Pair of Linear Equations in Two Variable Test

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Pair of Linear Equations in Two Variable Test - 2

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Weekly Quiz Competition

Question 1
1 / -0

Which of the following pairs of equations represents coincident lines?

A
2x + y = 4, 3x + y = 5

B
x + 2y = 7, 2x + 4y = 14

C
x – y = 5, 2x + 3y = 25

D
2x – 7y = 7, x + y = 8

Solution

For the pair of equations to represent coincident lines,

In lines x + 2y = 7 and 2x + 4y = 14,

So,
Question 2
1 / -0

Which of the following pairs of equations represents parallel lines?

A
x – y = 7, 2x – 2y = 15

B
x + 2y = 1, y + 3x = 4

C
2x + y = 4, x – 2y = 3

D
x + 2y = 4, 3x + 7y = 18

Solution

Two lines a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 are parallel when .
In lines x – y = 7 and 2x – 2y = 15,
Question 3
1 / -0

Under which of the following circumstances does a pair of linear equations have a unique solution?

A
When the lines representing the equations intersect at more than two points

B
When the lines representing the equations intersect at one point

C
When the lines representing the equations do not intersect

D
None of these

Solution

When 2 lines intersect at one point, the pair of linear equations has a unique solution.
Question 4
1 / -0

If x + y = 0 and x + 2y = 4, find the value of y.

A
0

B
4

C
8

D
16

Solution

a) x + y = 0 (subtract y from both sides)
x = - y
b) x + 2y = 4
Put x = - y
- y + 2y = 4
y = 4
Question 5
1 / -0

If a + b = 4 and a - b = 8, find the value of a b.

A
- 12

B
- 4

C
16

D
32

Solution

a + b = 4 (subtract b from both sides) ........(1)
a = 4 - b
Now, a - b = 8
Put a = 4 - b
4 - b - b = 8
4 - 2b = 8
- 2b = 8 - 4 (subtract 4 from both sides)
- 2b = 4
b = (divide both sides by -2)
b = - 2
Put b = - 2 in equation (1)
a = 4 - (- 2)
a = 4 + 2
a = 6
a b = 6 (- 2) = - 12
Question 6
1 / -0

If 7x + 4y = 18 and x - y = 0, then find the value of 11(x + y).

A

B

C
12

D
36

Solution

7x + 4y = 18 … (1)
Now, x - y = 0
Or x = y
Put x = y in equation (1).
7y + 4y = 18
11y = 18
Now, x = y
11x = 11y = 18
Now, 11(x + y) = 11x + 11y
= 18 + 18
= 36
Question 7
1 / -0

If 3a + b = 7 and 4a - 2b = 6, find the value of .

A
- 6

B
- 2

C
4

D
9

Solution

3a + b = 7 … (1)
4a - 2b = 6 … (2)
Multiplying equation (1) with 2
6a + 2b = 14 …. (3)
Adding equations (2) and (3)
6a + 2b = 14
4a - 2b = 6
10a = 20
a = (divide both sides by 10)
a = 2

Putting a = 2 in equation (1)
3 2 + b = 7
6 + b = 7
b = 7 - 6
b = 1
Now,
= 2 + 2
= 4
Question 8
1 / -0

If x + 5y = 0 and 2x - 5y = 0, find the value of x² + y².

A
- 4

B
0

C
7

D
13

Solution

x + 5y = 0 … (1)
2x - 5y = 0 … (2)

Adding equations (1) and (2)
x + 2x = 0
3x = 0
x = 0
Put x = 0 in equation (1)
0 + 5y = 0
5y = 0
y = 0
Now, x² + y² = 0² + 0² = 0
Question 9
1 / -0

Which of the following values of x and y represent the solution of the pair of equations a₁x + b₁y – c₁ = 0 and a₂x + b₂y – c₂ = 0?

A
x = , y =

B
x = , y =

C
x = , y =

D
None of these

Solution

a₁x + b₁ y – c₁ = 0 …. I] b₂
a₂x + b₂ y – c₂ = 0 ….. II] b₁
On subtracting II from I, we get
I a₁ b₂ x + b₁ b₂ y – c₁ b₂ = 0
II a₂ b₁ x + b₁ b₂ y – c₂ b₁ = 0
– – +

a₁ b₂ x – a₂ b₁ x – c₁ b₂ + c₂ b₁ = 0
Now, x (a₁ b₂ – a₂ b₁) = c₁ b₂ – c₂ b₁
x =
After multiplying I by a₂ and II by a₁ and then subtracting IV from III, we get
III a₁ a₂ x + a₂ b₁ y – c₁ a₂ = 0
IV a₁ a₂ x + a₁ b₂ y – c₂ a₁ = 0
– – +

(a₂ b₁ – a₁ b₂) y = c₁ a₂– c₂ a₁
Question 10
1 / -0

A total of 500 tickets to a show are sold. If an adult ticket costs £3, a child ticket costs £2 and a total amount of £1300 is collected, then how many tickets of each kind are sold?

