes 1/2 when 1 is subtracted from its denominator. Find the fraction.

Solution: Let the numerator = x, denominator = y, so fraction = x/y.

Condition 1: (x − 2)/y = 1/3

3(x − 2) = y

3x − 6   = y

3x − y   = 6   ...(1)

Condition 2: x/(y − 1) = 1/2

2x = y − 1

2x − y = −1    ...(2)

Subtract (2) from (1):

(3x − y) − (2x − y) = 6 − (−1)

 ⇒ x = 7

From (2):

2(7) − y = −1

y = 14 + 1 = 15

The fraction is 7/15.

Verify: (7−2)/15 = 5/15 = 1/3 and  7/(15−1) = 7/14 = 1/2

Q12: For what value of k will the pair of equations kx + 3y = k − 3 and 12x + ky = k have no solution?

Solution: For no solution: a₁/a₂ = b₁/b₂ ≠ c₁/c₂

a₁/a₂ = k/12

b₁/b₂ = 3/k

Setting a₁/a₂ = b₁/b₂:

k/12 = 3/k

k²   = 36

k    = ±6

Check k = 6:

c₁/c₂ = (6−3)/6 = 3/6 = 1/2

b₁/b₂ = 3/6 = 1/2

Here b₁/b₂ = c₁/c₂. This gives infinitely many solutions, not no solution.

Check k = −6:

a₁/a₂ = −6/12 = −1/2

b₁/b₂ = 3/(−6) = −1/2

c₁/c₂ = (−6−3)/(−6) = −9/−6 = 3/2

a₁/a₂ = b₁/b₂ = −1/2  ≠  c₁/c₂ = 3/2  

Therefore, k = −6 gives no solution.

Q13: The sum of digits of a two-digit number is 12. The number obtained by reversing the digits exceeds the original number by 18. Find the number.

Solution: Let the tens digit = x, units digit = y. 

So the number = 10x + y.

Condition 1 (sum of digits):

x + y = 12   ...(1)

Condition 2 (reversed number exceeds original by 18):

(10y + x) − (10x + y) = 18

 ⇒ 9y − 9x = 18

 ⇒ y − x   = 2    ...(2)

Adding (1) and (2):

2y = 14   

 ⇒    y = 7

 ⇒ x  = 12 − 7 = 5

Original number = 10(5) + 7 = 57

Section D: 5-Mark Questions (Long Answer & Word Problems)

Q14: A man travels 600 km partly by train and partly by car. It takes 8 hours 40 minutes if he travels 320 km by train and the rest by car. It would take 30 minutes more if he travels 200 km by train and the rest by car. Find the speed of the train and the car separately.

Solution: Let speed of train = x km/h, speed of car = y km/h.

Total distance = 600 km

Case 1: 320 km by train + 280 km by car = 8 hr 40 min = 26/3 hr

320/x + 280/y = 26/3   ...(1)

Case 2: 200 km by train + 400 km by car = 9 hr 10 min = 55/6 hr

200/x + 400/y = 55/6   ...(2)

Let 1/x = a and 1/y = b:

 ⇒ 320a + 280b = 26/3   ...(3)

 ⇒ 200a + 400b = 55/6   ...(4)

Multiply (3) by 5 and (4) by 8:

 ⇒ 1600a + 1400b = 130/3

 ⇒ 1600a + 3200b = 220/3

Subtract:

1800b = 90/3 = 30

 ⇒ b = 30/1800 = 1/60

From (3):

320a = 26/3 − 280 × (1/60)

 ⇒ 320a = 26/3 − 14/3

 ⇒  320a = 12/3 = 4

 ⇒ a = 4/320 = 1/80

Therefore, 

x = 1/a = 80 km/h

y = 1/b = 60 km/h

Hence, Speed of train = 80 km/h, Speed of car = 60 km/h

Q15: A father's age is three times the sum of the ages of his two children. After 5 years his age will be two times the sum of their ages. Find the present age of the father.

Solution: Let father's present age = x years.

Let sum of two children's present ages = y years.

Current condition:      x = 3y            ...(1)

After 5 years:

Father's age = x + 5

Sum of children's ages  = y + 5 + 5 = y + 10

Condition: x + 5 = 2(y + 10)

 ⇒ x + 5 = 2y + 20

 ⇒ x − 2y = 15       ...(2)

Substitute (1) in (2):

3y − 2y = 15

 ⇒ y = 15

From (1):

x = 3 × 15 = 45

Therefotre, Father's present age = 45 years.

Q16: A half perimeter of a rectangular garden is 36 m. If the length is 4 m more than its width, find the dimensions of the garden. Also find its area.

Solution: Let length = l metres, width = b metres.

Half-perimeter:   l + b = 36   ...(1)

Length condition: l − b = 4    ...(2)

Adding (1) and (2):

2l = 40   

 ⇒ l = 20 m

From (1):

b = 36 − 20 = 16 m

i.e., Length = 20 m, Width = 16 m

 ⇒ Area = 20 × 16 = 320 m²

Q17: 4 men and 6 boys can finish a piece of work in 5 days. 3 men and 4 boys can finish it in 7 days. Find the time taken by 1 man alone and 1 boy alone to finish the work.

