Pair of Linear Equation in Two Variables Test

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Pair of Linear Equation in Two Variables Test - 6

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Question 1
1 / -0

The product of two numbers is 54 and their sum is 21. Find the sum of the reciprocal of the numbers.

A

B

C

D

Solution

Let the two numbers be x and y.
According to the given conditions,
x × y = 54 ------------(i)
x + y = 21
x = 21 - y ----------(ii)

From (i) and (ii),
We have x × y = 54
(21 - y) × y = 54
21y - y² = 54
y² - 21y = -54
y² - 21y + 54 = 0
y² - 18y - 3y + 54 = 0
y (y - 18) - 3 (y - 18) = 0
(y - 3) (y - 18) = 0
Therefore, y = 3 or 18
So, the numbers are 3 and 18.

Reciprocal of 3 =

Reciprocal of 18 =
Sum of reciprocals =
=
=
Question 2
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The sum of the digits of a two-digit number is 6. Also, twice of this number is 6 more than the number obtained by reversing the order of the digits. Find the number.

A
21

B
23

C
24

D
25

Solution

Let the unit digit and tens digit of the number be x and y respectively.
Then, the number = 10y + x
Number after reversing the digits = 10x + y
According to the given information, x + y = 6 ..........(i)
2 ( 10y + x ) = 6 + ( 10x + y )
20y + 2x = 6 + 10x + y
20y - y + 2x - 10x = 6
19y - 8x = 6 ..........(ii)

From, eq (i), we get
x = 6 - y
Now putting this in eq (ii), we get
19y - 8 ( 6 - y ) = 6
19y - 48 + 8y = 6
27y = 6 + 48
27y = 54
y = 2

Thus, substituting it back in eq (i)
x + 2 = 6
x = 6 - 2
x = 4
Thus, the required number is 10y + x = 10 (2) + 4 = 20 + 4 = 24
Question 3
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The denominator of rational number is 4 more than its numerator. The number is obtained on subtracting 1 from the numerator and adding 5 to the denominator of its reciprocal. Find the original rational number.

A

B

C

D

Solution

Let the number be .

According to the question,
x = y - 4 -----------(i)

Also,
Using (i),
6y - 6 = 5y - 20 + 25
6y - 6 = 5y + 5
6y = 5y + 5 + 6
6y = 5y + 11
6y - 5y = 11
y = 11

Therefore, x = y - 4
x = 11 - 4
x = 7
Required number =
=
Question 4
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Find the value of k and m for which the given lines represent the pair of equations which are coincident.

9x + ky + 12 = 0
12x + 8y + m = 0

A
6, 16

B
8, 12

C
6, 18

D
12, 16

Solution

9x + ky + 12 = 0

12x + 8y + m = 0
On comparing these equations with a₁x + b₁y + c₁ = 0 and a₂x + b₂x + c₂ = 0,
We obtain, a₁ = 9, b₁ = k, c₁ = 12
a₂ = 12, b₂ = 8, c₂ = m
For the lines to be coincident; .
Now,

12 k = 72
k = 72 ÷ 12
k = 6

Also,

6m = 96
m = 96 ÷ 6
m = 16
Therefore, k = 6, m = 16
Question 5
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How many solutions do the following linear equations have?

7x – 8y + 6 = 0
9x + 9y – 9 = 0

A
A unique solution

B
No solution

C
Infinitely many solutions

D
May or may not have a solution

Solution

We know that if , then the equations have a unique solution.

Given equations,
7x – 8y + 6 = 0
9x + 9y – 9 = 0

Comparing these equations with a₁x + b₁y + c₁ = 0 and a₂x + b₂x + c₂ = 0;
We obtain, a₁ = 7, b₁ = -8, c₁ = 6
a₂ = 9, b₂ = 9, c₂ = -9

So,

Hence, the lines representing the given pair of equations have a unique solution.
Question 6
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If the following system of equation has no solution, which of the following can't be the values of z and k?