A
Adult tickets = 300
Child tickets = 200

B
Adult tickets = 200
Child tickets = 300

C
Adult tickets = 100
Child tickets = 400

D
Adult tickets = 400
Child tickets = 100

Solution

Let the number of adult tickets sold be x and the number of child tickets sold be y.
Total number of tickets sold = 500
x + y = 500
y = 500 - x
According to the question:
3x + 2y = 1300
3x + 2(500 - x) = 1300
3x + 1000 - 2x = 1300
x + 1000 = 1300
x = 1300 - 1000 = 300
So, number of adult tickets sold = 300
Number of child tickets sold = y = 500 - x
= 500 - 300
= 200
Question 11
1 / -0

A woman is now 20 years older than her son. After 5 years, the age of the woman will be twice the age of her son. What is the present age of the woman?

A
30 years

B
35 years

C
40 years

D
45 years

Solution

Let the present age of the son be x.
Then, present age of woman is x + 20.
After 5 years,
Son`s age = x + 5
Woman`s age = x + 20 + 5 = x + 25
According to question,
(x + 25) = 2(x + 5)
x + 25 = 2x + 10
x + 25 - 10 = 2x
x + 15 = 2x
15 = 2x - x
15 = x
So, son`s age = 15 years
Woman`s age = 15 + 20 = 35 years
Question 12
1 / -0

Cost of 2 pens and 8 pencils is 22p and the cost of 4 pens and 2 pencils is 16p. What will be the cost of one pen and one pencil?

A
Pen = 3p
Pencil = 4p

B
Pen = 4p
Pencil = 3p

C
Pen = 2p
Pencil = 3p

D
Pen = 3p
Pencil = 2p

Solution

Let cost of 1 pen = x
Let cost of 1 pencil = y
According to question:
2x + 8y = 22p ..........(1)
4x + 2y = 16p ...........(2)
Divide the second equation by 2,
2x + y = 8p
Also, 2x + 8y = 22p
Subtract both the equations,
2x + y - 2x - 8y = 8p - 22p
-7y = -14p
y = p = 2p
x = = = 3p
So, cost of 1 pen = 3p
Cost of 1 pencil = 2p
Question 13
1 / -0

The ratio of income of Aman to income of Raman is 4 : 3. Their expenditures are in the ratio 3 : 1. If their respective savings are Rs. 16,000 and Rs. 25,000, what is the income of Aman?

A
Rs. 35,400

B
Rs. 47,200

C
Rs. 11,800

D
Rs. 45,200

Solution

Income of Aman = Rs. 4x
Income of Raman = Rs. 3x
Expenditure of Aman = Rs. 3y
Expenditure of Raman = Rs. y
According to question:
4x - 3y = 16,000 … (i)
3x - y = 25,000 … (ii)

Multiplying (ii) by 3 and then subtracting (ii) from (i), we get

x = 11,800
Aman's income = Rs. 4x
= Rs. 4 11,800
= Rs. 47,200
Question 14
1 / -0

The numerator of a fraction is three more than the denominator. When 3 is added to the numerator and 10 is added to the denominator, the fraction becomes What is the product of the numerator and the denominator of the original fraction?

A
9

B
10

C
18

D
20

Solution

Let the denominator be x and the numerator be x + 3.
According to question:

3x + 18 = 2x + 20
x = 2
Then, numerator = x + 3 = 2 + 3 = 5
Fraction =
Product of numerator and denominator = 10
Hence, (2) is the correct option.
Question 15
1 / -0

A man rows a boat 4 km upstream and 5 km downstream in 3 hours. He also rows 3 km upstream and 6 km downstream in 4 hours. What are the equations formed if 'x' is the speed of the boat in still water and 'y' is the speed of the current?

A

B

C

D

Solution

'x' is the speed of the boat in still water.
'y' is the speed of the current.
Speed upstream = x - y
Speed downstream = x + y
Now, equation (1) becomes:

Put
4u + 5v = 3
Equation (2) becomes:

3u + 6v = 4
Hence, (2) is the correct option.