CBSE 10 Maths Important MCQs for Chapter 3 Pair of Linear Equations in Two Variables with Answers – 2023-24

CBSE Class 10 Maths MCQs for  Chapter 3 – Pair of Linear Equations in Two Variables are provided here to practice for the upcoming CBSE Exam 2023-24. These questions will make students familiarised with the most important concepts to prepare the objective type questions for their Board Exam. All these questions are provided with correct answers and a detailed explanation to help in understanding the logic behind each answer.

1. Graphically, the pair of equations

6x – 3y + 10 = 0

2x – y + 9 = 0

Represents two lines which are:

(A) Intersecting at exactly one point.

(B) Intersecting at exactly two points.

(C) Coincident.

(D) Parallel

Answer:  (D)

Explanation:

Here

Explanation:

Here,

Therefore, the pair of equations has infinitely many solutions.

3. If a pair of linear equations is consistent, then the lines will be:

(A) Parallel

(B) Always coincident

(C) Intersecting or coincident

(D) Always intersecting

Answer:  (C)

Explanation: If a pair of linear equations is consistent the two lines represented by these equations definitely have a solution, this implies that either lines are intersecting or coincident.

4. The pair of equations y = 0 and y = –7 has

(A) One solution

(B) Two solutions

(C) Infinitely many solutions

(D) No solution

Answer:  (D)

8. Two numbers are in the ratio 5 : 6. If 8 is subtracted from each of the numbers, the ratio becomes 4 : 5. Then the numbers are:

(A) 40, 42

(B) 42, 48

(C) 40, 48

(D) 44, 50

Answer: (C)

Explanation:

According to given information

9. The solution of the equations x – y = 2 and x + y = 4 is:

(A) 3 and 5

(B) 5 and 3

(C) 3 and 1

(D) –1 and –3

Answer: (C)

Explanation: Adding both equations, we have:

10. For which values of a and b, will the following pair of linear equations have infinitely many solutions?

x + 2y = 1

(a – b)x + (a + b)y = a + b – 2

(A) a = 2 and b = 1

(B) a = 2 and b = 2

(C) a =  ̶ 3 and b = 1

(D) a = 3 and b = 1

Answer: (D)

Explanation: For infinitely many solutions:

Solving equation (i) and (ii), we get a = 3 and b = 1.

11. The father’s age is six times his son’s age. Four years hence, the age of the father will be four times his son’s age. The present ages, in years, of the son and the father are, respectively

(A) 4 and 24

(B) 5 and 30

(C) 6 and 36

(D) 3 and 24

Answer:  (C)

Explanation: Let the age of father be x and of son is y.

Then according to question,

x = 6y …..(i)

Four years hence age of son will be y + 4 and age of father will be x + 4

Then according to question,

x + 4 = 4 (y + 4)

x – 4y = 12      …..(ii)

Solving equations (i) and (ii) we get:

y = 6  and x = 36

12. Rakshita has only Rs. 1 and Rs. 2 coins with her. If the total number of coins that she has is 50 and the amount of money with her is Rs 75, then the number of Rs.1 andRs.2 coins is, respectively

(A) 35 and 15

(B) 35 and 20

(C) 15 and 35

(D) 25 and 25

Answer:  (D)

Explanation:

Let her number of Rs.1 coins are x

Let the number of Rs.2 coins are y

Then

By the given conditions

x + y = 50   …..(i)

1 × x + 2 × y = 75

⇒ x + 2y = 75    …..(ii)

Solving equations (i) and (ii) we get:

(x + 2y) – (x + y) = 75 – 50

⇒ y = 25

Therefore, x = 50 – 25 = 25

So the number of coins are 25, 25 each.

13. In a competitive examination, one mark is awarded for each correct answer while 1/2 mark is deducted for every wrong answer. Jayanti answered 120 questions and got 90 marks. How many questions did she answer correctly?

(A) 100

(B) 95

(C) 90

(D) 60

Answer: (A)

Explanation: Let x be the number of correct answers of the questions in a competitive exam.

Then, 120 − x be the number of wrong answers

Then by given condition

14. The angles of a cyclic quadrilateral ABCD are:

Then value of x and y are:

(A) x = 20^o and y = 30^o                                              

(B) x = 40^o and y = 10^o

(C) x = 44^o and y = 15^o

(D) x = 15^o and y = 15^o

Answer: (A)

Explanation: In cyclic quadrilateral, sum of opposite angles is 180⁰

Therefore

6x + 10 + x + y = 180

⇒ 7x + y = 170           …..(i)

5x + 3y – 10 = 180

⇒ 5x + 3y = 190         …..(ii)

Multiplying equations (i) and (ii), we get:

x = 20^o and y = 30^o

15. A shopkeeper gives books on rent for reading. She takes a fixed charge for the first two days, and an additional charge for each day thereafter. Reema paid Rs. 22 for a book kept for six days, while Ruchika paid Rs 16 for the book kept for four days, then the charge for each extra day is:

(A) Rs 5

(B) Rs 4

(C) Rs 3

(D) Rs.2

Answer: (C)

Explanation: Let Rs. x be the fixed charge and Rs. y be the charge for each extra day.

Then by the given conditions

x + 4y = 22                  …..(i)

x + 2y = 16                  …..(ii)

Subtracting equation (ii) from (i), we get:

y = Rs. 3