2x + 3y + z = 0
12x + 18y + k = 0

A
z = 5, k = 30

B
z = 2, k = 10

C
z = 3, k = 9

D
z = 4, k = 18

Solution

We know that the system of equations has no solution when, .
Given equations:
2x + 3y + z = 0
12x + 18y + k = 0

On comparing these equations with a₁x + b₁y + c₁ = 0 and a₂x + b₂x + c₂ = 0,
We obtain, a₁ = 2, b₁ = 3, c₁ = z
a₂ = 12, b₂ = 18, c₂ = k

So,

So, the values of z and k can't be in the ratio 1:6, else, .
Question 7
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Arya starts his job with certain monthly salary and earns a certain increment every year. If his salary was Rs. 6,300 after 6 years of service and Rs. 7,200 after 12 years, find his initial salary and annual increment respectively.

A
5400, 150

B
5300, 200

C
5200, 150

D
5100, 200

Solution

Let the annual increment be Rs. y. And initial salary be Rs. x.
According to the question, x + 6y = 6300 ...............(i)And x + 12y = 7200 ...........(ii)
Subtracting (ii) from (i) we have x + 6y - x + 12 y = 6300 - 7200 -6y = - 900
6y = 900
y = 900 ÷ 6
= 150

Putting the value of y in (i) we have x + 6y = 6300
x + 6 (150) = 6300
x + 900 = 6300
x = 6300 - 900
x = 5400
Question 8
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The system of equations 1.5x + 2.5y + c = 0 and 2.25x + 3.75y + 10.8 = 0 can have many solutions, if

A
c = 7.2

B
c = 6

C
c = 4.8

D
c = 3.6

Solution

We know that the lines have many solutions when .

Given equations:
1.5x + 2.5y + c = 0
2.25x + 3.75y + 10.8 = 0

On comparing these equations with a₁x + b₁y + c₁ = 0 and a₂x + b₂x + c₂ = 0,
We get, a₁ = 1.5, b₁ = 2.5 , c₁ = c
a₂ = 2.25, b₂ = 3.75, c₂ = 10.8

Now,

Now,

2.5 /3.75 = c /10.8
27 = 3.75c
c = 27 ÷ 3.75
c = 7.2
Question 9
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After adding 5 to the numerator, a new fraction is formed which is equal to ; and if 3 is subtracted from the denominator, a new fraction is formed which is equal to . Find the fraction.

A

B

C

D

Solution

Let x be the numerator and y be the denominator.

According to the question;
4x + 20 = 5y ........(i)

Also,
x + 3 = y ........(ii)

Now, multiplying (i) by 1 and (ii) by 5 we have;
4x + 20 = 5y
5x + 15 = 5y
So, 4x + 20 = 5x + 15
x = 5

Putting the value of x in (ii) we have;
y = x + 3
= 5 + 3
= 8
Therefore, required fraction =
Question 10
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A square ABCD is drawn for the following pair of lines.
3x + 6y + 15 = 0
5x + 10y + c = 0

For equations to represent the sides AB and CD, the value of 'c' can't be

A
25

B
24

C
20

D
16

Solution

We know that the opposite sides of a square are parallel. So the system of equations should also be parallel.

We know that for the lines to be parallel:

Given equations:
3x + 6y + 15 = 0
5x + 10y + c = 0

On comparing these equations with a₁x + b₁y + c₁ = 0 and a₂x + b₂x + c₂ = 0

We get,
a₁ = 3, b₁ = 2, c₁ = 15
a₂ = 5, b₂ = 10, c₂ = c

Therefore,

Also,

3/5 = 15/c

3c = 75
c = 25

Therefore, value of c can't be 25 because then:
Question 11
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If 3x + 5y + 8 = 0 and 4.2x + 8y + 11.2 = 0, then which of the following is true?

