Pair of Linear Equation in Two Variables Test

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Pair of Linear Equation in Two Variables Test - 7

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Weekly Quiz Competition

Question 1
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The cost of 3 pencils boxes and 3 erasers boxes is Rs. 150. Four times the cost of pencil box is 20 more than the cost of eraser box. Find the respective costs of a pencil box and an eraser box.

A
Rs. 14, Rs. 36

B
Rs. 15, Rs. 30

C
Rs. 10, Rs. 15

D
Rs. 21, Rs. 13

Solution

Let x and y be the costs of a pencil box and an eraser box, respectively.
According to the question:
3x + 3y = 150
x + y = 50 ... (i)
Also, 4x - y = 20 ... (ii)
Putting the value of y from (i) in (ii), we get 4x - (50 - x) = 20
4x - 50 + x = 20
5x = 70
x = 14
So, x + y = 50
y = 50 - 14 = 36
Question 2
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The sum of the digits of a two-digit number is 9. Also, half of this number is 9 more than the number obtained by reversing the order of digits. Find the number.

A
72

B
63

C
54

D
36

Solution

Let the unit digit and the tens digit of the number be x and y, respectively.
Then,
Number = 10y + x
Number after reversing order of digits = 10x + y
According to the given information, x + y = 9 ... (i)
= 9 + (10x + y)

⇒ 10y + x = 18 + 20x + 2y
⇒ 10y - 2y + x - 20x = 18
⇒ 8y - 19x = 18 ... (ii)
From (i), we get
y = 9 - x
Now, putting this in (ii), we get
8(9 - x) - 19x = 18
⇒ 72 - 8x - 19x = 18
⇒ 72 - 18 = 8x + 19x
⇒ 54 = 27x
⇒ 27x = 54
⇒ x = 2
Thus, substituting it in (i),
2 + y = 9
y = 9 - 2
y = 7
Thus, the required number is:
10y + x = 10(7) + 2 = 70 + 2 = 72
Question 3
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Which of the following sets of equations does not have a unique solution?

A
2x + 3y = 12
3x + 7y = 15

B
x + 2y = 5
4x + 8y = 10

C
4x + 5y = 12
3x + 5y = 13

D
2x – 3y = 8
4x – 5y = 9

Solution

The condition for a unique solution is:

Putting the values for option 1,

For option 2,

For option 3;

For option 4,

So, only option 2 does not have a unique solution.
Question 4
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Which of the following values of p is not correct for the system of equations pm – 4n = 8 and 24m – 8n = 12 to have a unique solution?

A
12

B
24

C
6

D
8

Solution

We will compare the given equations with a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0.

This is the condition for the system of equations to have a unique solution.

≠

p ≠ 12
Question 5
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The pair of equations 15p – 45q = 24 and 6p – 18q =has

A
one solution

B
two solutions

C
infinite solutions

D
no solution

Solution

Given equations: 15p – 45q = 24 and 6p – 18q =
We will compare these equations with a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0.
= =
= =
This implies

So, the given pair of equations has infinite solutions.
Question 6
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Find the respective values of m and n from the following pair of linear equations:

m – n = 9

+ = 14

A
45, 36

B
44, 35

C
40, 31

D
42, 33

Solution

m – n = 9 ... (i)
+ = 14 ... (ii)
From (i), we get m = 9 + n
Putting this value of m in (ii), we get

+ = 14
Solving this further, we get
36 + 4n + 9n = 14 × 36
13n = 504 – 36
n =
n = 36
Putting this value in (i), we get m = 45
Question 7
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The ratio of incomes of Rama and Ravi is : and the ratio of their expenditures is to , respectively. If each of them saves Rs. 300 per month, find their expenditures.

