  CBSE Class 10 Pair of Linear Equations MCQ New Pattern

## CBSE Class 10 Pair of Linear Equations MCQ New Pattern By Clarify Knowledge

CBSE Class 10 Pair of Linear Equations MCQ New Pattern 2022

## CBSE Class 10 Pair of Linear Equations MCQ Table

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

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

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

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

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

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

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

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

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

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) x=20° and y = 30°

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°.

11. The pair of equations x = a and y = b graphically represents lines which are

(a) parallel

(b) intersecting at (b, a)

(c) coincident

(d) intersecting at (a, b)

Answer: (d) intersecting at (a, b)

The pair of equations x = a and y = b graphically represents lines which are intersecting at (a, b).

12. The pair of equations 5x – 15y = 8 and 3x – 9y = 24/5 has

(a) one solution

(b) two solutions

(c) infinitely many solutions

(d) no solution

Explanation:

The given pair of equations are 5x – 15y = 8 and 3x – 9y = 24/5.

Comparing with the standard form,

a1 = 5, b1 = -15, c1 = -8

a2 = 3, b2 = -9, c2 = -24/5

a1/a2 = 5/3

b1/b2 = -15/-9 = 5/3

c1/c2 = -8/(-24/5) = 5/3

Thus, a1/a2 = b1/b2 = c1/c2

Hence, the given pair of equations has infinitely many solutions.

13. The pair of equations x + 2y + 5 = 0 and –3x – 6y + 1 = 0 have

(a) a unique solution

(b) exactly two solutions

(c) infinitely many solutions

(d) no solution

Explanation:

Given pair of equations are x + 2y + 5 = 0 and –3x – 6y + 1 = 0.

Comparing with the standard form,

a1 = 1, b1 = 2, c1 = 5

a2 = -3, b2 = -6, c2 = 1

a1/a2 = -1/3

b1/b2 = 2/-6 = -1/3

c1/c2 = 5/1

Thus, a1/a2 = b1/b2 ≠ c1/c2

Hence, the given pair of equations has no solution.

14. 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

Explanation:

Given pair of equations are cx – y = 2 and 6x – 2y = 3.

Comparing with the standard form,

a1 = c, b1 = -1, c1 = -2

a2 = 6, b2 = -2, c2 = -3

a1/a2 = c/6

b1/b2 = -1/-2 = 1/2

c1/c2 = -2/-3 = ⅔

Condition for having infinitely many solutions is

a1/a2 = b1/b2 = c1/c2

c/6 = ½ = ⅔

Therefore, c = 3 and c = 4

Here, c has different values.

Hence, for no value of c the pair of equations will have infinitely many solutions.

15. If the lines representing the pair of linear equations a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 are coincident, then

(a) a1/a2 = b1/b2

(b) a1/a2 = b1/b2 = c1/c2

(c) a1/a2 ≠ b1/b2

(d) a1/a2 = b1/b2 ≠ c1/c2

Answer: (b) a1/a2 = b1/b2 = c1/c2

If the lines representing the pair of linear equations a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 are coincident, then a1/a2 = b1/b2 = c1/c2.

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