Class 10 Maths MCQs for Chapter 3 (Pair of Linear Equations in Two Variables) are provided here online with answers. These multiple choice questions will help students to score good marks in the board exam. These objective questions are prepared as per the latest CBSE syllabus and NCERT textbook. Practising these questions will help them to develop problem-solving skills.

## Class 10 Maths MCQs for Pair of Linear Equations in Two Variables

Students are required to solve these questions here and choose the appropriate answer for it. They can verify their answers with the given answers here.

#### Below are the MCQs for Chapter 3

**1.The pairs of equations x+2y-5 = 0 and -4x-8y+20=0 have:**

(a)Unique solution

(b)Exactly two solutions

(c)Infinitely many solutions

(d)No solution

Answer:** (c)**

Explanation:

a1/a2 = 1/-4

b1/b2 = 2/-8 = 1/-4

c1/c2 = -5/20 = -¼

This shows:

a1/a2 = b1/b2 = c1/c2

Therefore, the pair of equations has infinitely many solutions.

**2. If a pair of linear equations is consistent, then the lines are:**

(a)Parallel

(b)Always coincident

(c)Always intersecting

(d)Intersecting or coincident

Answer:** d**

Explanation: Because the two lines definitely have a solution.

**3. The pairs of equations 9x + 3y + 12 = 0 and 18x + 6y + 26 = 0 have**

(a)Unique solution

(b)Exactly two solutions

(c)Infinitely many solutions

(d)No solution

Answer:** d**

Explanation: Given, 9x + 3y + 12 = 0 and 18x + 6y + 26 = 0

a1/a2 = 9/18 = 1/2

b1/b2 = 3/6 = 1/2

c1/c2 = 12/26 = 6/13

Since, a1/a2 = b1/b2 ≠ c1/c2

So, the pairs of equations are parallel and the lines never intersect each other at any point, therefore there is no possible solution.

**4. If the lines 3x+2ky – 2 = 0 and 2x+5y+1 = 0 are parallel, then what is the value of k?**

(a)4/15

(b)15/4

(c)⅘

(d)5/4

Answer:** (b)**

Explanation: The condition for parallel lines is:

a1/a2 = b1/b2 ≠ c1/c2

Hence, 3/2 = 2k/5

k=15/4

**5. If one equation of a pair of dependent linear equations is -3x+5y-2=0. The second equation will be:**

(a)-6x+10y-4=0

(b)6x-10y-4=0

(c)6x+10y-4=0

(d)-6x+10y+4=0

Answer:** a**

Explanation: The condition for dependent linear equations is:

a1/a2 = b1/b2 = c1/c2

For option a,

a1/a2 = b1/b2 = c1/c2= ½

**6.The solution of the equations x-y=2 and x+y=4 is:**

(a)3 and 1

(b)4 and 3

(c)5 and 1

(d)-1 and -3

Answer: **a**

Explanation: x-y =2

x=2+y

Substituting the value of x in the second equation we get;

2+y+y=4

2+2y=4

2y = 2

y=1

Now putting the value of y, we get;

x=2+1 = 3

Hence, the solutions are x=3 and y=1.

**7. A fraction becomes 1/3 when 1 is subtracted from the numerator and it becomes 1/4 when 8 is added to its denominator. The fraction obtained is:**

(a)3/12

(b)4/12

(c)5/12

(d)7/12

Answer: **c**

Explanation: Let the fraction be x/y

So, as per the question given,

(x -1)/y = 1/3 => 3x – y = 3…………………(1)

x/(y + 8) = 1/4 => 4x –y =8 ………………..(2)

Subtracting equation (1) from (2), we get

x = 5 ………………………………………….(3)

Using this value in equation (2), we get,

4×5 – y = 8

y= 12

Therefore, the fraction is 5/12.

**8. The solution of 4/x+3y=14 and 3/x-4y=23 is:**

(a)⅕ and -2

(b)⅓ and ½

(c)3 and ½

(d)2 and ⅓

Answer:** a**

Explanation: Let 1/x = m

4m + 3y = 14

3m – 4y = 23

By cross multiplication we get;

m/(-69-56) = y/(-42-(-92)) = 1/(-16-9)

m/-125=y/50=-1/25

m/-125 = -1/25 and y/50=-1/25

m=5 and y=-2

m=1/x or x=1/m = ⅕

**9. Ritu can row downstream 20 km in 2 hours, and upstream 4 km in 2 hours. Her speed of rowing in still water and the speed of the current is:**

(a)6km/hr and 3km/hr

(b)7km/hr and 4km/hr

(c)6km/hr and 4km/hr

(d)10km/hr and 6km/hr

Answer: **c**

Explanation: Let, Speed of Ritu in still water = x km/hr

Speed of Stream = y km/hr

Now, speed of Ritu, during,

Downstream = x + y km/h

Upstream = x – y km/h

As per the question given,

2(x+y) = 20

Or x + y = 10……………………….(1)

And, 2(x-y) = 4

Or x – y = 2………………………(2)

Adding both the equations, we get,

2x=12

x = 6

Putting the value of x in eq.1, we get,

y = 4

Therefore,

Speed of Ritu is still water = 6 km/hr

Speed of Stream = 4 km/hr

**10. The angles of cyclic quadrilaterals ABCD are: A = (6x+10), B=(5x)°, C = (x+y)° and D=(3y-10)°. The value of x and y is:**

(a)x=20° and y = 10°

(b)x=20° and y = 30°

(c)x=44° and y=15°

(d)x=15° and y=15°

Answer: **b**

Explanation: We know, in cyclic quadrilaterals, the sum of the opposite angles are 180°.

Hence,

A + C = 180°

6x+10+x+y=180 =>7x+y=170°

And B+D=180°

5x+3y-10=180 =>5x+3y=190°

By solving the above two equations we get;

x=20° and y = 30°.