A
The pair of equations will have infinite solutions.

B
The pair of equations will have no solution.

C
The graph of these equations will consist of two parallel lines.

D
The pair of equations will have only one solution.

Solution

We have two linear equations in two variables in a plane,
a₁x + b₁y + c₁ = 0
a₂x + b₂y + c₂ = 0
For the pair of equations to have one solution,

3x + 5y + 8 = 0 --------- line I
4.2x + 8y + 11.2 = 0-------- line II

It can be observed that (≠ )

And, if we draw lines representing the equations, then the lines intersect at a single point. So, the pair of equations has a unique solution or only one solution. The pair of linear equations is consistent.
Question 12
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What will be the value of 34x + y, if 17x + 68y = 5, 102x + 340y = 166, and x ≠ 0 and y ≠ 0?

A
280

B
282

C
284

D
286

Solution

17x + 68y = 5 -------(1)
102x + 340y = 166 ------(2)
Multiplying equation (1) by 6, we get
102x + 408y = 30 -------(3)
Subtracting equation (2) from (3),
68y = -136
y = -2
Using the value of y in equation (1),
17x + 68(-2) = 5
17x = 141
34x = 141 × 2 = 282
34x + y = 282 - 2 = 280
Question 13
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If the subtraction of a two digit number from the number formed by interchanging the positions of the digits of the same number gives the answer as 45, then what is the difference between the two digits of that number?

A
7

B
6

C
5

D
4

Solution

Let the tens digit be x and units digit be y.
Then, (10x + y) - (10y + x) = 45
10x + y – 10y - x = 45
9x – 9y = 45
9 (x – y) = 45
x – y = 5
Question 14
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What is the value of M + N, if the following pair of equations has infinite solutions?

15x + 28y = N
22.5x + My = 19.5

A
55

B
50

C
45

D
40

Solution

If we have two linear equations in two variables in a plane;
a₁x + b₁y + c₁ = 0
a₂x + b₂y + c₂ = 0
For infinite solutions:

15x + 28y = N
22.5x + My = 19.5

Thus, M = = 42

N = = 13

M + N = 42 + 13 = 55
Question 15
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The cost of 17 pencils(p) and 136 erasers(e) is Rs. 197. Which of the following equations can represent the given situation?

A
17p + 163e = 197

B
8.5p + 68e = 98.5

C
85p + 68e = 985

D
170p + 1630e = 1790

Solution

According to the question,
17p + 136e = 197
Multiplying the whole equation by 0.5:
8.5p + 68e = 98.5
Question 16
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Jamie bought 2 bottles of lemonade and 1 packet of crisps for Rs. 140. Jennifer bought 1 bottle of lemonade and 3 packets of crisps for Rs. 170. Find the cost of a bottle of lemonade and a packet of crisps.

A
Rs. 70; Rs. 60

B
Rs. 40; Rs. 70

C
Rs. 50; Rs. 40

D
Rs. 50; Rs. 60

Solution

Let the cost of 1 bottle of lemonade = x
Cost of 1 packet of crisps = y
As given, 2x + y = 140 ..(i)
x + 3y = 170 ..(ii)

Multiply (ii) by (-2) and add with (i),
2x + y = 140
-2x - 6y = -340
-5y = -200
y = 40

Substitute the value of y in (ii).
x + 3(40) = 170
x + 120 = 170
x = 50
x = Rs. 50, y = Rs. 40
Question 17
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A cab charges a fixed amount and some additional amount for every kilometre of distance travelled. Charges for a distance of 9 km are Rs. 118 and for 16 km are Rs. 202. Find the fixed charges.