A
Rs. 1800 and Rs. 2400, respectively

B
Rs. 2000 and Rs. 2200, respectively

C
Rs. 2000 and Rs. 2500, respectively

D
Rs. 2000 and Rs. 2800, respectively

Solution

Let the incomes of Rama and Ravi be x and x, respectively.
Let the expenditures of Rama and Ravi be y and y, respectively.
Saving of Rama = x -y = 300 (given) ... (i)
Saving of Ravi =x -y = 300 (given) ... (ii)
Multiplying (i) by , we get
x – y = ... (iii)
Subtracting (ii) from (iii), we get
-y +y =- 300
Solving, y = 7200
Expenditure of Rama = × 7200 = Rs. 1800
Expenditure of Ravi = × 7200 = Rs. 2400
Question 8
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Solve for x:

+= 1 and+=

A
18

B
19

C
20

D
21

Solution

+ = 1 ... (i)

And + = ... (ii)

Multiplying (ii) by 2, we get

+ = ... (iii)

Subtracting (i) from (iii), we get

= - 1

=

x = 18
Question 9
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Which of the following linear equations is parallel to the line 18x + 36y - 108 = 0?

A
9x + 18y - 36 = 0

B
4x + 3y - 9 = 0

C
x - 2y + 1 = 0

D
3x + 4y - 20 = 0

Solution

We will compare the given equations with a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0.
For parallel lines,

Comparing equations 18x + 36y -108 = 0 and 9x + 18y - 36 = 0 with standard equations,

= ≠

Or = ≠ Hence, the two taken lines are parallel.
Question 10
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The sum of the numerator and the denominator of a fraction is 7. If we add 2 to the numerator and subtract 1 from the denominator, then the fraction becomes . What is the value of three times the original fraction?

A

B

C

D

Solution

Let the numerator be x and the denominator be y.
According to question:
x + y = 7 … (i)

Or 3x + 6 = 5y - 5
3x - 5y = -11 … (ii)

Solving (i) and (ii), we get

y = 4
x = 3
The fraction becomes .
Three times the fraction =
Question 11
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The numerator of a rational number is greater than its denominator by 12. If 9 subtracted from the numerator and 5 is added to the denominator, the new rational number becomes. Find the original rational number.

A

B

C

D

Solution

Let x and y be the numerator and denominator, respectively.
According to the question,
x = y + 12 ... (1)
Also, =
3x – 27 = 2y + 10 ... (2)
Putting the value of x from (1) in (2), we get
3(y + 12) – 27 = 2y + 10
3y + 36 – 27 = 2y + 10
3y – 2y = 10 – 9
y = 1
x = 1 + 12 = 13

Required fraction =
Question 12
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Which of the following can be the values of x and y if and?

A
x = 1, y = 3

B
x = 2, y = 4

C
x = 2, y = 3

D
x = 3, y = 2

Solution

Given equations:

i.
3x + 4y = 11x
4y = 8x

ii.
4x + 5y = 14y – 14x
18x = 9y
So clearly, the value of y is double the value of x.
Out of the given options, only the values given in option 2 satisfy this condition.
Question 13
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Find the value of 11x - 7y, if 17x + 19y = 650 and 23x - 13y = 144.

A
34

B
44

C
54

D
64

Solution

Given equations:
17x + 19y = 650 ... (i)
23x - 13y = 144 ... (ii)
Multiplying (i) by 23 and (ii) by 17, we have
391x + 437y = 14950
391x - 221y = 2448
Now, subtracting these fractions, we have
658y = 12502
y = 12502 ÷ 658
y = 19
Putting the value of y in (i),
17x + 19y = 650
17x + 19(19) = 650
17x + 361 = 650
17x = 650 – 361
17x = 289
x = 289 ÷ 17
x = 17
Now, 11x - 7y
= 11(17) - 7(19)
= 187 - 133
= 54
Question 14
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If the following lines are coincident, then what will be the value of k?

5x + 7y + 15 = 12
15x + 21y + k = 10

A
17

B
18

C
19

D
21

Solution

Reference equations of the line are a₁x + b₁y + c₁= 0 and a₂x + b₂y + c₂ = 0.
If the lines are coincident, then

Given equations:
5x + 7y + 15 = 12 can be written as 5x + 7y + 3 = 0.
15x + 21y + k = 10 can be written as 15x + 21y + k - 10 = 0.

Comparing these equations with the reference equations, we get
a₁ = 5, b₁ = 7, c₁ = 3
a₂ = 15, b₂ = 21, c₂ = k - 10

We know that

So,

7k – 70 = 63
7k = 63 + 70
7k = 133
k = 133 ÷ 7
k = 19
Question 15
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Harmeet and Manmeet were playing with marbles. When they started playing, Harmeet lost 10 of his marbles to Manmeet and the ratio of marbles of Harmeet to those of Manmeet became 5 : 4. After some time, Manmeet lost 20 of his marbles to Harmeet and the ratio of marbles of Harmeet to those of Manmeet became 7 : 2. Find the difference between the marbles they had at the beginning.

