# CBSE 10th Maths Exam 2021: Important MCQs from Chapter 3 Pair of Linear Equations in Two Variables with Answers

Class 10 MCQs on Maths Chapter 3 - Pair of Linear Equations in Two Variables are provided here. All questions are important for CBSE Maths Exam 2021.

*CBSE Class 10 Maths MCQs Chapter 3*

MCQs on Class 10 Maths Chapter 3- Pair of Linear Equations in Two Variables are provided here to practice for the upcoming CBSE Exam. These questions will make students familiarised with the important concepts to prepare the objective type questions for the exam. All these questions are provided with correct answers and a detailed explanation to understand the logic behind each answer. PDF of all the MCQs is also made available for students to download the questions and practice in offline mode.

**Note - **Students must note that **Cross Multiplication Method has been excluded from the revised CBSE syllabus of Class10 Maths**. So, students should prepare according to the new syllabus only.

**Check below the solved MCQs from Class 10 Maths Chapter 3 ****Pair of Linear Equations in Two Variables:**

**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

Therefore, lines are parallel.

**2.** The pair of equations x + 2y – 5 = 0 and −3x – 6y + 15 = 0 have:

(A) A unique solution

(B) Exactly two solutions

(C) Infinitely many solutions

(D) No solution

**Answer: (C)**

**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)**

**Explanation: **The graph of equations will be parallel lines. So the equations have no solution.

**5.** If the lines given by

3x + 2ky = 2

2x + 5y + 1 = 0

are parallel, then the value of k is

(A) 5/4

(B) 2/5

(C) 15/4

(D) 3/2

**Answer: (C)**

**Explanation:**

For parallel lines

** **

**6.** The value of c for which the pair of equations cx – y = 2 and 6x – 2y = 3 will have infinitely many solutions is

(A) 3

(B) – 3

(C) –12

(D) no value

**Answer: (A)**

**Explanation:** For infinitely many solutions:

**7.** One equation of a pair of dependent linear equations is –5x + 7y – 2 = 0. The second equation can be

(A) 10x + 14y + 4 = 0

(B) –10x – 14y + 4 = 0

(C) –10x + 14y + 4 = 0

(D) 10x – 14y = –4

**Answer: (D)**

**Explanation:** For dependent pair, the two lines must have

**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 180^{0}

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

**All the above questions can also be downloaded in PDF from the following link:**

**Important Articles for the Preparations of Class 10 Maths Exam 2020:**

We have prepared some articles which will help the class 10 students in preparations for their Maths exam by summarising all important resources in one place. Students can check the following links to explore the important articles to prepare in an effective manner and perform well in the exam:

**CBSE Class 10 Maths Exam Pattern 2021 with Blueprint & Marking Scheme**

**CBSE Class 10 Maths Important Questions and Answers for Board Exam 2021**

**CBSE Class 10 Maths Solved Previous Year Question Papers**

**CBSE Class 10th Maths Chapter-wise Important Formulas, Theorems & Properties**