HOTS Questions on Class 10 Maths Chapter 3 ‘Pair of Linear Equations in Two Variables’ with Answers

Higher-Order Thinking Skills (HOTS) question set for Chapter 3 ‘Pair of Linear Equations in Two Variables’ for Class 10 is crafted to move students beyond routine solution techniques and develop strong, flexible problem-solving skills in linear systems. Aligned with CBSE and NCERT objectives, the questions emphasise conceptual understanding and creative application of ideas such as graphical and algebraic methods of solving systems, interpretation of solutions, consistency and dependence, and modelling real-world situations with linear equations. Each question includes a clear answer and brief reasoning to support classroom discussion and self-study. Download the PDF to access the same HOTS questions in a printable format, making offline learning and exam revision easier.

Solved HOTS Questions on Pair of Linear Equations in Two Variables

Question 1: For what value of k will the pair of equations x − ky + 4 = 0 and 2x − 4y − 8 = 0 be inconsistent?

Solution:

For the system to be inconsistent (no solution):

a₁/a₂ = b₁/b₂ ≠ c₁/c₂

Here: a₁/a₂ = 1/2, b₁/b₂ = −k/−4 = k/4

Setting a₁/a₂ = b₁/b₂:

1/2 = k/4 ⇒ k = 2

Answer: k = 2

Question 2: For what value of m will the system mx + 3y = m − 3 and 12x + my = m have no solution?

Solution:

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

m/12 = 3/m ⇒m² = 36 ⇒ m = ±6

Check m = 6: c₁/c₂ = (6−3)/6 = 3/6 = 1/2; b₁/b₂ = 3/6 = 1/2 

⇒ Equal, so this gives infinite solutions, not no solution.

Check m = −6: c₁/c₂ = (−6−3)/(−6) = −9/−6 = 3/2; b₁/b₂ = 3/−6 = −1/2 

⇒ 3/2 ≠ −1/2 

Answer: m = −6

Question 3: The sum of two numbers is 8. If their sum is four times their difference, find the numbers.

Solution:

Let the two numbers be x and y (x > y).

x + y = 8 … (1)

x + y = 4(x − y)

8 = 4x − 4y

4x − 4y = 8 ⇒  x − y = 2 … (2)

Adding (1) and (2): 2x = 10 ⇒  x = 5

From (1): y = 8 − 5 = 3

The two numbers are 5 and 3.

Question 4: In a deer park, the number of heads and legs of deer and human visitors were counted. There were 39 heads and 132 legs. Find the number of deer and the number of human visitors.

Solution:

Let number of deer = x, number of humans = y.

x + y = 39 … (1) [Total heads]

4x + 2y = 132 … (2) [Total legs]

From (1): y = 39 − x

Substituting in (2): 4x + 2(39 − x) = 132

4x + 78 − 2x = 132

2x = 54 ⇒ x = 27

y = 39 − 27 = 12

Number of deer = 27; Number of human visitors = 12.

Question 5: Solve the following pair of equations:

103x + 51y = 617

97x + 49y = 583

Solution:

Adding both equations:

200x + 100y = 1200

⇒ 2x + y = 12 … (3)

Subtracting the second from the first:

6x + 2y = 34

⇒ 3x + y = 17 … (4)

Subtracting (3) from (4):

x = 5

Substituting in (3): 2(5) + y = 12 ⇒ y = 2

Answer: x = 5, y = 2

Question 6: 8 men and 12 women can finish a piece of work in 10 days, while 6 men and 8 women can finish it in 14 days. Find the time taken by one man alone and one woman alone to finish the work.

Solution:

Let one man alone take 'a' days and one woman alone take 'b' days.

Work done by one man in 1 day = 1/a

Work done by one woman in 1 day = 1/b

Equation 1: 8/a + 12/b = 1/10 … (1)

Equation 2: 6/a + 8/b = 1/14 … (2)

Let 1/a = p and 1/b = q:

8p + 12q = 1/10 

⇒ 80p + 120q = 1 … (1')

6p + 8q = 1/14 

⇒ 84p + 112q = 1 … (2')

Multiply (1') by 84 and (2') by 80:

6720p + 10080q = 84

6720p + 8960q = 80

Subtracting: 1120q = 4 

⇒ q = 1/280

Substituting: 80p + 120(1/280) = 1

80p = 1 − 3/7 = 4/7

p = 1/140

One man alone can finish the work in 140 days; one woman alone in 280 days.

Question 7:

Aftab tells his daughter, "Seven years ago, I was seven times as old as you were then. Also, three years from now, I shall be three times as old as you will be." Represent this situation algebraically and find their current ages.

Solution:

Let Aftab's present age = x and daughter's present age = y.

Seven years ago:

x − 7 = 7(y − 7)

x − 7 = 7y − 49

x − 7y = −42 … (1)

Three years from now:

x + 3 = 3(y + 3)

x + 3 = 3y + 9

x − 3y = 6 … (2)

Subtracting (1) from (2):

4y = 48 

⇒ y = 12

From (2): x = 6 + 3(12) = 6 + 36 = 42

Aftab's present age = 42 years; Daughter's present age = 12 years.

Question 8: Cars are parked in a parking lot in equal rows. If 3 more cars are added per row, there would be 1 row fewer. If 3 fewer cars are placed in a row, there would be 2 more rows. Find the total number of cars.

Solution:

Let the number of cars per row = x and number of rows = y.

Total cars = xy

Condition 1: (x + 3)(y − 1) = xy

xy − x + 3y − 3 = xy

3y − x = 3 … (1)

Condition 2: (x − 3)(y + 2) = xy

xy + 2x − 3y − 6 = xy

2x − 3y = 6 … (2)

Adding (1) and (2):

x = 9

From (1): 3y − 9 = 3 ⇒ 3y = 12 ⇒ y = 4

Total cars = xy = 9 × 4 = 36 cars

Questions PDF with worked-out examples for Class 10 Chapter 3: Pair of Linear Equations in Two Variables, perfect for last-minute CBSE exam revision.

Class 10 Chapter 3: Pair of Linear Equations in Two Variables HOTS PDF