Solution: Let 1 man take x days and 1 boy take y days to finish the work alone.

In 1 day, 1 man does 1/x of work, 1 boy does 1/y of work.

Condition 1 (4 men + 6 boys finish in 5 days):

5(4/x + 6/y) = 1

⇒ 20/x + 30/y  = 1   ...(1)

Condition 2 (3 men + 4 boys finish in 7 days):

7(3/x + 4/y) = 1

⇒  21/x + 28/y  = 1   ...(2)

Let 1/x = a, 1/y = b:

20a + 30b = 1   ...(3)

21a + 28b = 1   ...(4)

Multiply (3) by 21 and (4) by 20:

⇒ 420a + 630b = 21

⇒ 420a + 560b = 20

Subtract:

⇒ 70b = 1   

⇒ b = 1/70

⇒ y = 70 days

From (3):

⇒ 20a = 1 − 30/70 

= 1 − 3/7 = 4/7

⇒ a   = 4/(7 × 20) = 1/35

⇒ x = 35 days

1 man alone can finish the work in 35 days and 1 boy alone can finish the work in 70 days.

Section E: Previous Year Questions (PYQs) with Solutions

These questions have appeared in actual CBSE board exams. Practising PYQs is one of the most effective things you can do in the last few weeks before your exam. Pay close attention to the type of problem and the method used.

Q 18: Solve the pair: 3x + 4y = 10 and 2x − 2y = 2 (CBSE 2019)

Solution: 3x + 4y = 10 ...(1)

2x − 2y = 2 ...(2)

Multiply (2) by 2:  4x − 4y = 4   ...(3)

Add (1) and (3):    7x = 14  

 ⇒ x = 2

From (2):  2(2) − 2y = 2  

⇒   y = 1

⇒ x = 2, y = 1

Q19: The larger of two supplementary angles exceeds the smaller by 18°. Find both angles. (CBSE 2019)

Solution: x + y = 180,  x − y = 18

2x = 198  

⇒  x = 99°

⇒ y = 81°

Angles are 99° and 81°.

Q20: A father's age is three times the sum of the ages of his two children. After 5 years his age will be twice the sum of their ages. Find the father's present age. (CBSE 2019)

Solution: x = 3y  and  x − 2y = 15

Substituting: 3y − 2y = 15  

⇒   y = 15

⇒ Father's age x = 45 years.

Q 21: A fraction becomes 1/3 when 2 is subtracted from its numerator. It becomes 1/2 when 1 is subtracted from its denominator. Find the fraction. (CBSE 2019)

Solution: 3x − y = 6   ...(1)

2x − y = −1  ...(2)

Subtract: x = 7,  y = 15

Fraction = 7/15

Q 22: In rectangle ABCD, AB = (x + y) cm and BC = (x − y) cm. Given AB = 30 cm and BC = 14 cm. Find x and y. (CBSE 2018)

Solution: x + y = 30   ...(1)

x − y = 14   ...(2)

Adding: 2x = 44  

⇒   x = 22

From (1): y = 30 − 22 = 8

⇒ x = 22 cm, y = 8 cm

Q 23: At a sports event, prize money per student for Hockey = ₹x and for Cricket = ₹y. Given: 5x + 4y = 9500 and 4x + 3y = 7370. (i) Find prize for Hockey and Cricket separately. (ii) Which game's prize is more and by how much? (iii) Total prize if 2 students each from both games? (CBSE 2023)

Solution: 

5x + 4y = 9500   ...(1)

4x + 3y = 7370   ...(2)

Multiply (1) by 3 and (2) by 4, then subtract:

15x + 12y − (16x + 12y) = 28500 − 29480

−x = −980   

⇒ x = ₹980 (Hockey)

Multiply (1) by 4 and (2) by 5, then subtract:

20x + 16y − (20x + 15y) = 38000 − 36850

y = 1150   

⇒ y = ₹1150 (Cricket)

(i) Hockey = ₹980, Cricket = ₹1150

(ii) Cricket prize is more by ₹170

(iii) Total = 2 × 980 + 2 × 1150 = 1960 + 2300 = ₹4260

Q 24: For what value of k will pair of equations have no solution? 3x + y = 1 and (2k − 1)x + (k − 1)y = 2k + 1. (CBSE 2012)

Solution: For no solution: a₁/a₂ = b₁/b₂ ≠ c₁/c₂

a₁/a₂ = 3/(2k−1)

b₁/b₂ = 1/(k−1)

Setting equal: 3/(2k−1) = 1/(k−1)

3(k−1)    = 2k − 1

⇒ 3k − 3    = 2k − 1

⇒ k = 2

Q 25: Solve graphically: x + 3y = 6 and 2x − 3y = 12. Also find the area of the triangle formed by the two lines and the y-axis.(CBSE 2011)