A
10

B
20

C
30

D
40

Solution

Let the number of marbles that Harmeet and Manmeet initially had be x and y.
According to the question,

4x - 40 = 5y + 50
4x - 5y = 90 --- (i)
Also,

2x + 20 = 7y - 70
2x - 7y = -90 --- (ii)
Multiplying (ii) by 2, we get
4x - 14 y = -180 --- (iii)
Now subtracting (iii) from (i), we get
9y = 270
y = 270 ÷ 3
y = 30
Putting the value of y in (i), we get
4x - 5y = 90
4x - 5(30) = 90
4x - 150 = 90
4x = 240
x = 240 ÷ 4
x = 60
Number of marbles that Manmeet initially had = y = 30
Number of marbles that Harmeet initially had = x = 60
Required difference = x - y
= 60 – 30
= 30
Question 16
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In a box, there are 10 pens in total, out of which some are blue and some are red. When twice the number of blue pens is added to three times the number of red pens, the result obtained is 23. How many red pens are there in the box?

A
7

B
3

C
4

D
2

Solution

Let the number of blue pens be x and the number of red pens be y.
According to question:
x + y = 10 … (i)
and 2x + 3y = 23 … (ii)
Multiply (i) by 2 and subtract (ii) from (i),

y = 3
Number of red pens = 3
Question 17
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A boat can cover 30 km going upstream in 20 minutes and 40 km going downstream in 24 minutes. How much time will the boat take to travel 84 km downstream if the speed of the stream is doubled?

A
48 minutes

B
46 minutes

C
36 minutes

D
24 minutes

Solution

Let the speed of the boat be x km/hr and the speed of the stream be y km/hr.

Speed when going upstream = (x - y) kmph
Speed when going downstream = (x + y) kmph
A.T.Q,

30 × 60 = 20x – 20y
1,800 = 20x – 20y
Dividing both sides by 20 we get,
90 = x – y --- (i)
Also,

60 × 40 = 24x + 24y
2,400 = 24x + 24y
Dividing both sides by 24 we get,
100 = x + y --- (ii)
Adding (i) and (ii), we get
x – y + x + y = 90 + 100
2x = 190
x = 190 ÷ 2
x = 95
Therefore, speed of the boat = 95 km/hr
Speed of the stream:
x – y = 90
95 – y = 90
y = 95 – 90
y = 5 km/hr
Time taken by the boat to travel 84 km when the speed of the stream is doubled = ^₌ x 60 (because 1 hour = 60 minutes)
=
= 48 minutes
Question 18
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The difference between the age of Guri and Shuri is 4 years. Five years ago, the sum of their ages was 36 years. What will be the ratio of the age of Guri to that of Shuri 7 years from now, if Shuri is younger than Guri?

A
6 : 7

B
7 : 6

C
8 : 7

D
7 : 8

Solution

Let the age of Guri and Shuri five years ago be x and y, respectively.
According to the question,
x + y = 36
x = 36 - y ... (i)
Also,
x - y = 4
Putting the value of x as given in (i) we get,
36 - y - y = 4
36 - 2y = 4
-2y = 4 - 36
-2y = -32
2y = 32
y = 32 ÷ 2
y = 16
Therefore, the age of Shuri, 5 years ago = 16 years
Putting y = 16 in the equation x - y = 4,
x - 16 = 4
x = 4 + 16
x = 20
Hence, age of Guri, 5 years ago = 20 years
Present age of Guri = 20 + 5 = 25 years
Present age of Shuri = 16 + 5 = 21 years
Age of Guri and Shuri, 7 years from now, will be 32 years and 28 years, respectively.

Required ratio = 32 : 28

= 8 : 7
Question 19
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The cost of 3 mangoes and 4 oranges is Rs. 29, and that of 4 mangoes and 3 oranges is Rs. 27. If 5 mangoes and 5 oranges are sold at a profit of 25% in total, what will be the total selling price of 5 mangoes and 5 oranges?