A
Rs. 10

B
Rs.15

C
Rs. 20

D
Rs. 25

Solution

Let the fixed charges of the cab = Rs. x
Additional charges per km = y
According to the question,
x + 9y = 118 ..........(i)
x + 16y = 202 ............(ii)
From eq (i), we get
x = 118 – 9y ..............(iii)

Substituting this in equation (ii), we get
118 – 9y + 16y = 202
7y = 84
y = 12 ..........(iv)

Thus, putting it in eq (i), we get
x + 9(12) = 118
x + 108 = 118
x = 10
Thus, fixed charges = Rs. 10
Question 18
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5 boys and 3 girls can finish the experiment in 6 days, while 3 boys and 6 girls can finish it in 4 days. Find the time taken by 1 boy alone to finish the whole experiment.

A
90

B
87

C
84

D
64

Solution

Let the work done by 1 girl = y days; and work done by 1 boy = x days

Thus, work done by a boy in 1 day =
Work done by a girl in 1 day =

Thus, according to the question,

.................(i)

..................(ii)

Thus, putting and in the above equations, we get

5p + 3q = ............(iii)

3p + 6q = ..................(iv)

From eq (iv), we get

12p + 24q = 1

p =

Now, putting this value in eq (iii), we get

30 - 720q + 216q = 12

504q = 18

q =

Now putting back this in eq (iii), we get

420p + 9 = 14

420p = 5

p =
Thus, work done by 1 boy = p = =
Therefore, one boy can complete the whole experiment in 84 days.
Question 19
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Ashley got 100 marks in his maths test. He got 4 marks for each right answer and lost 1 mark for each wrong answer. If 2 marks are deducted for each wrong answer instead of 1 mark, then his total score would become 80. Find the total number of questions in the test.

A
50

B
45

C
30

D
20

Solution

Let the number of right answers = x
Number of wrong answers = y
According to the question,
4x – y = 100 ...........(i)
4x – 2y = 80 ...........(ii)
Thus, from eq (i), we get
y = 4x – 100

Putting this in eq (ii), we get
4x – 2(4x – 100) = 80
4x – 8x + 200 = 80
120 = 4x
x = 30

Then, substituting back this in eq (i)
4(30) – y = 100
120 – y = 100
y = 20
Thus, total questions in the test = x + y = 30 + 20 = 50
Question 20
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Sonal can row a boat to downstream 30 km in 3 hours, and to upstream 8 km in 4 hours. Find her speed of rowing in still water.

A
2 kmph

B
6 kmph

C
8 kmph

D
10 kmph

Solution

Let the speed of boat in still water = x km/h
Speed of stream = y km/hr
As we know,
Speed while rowing upstream = (x – y) km/h
Speed while rowing downstream = (x + y) km/h

Thus,
3(x + y) = 30
x + y = 10 ..............(i)
Similarly,
4(x – y) = 8
x – y = 2 ............(ii)

Now, adding (i) and (ii), we get
2x = 12
x = 6
Hence, the speed of boat in still water is 6 kmph.
Question 21
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Solve for a and b in the following question.

A

B

C

D

Solution

We have,

Let

---(1)

---(2)

Replace with X and Y

6X + 5Y = 10

X = ---(3)

4X – 10Y = 12 ---(4)

Putting the value of X in (4), we get

4 {} - 10Y = 12

40 - 20 Y - 60Y = 72
-80Y = 72 - 40
-80Y = 32
Y = -32/80
Y = - 0.4

Putting the value of Y in (3), we get

X =

X =

X =

X = 2

Now, put the value of X and Y in equations (1) and (2).

We get,

=

4 b = { -}

By taking L.C.M;

4b =

b = ---(5)

Put the value of b in (2), which is:

By taking L.C.M, we get

a =

Putting the value of a in (5), we get

b =

b =
b =

b =

So, the value of a = , b =

Question 22

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Match the following.

