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

*CBSE Class 10 Pair of Linear Equations MCQ New Pattern 2022*

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## 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

Answer:** (c) Infinitely many solutions**

Explanation:

a_{1}/a_{2} = 1/-4

b_{1}/b_{2} = 2/-8 = 1/-4

c_{1}/c_{2} = -5/20 = -¼

This shows:

a_{1}/a_{2} = b_{1}/b_{2} = c_{1}/c_{2}

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) 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

Answer:** (d) No solution**

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

a_{1}/a_{2} = 9/18 = 1/2

b_{1}/b_{2} = 3/6 = 1/2

c_{1}/c_{2} = 12/26 = 6/13

Since, a_{1}/a_{2} = b_{1}/b_{2} ≠ c_{1}/c_{2}

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) 15/4**

Explanation: The condition for parallel lines is:

a_{1}/a_{2} = b_{1}/b_{2} ≠ c_{1}/c_{2}

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) -6x+10y-4=0**

Explanation: The condition for dependent linear equations is:

a_{1}/a_{2} = b_{1}/b_{2} = c_{1}/c_{2}

For option a,

a_{1}/a_{2} = b_{1}/b_{2} = c_{1}/c_{2}= 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) 3 and 1**

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) 5/12**

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

Answer:** (a) ⅕ and -2**

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) 6km/hr and 4km/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

Answer:** (c) infinitely many solutions**

Explanation:

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

Comparing with the standard form,

a_{1} = 5, b_{1} = -15, c_{1} = -8

a_{2} = 3, b_{2} = -9, c_{2} = -24/5

a_{1}/a_{2} = 5/3

b_{1}/b_{2} = -15/-9 = 5/3

c_{1}/c_{2} = -8/(-24/5) = 5/3

Thus, a_{1}/a_{2} = b_{1}/b_{2} = c_{1}/c_{2}

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

Answer: **(d) no solution**

Explanation:

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

Comparing with the standard form,

a_{1} = 1, b_{1} = 2, c_{1} = 5

a_{2} = -3, b_{2} = -6, c_{2} = 1

a_{1}/a_{2} = -1/3

b_{1}/b_{2} = 2/-6 = -1/3

c_{1}/c_{2} = 5/1

Thus, a_{1}/a_{2} = b_{1}/b_{2} ≠ c_{1}/c_{2}

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

Answer: **(d) no value**

Explanation:

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

Comparing with the standard form,

a_{1} = c, b_{1} = -1, c_{1} = -2

a_{2} = 6, b_{2} = -2, c_{2} = -3

a_{1}/a_{2} = c/6

b_{1}/b_{2} = -1/-2 = 1/2

c_{1}/c_{2} = -2/-3 = ⅔

Condition for having infinitely many solutions is

a_{1}/a_{2} = b_{1}/b_{2} = c_{1}/c_{2}

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 a _{1}**

**x + b**

_{1}**y + c**

_{1}**= 0 and a**

_{2}**x + b**

_{2}**y + c**

_{2}**= 0 are coincident, then**

(a) a_{1}/a_{2} = b_{1}/b_{2}

(b) a_{1}/a_{2} = b_{1}/b_{2} = c_{1}/c_{2}

(c) a_{1}/a_{2} ≠ b_{1}/b_{2}

(d) a_{1}/a_{2} = b_{1}/b_{2} ≠ c_{1}/c_{2}

Answer: **(b) a _{1}**

**/a**

_{2}**= b**

_{1}**/b**

_{2}**= c**

_{1}**/c**

_{2}If the lines representing the pair of linear equations a_{1}x + b_{1}y + c_{1} = 0 and a_{2}x + b_{2}y + c_{2} = 0 are coincident, then a_{1}/a_{2} = b_{1}/b_{2} = c_{1}/c_{2}.