A
Rs. 48

B
Rs. 50

C
Rs. 52

D
Rs. 56

Solution

Let the cost of one orange be Rs. y and that of one mango be Rs. x.
A.T.Q,
3x + 4y = 29 --- (i)
4x + 3y = 27 --- (ii)
Now multiplying (i) by 4 and (ii) by 3 we get,
12x + 16y = 116
12x + 9y = 81
Now subtracting these fractions we get,
12x + 16y – 12x - 9y = 116 – 81
7y = 35
y = 35 ÷ 7
y = 5
Therefore, cost of one orange = Rs. 5
Now putting the value of y in eq, 3x + 4y = 29,
3x + 4(5) = 29
3x + 20 = 29
3x = 29 - 20
3x = 9
x = 9 ÷ 3
x = 3
Therefore, the cost of one mango = Rs. 3
Cost price of 5 mangoes and 5 oranges = Rs. (3 × 5) + (5 × 5)
= Rs. (15 + 25)
= Rs. 40
We know that profit = 25%
Selling price of 5 oranges and 5 mangoes = Rs.
=
= Rs. 50
Question 20
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Which of the following values of x and y represent the solution of the pair of equations 2a₁x + b₁y – c₁ = 0 and 2a₂x + b₂y – c₂ = 0?

A
x = , y =

B
x = , y =

C
x = , y =

D
x = , y =

Solution

2a₁x + b₁y – c₁= 0 --- (1)
2a₂x + b₂y – c₂ = 0 --- (2)
By multiplying eq (1) by b₂and eq (2) by b_1,
(2a₁x + b₁y – c₁ = 0) × b₂
(2a₂x + b₂y – c₂ = 0) × b₁
2a₁b₂x + b₁b₂y – c₁b₂ = 0 --- (3)
2a₂b₁x + b₁b₂y – c₂b₁ = 0 --- (4)

Now, eq (3) – eq (4):
2a₁b₂x + b₁b₂y – c₁b₂ - (2a₂b₁x + b₁b₂y – c₂b₁) = 0
2a₁b₂x + b₁b₂y – c₁b₂ - 2a₂b₁x - b₁b₂y + c₂b₁ = 0
2x(a₁b₂ - a₂b₁) – c₁b₂ + c₂b₁ = 0
2x(a₁b₂ - a₂b₁) = c₁b₂ - c₂b₁

x =

2a₁x + b₁y – c₁ = 0 --- (1)
2a₂x + b₂y – c₂ = 0 --- (2)
By multiplying eq (1) by a₂ and eq (2) by a_1,
(2a₁x + b₁y – c₁ = 0) × a₂
(2a₂x + b₂y – c₂ = 0) × a₁
2a₁a₂x + a₂b₁y – c₁a₂ = 0 --- (5)
2a₁a₂x + a₁b₂y – c₂a₁= 0 --- (6)
Now, eq (5) – eq (6):
2a₁a₂x + a₂b₁y – c₁a₂ - (2a₁a₂x + a₁b₂y – c₂a₁) = 0
2a₁a₂x + a₂b₁y – c₁a₂ - 2a₁a₂x - a₁b₂y + c₂a₁ = 0
a₂b₁y - a₁b₂y + c₂a₁ - c₁a₂= 0
y(a₂b₁ - a₁b₂) = c₁a₂ - c₂a₁
Question 21
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Fill in the blanks:

(i) When the lines representing the equations are intersecting at a single point, then it has _____ solutions.
(ii) If of the pair of linear equations and , then the pair of linear equations has _____ solutions.
(iii) If the pair of linear equations is inconsistent, then the lines are _____.

A
(i) unique, (ii) infinite, (iii) intersecting

B
(i) unique, (ii) infinite, (iii) parallel

C
(i) no, (ii) unique, (iii) parallel

D
(i) infinite, (ii) no, (iii) parallel

Solution

(i) If the lines are intersecting at a single point, then the pair of equations has a unique solution and the pair of linear equations is consistent.
(ii) If of the pair of linear equations and , then the pair of linear equations has infinitely many solutions and the equations are consistent and dependent.
(iii) If the lines are parallel to each other, then there is no solution and the pair of linear equations is inconsistent.
Question 22
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Which of the following statements is false?

