Practice Test: Pair of Linear Equations in Two Variables - Class 10 MCQ

Given :

2 k x + 5 y – 7 = 0  ...( i )
6 x – 5 y – 1 = 0   ... ( ii )
Pair of linear equations has a unique solution.
We know for unique solution.

Comparing from ( i ) and ( ii ) we have

Put these values in formula.

Thus we get answer many values of k but leaving k ≠ -3.

An equation has infinitely many solutions when the lines are coincident.
The lines are coincident when 
So 3x + 4y = k, 9x + 12y = 6 are coincident when

An equation has infinitely many solutions when the lines are coincident.
The lines are coincident when 
So 2x + 5y = k, kx + 15y = 18 are coincident when

3/12 = 7/2k  [ applying a1/a2=b1/b2 ]
3 x 2k = 7 x 12
k=14

Work done by 1 girl and 1 boy in x and y days respectively.

work done by 1 girl and 1 boy in 1 day is (1/x) and (1/y).

so, work done by 8 girls and 12 boys in 1 day is (8/x) + (12/y) = 1/10

let (1/x) = a and (1/y) = b

so, 8a + 12b = 1/10

→ 80a + 120b = 1 ---- (1)

work done by 6 girls and 8 boys in 1 day is (6/x) + (8/y) = 1/14

6a + 80 = 1/14

→ 84a + 112b = 1 ---- (2)

By elimination method, Multiple equation 1 by 21 on both sides, we get

1680a + 2520b = 21 ---- (3)

Multiply equation 2 by 20 on both sides, we get

1680a + 2240b = 20 ---- (4)

On solving equation 3 and 4, we get

2520 b - 2240b = 21 - 20

→ 280 b = 1

→ b = 1/280

b =1/y

→ 1/280 = 1/y

→ y = 280

80a + 120 x (1/280) = 1 (From 1)

→ 80a + (3/7) = 1

→ 80a = 1 - (3/7)

→ 80a = (7 - 3)/7

→ 80a = 4/7

→ a = 4/(7 × 80)

→ a = 1/140

→ a = 1/x

→ 1/140 = 1/x

→ x = 140

1 girl and 1 boy alone take 140 days and 280 days to complete a work.

kx + 4y = 5, 3x + 2y = 5
Here, a1=k,b1=4,c1​=−5
and a2=3,b2=2,c2=-5

So , The equation is consistent when 

k ≠ 6

Since ABCD is a rectangle
⇒ AB = CD and BC = AD
x + y = 30 …………….. (i)
x – y = 14 ……………. (ii)
(i) + (ii) ⇒ 2x = 44
⇒ x = 22
Plug in x = 22 in (i)
⇒ 22 + y = 30
⇒ y = 8

Let the present age of man is x and of son is y.
Six years hence,
Man’s age =x+6
Son’s age=y+6
Man’s age is 3 times son’s age
x+6=3(y+6)
x+6=3y+18
x=3y+12    …...1
Three years ago,
Man’s age =x-3
Son’s age=y-3
Man’s age was 9 times as of son
x-3=9(y-3)
x-3=9y-27
x=9y-24   ….2
From 1 and 2
3y+12=9y-24
6y=36
y=6
x=3*6+12=18+12=30 years

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.

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.

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.

The condition for parallel lines is:
a₁/a₂ = b₁/b₂ ≠ c₁/c₂
Hence, 3/2 = 2k/5
k = 15/4

The condition for dependent linear equations is:
a₁/a₂ = b₁/b₂ = c₁/c₂
For option a,
a₁/a₂ = b₁/b₂ ≠ c₁/c₂ =  ½