Systems of Equations	Solutions
1. ,	(i) ,
2. ,	(ii) x = 20, y = 2
3. ,	(iii) ,
4. 2x - 5y = 10 3x - 6y = 12	(iv) ,

1 - (iii), 2 - (iv), 3 - (iv), 4 - (iii)

1 - (i), 2 - (ii), 3 - (iii), 4 - (iv)

1 - (ii), 2 - (i), 3 - (iv), 4 - (iii)

1 - (i), 2 - (ii), 3 - (iv), 4 - (iii)

Solution

Solutions of all the four pairs of linear equations in two variables:

1.

Taking L.C.M, we get

x + 20y = 60 --------------(a)

3x + 30y = 120

x =

Put the value of x in (a).

+ 20y = 60

By taking L.C.M, we get

120 – 30y + 60y = 180

30y = 180 - 120 = 60

y = = 2

Put the value of (y) in x to find the value of x.

So,

x = = = 20

So, the values of x and y are: x = 20 and y = 2 for the above two equations.

2. --------------(1)

By taking L.C.M, we get

3x + 4y = 2

-----------------(2)

----------------(3)

By taking L.C.M, we get

x =

Put the value of x in (2).

14x + 15y = 5 ---------(2)

So, we get

28 – 56 y + 45y = 15

11y = 13

Put the value of (y) in x to find it's value. So,

By taking L.C.M, we get,

So, the values of x and y are ,

. ,

By taking L.C.M, we get

We can in this way:

--------------------------(1)
Now, the second equation is:

Taking L.C.M, we get

--------------------(2)

Put the value of x in equation (2).

----------------(3)

Put the value of (y) in equation 1.

So, the value of

and

First equation can be written as

------------(1)

Second equation is

--------------------(2)

Put the value of x in equation 2.

Taking L.C.M, we get

3 y = - 6

y = - 2

Put the value of y in equation 1.

So,

x =

= 0

So, the value of x = 0, y = -2

Question 23
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There are four pairs of equations given below. Identify the solutions for each equation as consistent or inconsistent.

(1) 2x + 10y = 32
6x – 15y = 34

(2)

(3)

(4) 0.4x + 0.3y = 2
0.8x + 0.6y = 4

A
Consistent, consistent, inconsistent, inconsistent

B
Consistent, consistent, consistent, consistent

C
Inconsistent, inconsistent, inconsistent, consistent

D
Consistent, inconsistent, inconsistent, consistent

Solution

We have four pairs of equations in two variables.
(I) 2x + 10y = 32
6x – 15y = 34

These two pairs have consistent solutions because they are making intersecting lines.

(2)

These two pairs have inconsistent solutions because they are making parallel lines.

(3)

These two pairs have inconsistent solutions because they are making parallel lines.

(4) 0.4x + 0.3y = 2
0.8x + 0.6y = 4

These two pairs have consistent solutions because they are making coinciding lines.
Question 24
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Which of the following graphs represent the given pair of lines.

4 x – y = 4 and 6 x – 2 y = 6

A

B

C

D

Solution

Table for 4x – y = 4

x 0 1

y - 4 0

Table for 6x – 2y = 6

x 0 1

y - 3 0
Question 25
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Read the statements carefully and choose the correct option.
(a) Only Statement I is correct.
(b) Only Statement II is correct.
(c) Both statements I and II are correct.
(d) Both statements I and II are incorrect.

Statement I: 7x - 3y = 26
(a + b )x – ( -a + b)y = 5a + b
These two equations have unique solutions, if a = 2, b = 5.

Statement II:

These two equations have infinitely many solutions.

A
(a)

B
(b)

C
(d)

D
(c)

Solution

In statement I: 7x - 3y = 26
(a + b)x – (-a + b)y = 5a + b
We have,
a = 2, and b = 5
Statement I is incorrect because the pair of equations has no solutions, i.e. having parallel lines.

Condition for parallel lines:

So, this statement has no unique solutions because it does not satisfy the condition of unique solutions.

In statement II:

Statement II is also incorrect because the pair of linear equations has unique solutions, i.e these two equations will make intersecting lines. Condition for intersecting lines:

So, both the statements are incorrect.