A. y + 3x = 0 and 12y + 23x = 0, both the lines pass through the origin.
B. The pair of equations 8x + 6y = 15 and 3x – 4y = 25 has a unique solution.
C. 2x + 6y = 12 and 8x + 24y = 48 are an inconsistent pair of equations.
D. x + 6y = 12 and x + 6y = 16 are an inconsistent pair of equations.

A
A

B
B

C
C

D
D

Solution

Statement A:
y + 3x = 0 and 12y + 23x = 0
Putting x and y as zero in y + 3x = 0,
0 + 3(0) = 0 + 0 = 0
Putting x and y as zero in 12y + 23x = 0,
12(0) + 23(0) = 0
0 = 0
The coordinates of the origin are (0, 0).
Put x and y as zero in both equations satisfies the equations.
Hence, both the equations should pass the origin.
Statement A is true.

Statement B:
Given equations can be written as:
8x + 6y - 15 = 0 and 3x – 4y - 25 = 0
Comparing these with a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0,
For a unique solution:

So,

Hence, statement B is true.

Statement C:
Equations can be written as:
2x + 6y - 12 = 0 and 8x + 24y - 48 = 0
Comparing these with a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0,
For inconsistent lines:

Hence, statement C is false as the pair of equations is dependent and consistent.

Statement D:
Equations can be written as:
x + 6y - 12 = 0 and x + 6y - 16 = 0
Comparing these with a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0,
For inconsistent solution:

Putting the values:

So, statement D is true.
Question 23
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Which of the following pairs of linear equations has/have solution (x, y) that are positive rational number?

(i) 3x + 2y = 10 and 4x + 3y – 9 = 0
(ii) 5x + 2y + 4 = 0 and 2x + 3y + 6 = 0
(iii) 2x +y = 9 and x + 5y – = 0
(iv) and

A
(iii) only

B
(i) and (iii)

C
(iv) only

D
(ii) and (iv)

Solution

Given pairs of equations are:
(i) 3x + 2y = 10 and 4x + 3y – 9 = 0
3x + 2y = 10 ... eq1
And,
4x + 3y = 9 ... eq2
After multiplying equation 1 by 4, and equation 2 by 3 we get:
4 × (3x + 2y = 10) and 3 × (4x + 3y = 9)
12x + 8y = 40 ... eq3
And,
12x + 9y = 27 ... eq4
Subtracting equation 4 from equation 3 we get:

y = -13
Putting y = -13 in equation 3, we get:
12x + 8(-13) = 40
12x – 104 = 40
12x = 40 + 104
12x = 144
x = 12
This is not the required set of equations as the value of y is negative.
(ii) 5x + 2y + 4 = 0 and 2x + 3y + 6 = 0
5x + 2y = -4 ... eq 1
And,
2x + 3y = -6 ... eq 2
After multiplying equation 1 by 2 and equation 2 by 5 we get:
2 × (5x + 2y = -4) and 5 × (2x + 3y = -6)
10x + 4y = -8 ... eq 3
And,
10x + 15y = -30 … eq 4
Subtracting equation 3 from equation 4, we get:

y = -2
Putting y = -2 in equation 3, we get:
10x + 4(-2) = -8
10x – 8 = -8
10x = -8 + 8
10x = 0
x = 0
This is not the required set of equations as value of y is negative.
(iii) 2x +y = 9 and x + 5y –= 0
= 9 and x + 5y =
4x + 3y = 2 × 9 and 2(x + 5y) = 3
4x + 3y = 18 ... eq1
And ,
2x + 10y = 3 ... eq2
After multiplying equation 1 by 2 and equation 2 by 4 we get:
2 × (4x + 3y = 18) and 4 × (2x + 10y = 3)
8x + 6y = 36 ... eq 3
And,
8x + 40y = 12 ... eq4
Subtracting equation 3 from equation 4, we get:

y =
y =
Putting y = in equation 3, we get:
8x + 6() = 36
= 36
136x – 72 = 612
136x = 612 + 72
136x = 684
x =
This is not the required set of equations as the value of y is negative.
(iv) and
5(x – 3) = y + 2 and 1(10y + x) = 4(y + x)
5x – 15 = y + 2 and 10y + x = 4y + 4x
5x – y = 2 + 15 and 10y – 4y = 4x – x
5x – y = 17 and 6y = 3x
5x – y = 17 and x = 2y
Putting x = 2y in (5x – y = 17), we get:
5(2y) – y = 17
10y – y = 17
9y = 17
y =
We know that,
x = 2y
Therefore, x = 2 ×
x =
^{For this pair of equations, the solutions for 'x' and 'y' are positive rational numbers.
Hence, option 3 is the answer.}

Question 24

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Match the following.