Solution: Table of values for x + 3y = 6  (y = (6−x)/3):

x = 0 ⇒  y = 2;   

x = 6 ⇒  y = 0;  

x = −3 ⇒  y = 3

Table of values for 2x − 3y = 12  (y = (2x−12)/3):

x = 0 ⇒  y = −4;  

x = 6 ⇒  y = 0;   

x = 3 ⇒  y = −2

Intersection point (solve algebraically):

x + 3y = 6    ...(1)

2x − 3y = 12  ...(2)

Adding: 3x = 18  ⇒   x = 6, y = 0

Intersection = (6, 0)

y-axis intercepts:

Line 1 at x = 0: y = 2  ⇒   point A(0, 2)

Line 2 at x = 0: y = −4 ⇒   point B(0, −4)

Triangle vertices: (6, 0), (0, 2), (0, −4)

Base AB on y-axis = |2 − (−4)| = 6 units

Height = perpendicular distance from (6,0) to y-axis = 6 units

Area = ½ × base × height = ½ × 6 × 6 = 18 sq units

Section F: HOTS & Case-Study Based Questions

Q 26: 2 women and 5 men can together finish a piece of embroidery in 4 days, while 3 women and 6 men can finish it in 3 days. Find the time taken by 1 woman alone and 1 man alone.

Solution: Let 1 woman take x days and 1 man take y days alone. In 1 day, 1 woman does 1/x work, 1 man does 1/y work.

4(2/x + 5/y) = 1  

⇒ 8/x + 20/y = 1   ...(1)

3(3/x + 6/y) = 1  

⇒  9/x + 18/y = 1   ...(2)

Let a = 1/x, b = 1/y:

8a + 20b = 1   ...(3)

9a + 18b = 1   ...(4)

Multiply (3) by 9 and (4) by 8:

72a + 180b = 9

72a + 144b = 8

Subtract: 36b = 1  

⇒  b = 1/36

⇒ y = 36 days

From (3): 8a = 1 − 20/36 = 16/36 = 4/9

⇒ a = 1/18  

⇒  x = 18 days

1 woman alone can finish the work in 18 days, and 1 man alone can finish the work in 36 days.

Q 27: The sum of the numerator and denominator of a fraction is 12. If 1 is added to both the numerator and denominator, the fraction becomes 3/4. Find the fraction.

Solution: Let numerator = x, denominator = y.

Condition 1:  x + y = 12         ...(1)

Condition 2: (x+1)/(y+1) = 3/4:

⇒ 4(x+1) = 3(y+1)

⇒ 4x − 3y = −1         ...(2)

From (1): x = 12 − y:

⇒ 4(12−y) − 3y = −1

⇒ 48 − 7y = −1

⇒ 7y = 49  

⇒ y = 7, x = 5

⇒ Fraction = 5/7.  [Condition: (5+1)/(7+1) = 6/8 = 3/4 ]

Q 28: Case Study

A coaching institute offers a 'Group Discount' scheme. 5 students from Class 10 and 3 students from Class 9 paid ₹15,700 total. If 3 students from Class 10 and 5 students from Class 9 paid ₹14,300 total, find the fee per student for each class. Also find the total fees if the ratio of Class 10 to Class 9 students is 2:3 and there are 10 students total.

Solution: Let Class 10 fee per student = ₹x, Class 9 fee per student = ₹y.

5x + 3y = 15700   ...(1)

3x + 5y = 14300   ...(2)

Add (1) and (2):   8x + 8y = 30000

⇒ x + y   = 3750   ...(3)

Subtract (2) from (1): 2x − 2y = 1400

⇒ x − y   = 700    ...(4)

From (3) and (4):

2x = 4450  

⇒  x = ₹2225

⇒ y  = 3750 − 2225 = ₹1525

If ratio Class 10 : Class 9 = 2:3 and total = 10 students:

Class 10 has 4 students; Class 9 has 6 students

Total fees = 4 × 2225 + 6 × 1525

= 8900 + 9150 = ₹18,050

Exam Tips & Common Mistakes to Avoid

Given below are exam tips for Pair of Linear Equations in Two Variables help students approach problems in a structured way, avoid common mistakes, and improve accuracy:

• Choose Your Method Wisely: If one variable has coefficient 1: Substitution. If coefficients can be matched easily: Elimination. Under time pressure, elimination is almost always faster.

• Always Verify Your Answer: After solving, substitute x and y back into BOTH original equations. 

• Define Variables Clearly: In word problems, always write ‘Let x = …’and ‘Let y = …’ explicitly. CBSE awards 1 mark for this step in 5-mark word problems.

• Graphical Questions: Use a ruler. Plot at least 3 points per line. Mark the intersection clearly. For area questions, identify the triangle's vertices precisely before applying the formula.

• Consistency Conditions: Memorise: a₁/a₂ ≠ b₁/b₂ ⇒  unique. Equal but ≠ c ratio ⇒  no solution. All three equal ⇒  infinite solutions. A 1-mark MCQ on this appears in almost every board exam.

• Word Problems: Translate First. Before doing any arithmetic, read the entire problem and write both equations. Many students write one equation and start solving, then realise they misread the second condition.
  
  

Common mistakes to avoid