System of equations	Solutions
(a) 2x + y = 7 and y = 8 + x	(i) x = -2, y = 1
(b) x + 7y = 9 and 3x + y = 21	(ii) No solution
(c) x + 4 = 2y and x + 3 = y	(iii) x = , y =
(d) x + 8 = y and 2y = 12 + 2x	(iv) x = , y =

(a) - iii, (b) - ii, (c) - i, (d) - iv

(a) - iii, (b) - iv, (c) - i, (d) - ii

(a) - ii, (b) - iv, (c) - i, (d) - iii

(a) - iv, (b) - iii, (c) - i, (d) - ii

Solution

(a) Solving
2x + y = 7 and y = 8 + x
2x + y = 7
2x = 7 - y
Dividing both sides by 2,

x =

Using this value of x in y = 8 + x,

y = 8 +

3y = 23

y =

Using the value of y in y = 8 + x,
= 8 + x
x =

Correct answer is (iii)

(b) x + 7y = 9 and 3x + y = 21
x = 9 - 7y (From x + 7y = 9)

Using this value in 3x + y = 21,
3 (9 - 7y) + y = 21
27 - 21y + y = 21
27 - 21 = 21y - y
6 = 20y

y =

x + 7y = 9

x + 7() = 9

x + = 9

x =

x =

Hence, correct answer is iv.

(c) x + 4 = 2y and x + 3 = y
This can be written as,
x = 2y - 4

Putting the above value of x in the equation x + 3 = y,
(2y - 4) + 3 = y
2y - 1 = y
2y - y = 1
y = 1
Now,
x = 2y - 4 = 2(1) - 4 = 2 - 4 = -2
Correct answer is i.

(d) x + 8 = y and 2y = 12 + 2x

The equations can be written as,
x - y + 8 = 0 and 2x - 2y + 12 = 0
Comparing these with
a₁x + b₁y + c₁ = 0
a₂x + b₂y + c₂ = 0
We get,

Hence, this set of equations has no solution.
Correct answer is ii.

Question 25
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Which of the following graphs represent the parallel lines of equations, 2x - 5 + 3y = 0 and 4x - 12 = -6y?

A

B

C

D

Solution

2x + 3y = 5

x 1 -2

y 1 3

4x + 6y = 12

x 0 3

y 2 0

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Pair of Linear Equation in Two Variables Test - 7

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Pair of Linear Equation in Two Variables Test - 7

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Question 1

Rs. 14, Rs. 36

Rs. 15, Rs. 30

Rs. 10, Rs. 15

Rs. 21, Rs. 13

Solution

Question 2

72

63

54

36

Solution

Question 3

2x + 3y = 123x + 7y = 15

x + 2y = 54x + 8y = 10

4x + 5y = 123x + 5y = 13

2x – 3y = 84x – 5y = 9

Solution

Question 4

12

24

6

8

Solution

Question 5

one solution

two solutions

infinite solutions

no solution

Solution

Question 6

45, 36

44, 35

40, 31

42, 33

Solution

Question 7

Rs. 1800 and Rs. 2400, respectively

Rs. 2000 and Rs. 2200, respectively

Rs. 2000 and Rs. 2500, respectively

Rs. 2000 and Rs. 2800, respectively

Solution

Question 8

18

19

20

21

Solution

Question 9

9x + 18y - 36 = 0

4x + 3y - 9 = 0

x - 2y + 1 = 0

3x + 4y - 20 = 0

Solution

Question 10

Solution

Question 11

Solution

Question 12

x = 1, y = 3

x = 2, y = 4

x = 2, y = 3

x = 3, y = 2

Solution

Question 13

34

44

54

64

Solution

Question 14

17

2x + 3y = 12
3x + 7y = 15

x + 2y = 5
4x + 8y = 10

4x + 5y = 12
3x + 5y = 13

2x – 3y = 8
4x – 5y = 9