Class 10 Maths Chapter 3 Previous Year Questions - Pair of Linear Equations in Two Variables

Ans: (d)
x + 2y + 5 = 0 
On comparing with 
a₁x + b₁y + c₁ = 0, we get a₁ = 1, b₁ = 2, c₁ = 5 – 3x = 6y – 1 
3x + 6y – 1 = 0 
On comparing with a₂x + b₂y + c₂ = 0, we get a₂ = 3, b₂ = 6, c₂ = – 1

a₁a₂ = 13 , b₁b₂ = 26 = 13 ,

c₁c₂ = 5-1 = -5

Here, we get ( a₁a₂ = b₁b₂ ≠ c₁c₂ )

Ans:
2x + y = 13 ...(i)  
4x – y = 17 ...(ii)
On adding eqn.(i) and eqn.(ii)
6x = 30
x = 5
Put the value of x in eqn.(i)  
2 × 5 + y = 13
⇒10 + y = 13 
∴ y = 3
So,  x – y = 5 – 3 = 2

Ans: (b) 4/5
Given that,
Pair of linear equations as 5x + 2y − 7 = 0  and 2x + ky + 1 = 0 
For no solution,

→ 52 = 2k ≠ -71

→ 52 = 2k

→ k5 = 45

Ans: Given the pair of linear equations as,
⇒ x + 2y = 9     ...(i)
⇒ y − 2x = 2
⇒ −2x + y = 2    ...(ii)
Multiplying eqn (i) by 2 and adding to eqn (ii), we get
⇒ (−2x + y) + ( 2x + 4y) = 2 + 18
⇒ 5y = 20
⇒ y = 4
Putting in eqn (i),
⇒ x + 2(4) = 9
⇒ x = 9 − 8
⇒ x = 1
So, the required solution is x = 1 and y = 4

Ans: Given the equations of line are
⇒ x + y = −1    ...(i)
⇒ x − y = 1    ...(ii)
The intersection point of both the lines will be the point that lies on both the lines,
So, adding eqn (i) and (ii),
⇒ (x + y) + (x − y) = −1 + 1
⇒ 2x = 0
⇒ x = 0
Putting in eqn (i),
⇒ 0 + y = −1
⇒ y = −1
So, the point will be (x, y) = (0, −1)
Hence, the point (−4, 3) does not lie on both the lines.

Ans: (a) consistent with unique solution.

Ans: Given system of linear equations are 
7x − 2y = 5   ...(i)
8x + 7y = 15   ...(ii)
By multiplying eq. (i) by 7 and eq. (ii) by 2, we get
49x − 14y = 35
16x + 14y = 30
65x = 65
∴ x = 1
Substituting the value of x is eq (i), we get
7(1) − 2y = 5
or, 7 − 2y = 5
or −2y = −2, or y = 1
Therefore, x = 1 and y = 1

Ans: Let the age of Rashmi = x years
and the age of Nazma = y years
Three years ago,
Rashmi's age = (x − 3) years
Nazma's age = (y − 3) years
According to the question,
(x − 3) = 3(y − 3)
⇒ x − 3 = 3y − 9
⇒ x = 3y − 6      ...(i)
Ten years later,
Rashmi's age = x + 10
Nazma's age= y + 10
According to the question,
(x + 10) = 2(y + 10)
x + 10 = 2y + 20
x = 2y + 10      ...(ii)
From eq. (i) and (ii), we get
3y − 6 = 2y + 10
y = 16
Substituting the value of y in eq. (i), we get
x = 3 × 16 − 6 = 48 − 6 = 42
Thus, Rashmi is 42 years old, and Nazma is 16 years old.

Ans: (c)
Sol: The given pair of linear equations is 2x = 5y+ 6 and 15y = 6x - 18
i.e ., 2x - 5y - 6 = 0  and 6x- 15y- 18 = 0
As, 2/6 = -5/-15 = -6/-18
i.e.. 1 /3 = 1 / 3 = 1/3
Since, a₁/a₂=b₁/b₂=c₁/c₂
Therefore, the lines are coincident.

Ans: (c)
Sol: x + ky = 5
At x = 2,  y = 1
2 + k(1) = 5
∴ k = 3

Ans: (d)
x + 2y + 5 = 0 
On comparing with 
a₁x + b₁y + c₁ = 0, In  x + 2y + 5 = 0, we get a₁ = 1, b₁ = 2, c₁ = 5 
Rearranging – 3x = 6y – 1 , We get 3x + 6y – 1 = 0 
On comparing with a₂x + b₂y + c₂ = 0, we get a₂ = 3, b₂ = 6, c₂ = – 1

Ans: Given equations are

x = 5 ---------------(i)
y = 7  ---------------(ii)
Draw the line x = 5 parallel to the y-axis and y= 7 parallel to the x-axis.
∴ The graph of equation (i) and (ii) is as follows

The lines x = 5 and y = 7 intersect each other at (5, 7).

Ans: Given pair of equations are

x = 0    ------(i)
and y = -3    ------(ii)
x = 0 means y-axis and draw a line y = -3 parallel to x-axis. The graph of given equations (i) and (ii) is

The lines intersect each other at (0, -3). Therefore, the given pair of equations Is consistent.

Ans: Let x and y be two numbers such that x> y
According to the question,

and x + 2y = 13 ---- (ii)
Subtracting (i) from (ii), we get
3y = 9
⇒ y = 3
Substitute y = 3 in (i) we get
x - 3 = 4
⇒  x = 7

Ans: (A) Let required fraction be x/y
According to question,

x + 1y - 1 = 1

⇒ x + 1 = y - 1

⇒ x = y - 2 ... (i)

Also, xy + 1 = 12

⇒ 2x = y + 1              ...(ii)
From equations (i) and (ii), we get
2y — 4 = y + 1
y = 5
∴ x = 3
Required fraction x/y is 3/5
OR
(B) 3x + y = 1
(2k - 1 )x + (k - 1 )y = 2k + 1

For no solution;

⇒ 32k - 1 = 1k - 1 ≠ 12k + 1

2k - 1 = 3k - 3

⇒ k = 2

Also, 1k - 1 ≠ 12k + 1

2k + 1 ≠ k - 1

⇒ k ≠ -2

Ans: (i) For Hockey, the amount given to per student =  x
For cricket, the amount given to per student = y
From the question,
5x + 4y =9500    (i)
4x + 3y = 7370   (ii)

(ii) (a) Multiply (1) by 3 and (2) by 4 and then subtracting, we get

15x + 12y- (16x + 12y) = 28500 - 29480
⇒ - x = - 980
⇒ x = ₹980
The prize amount given for hockey is Rs. 980 per student
(b) Multiply (1) by 4 and (2) by 5 and then subtracting, we get
20x + 16y- 20x - 15y = 38000 - 36850
⇒ y = 1150
The prize amount given for cricket is more than hockey by (1150 - 980) = 170.
(iii) Total prize amount = 2 x 980 + 2 x 1150
= Rs. (1960 + 2300) = Rs. 4260

Ans: (c)
Sol: 

• Calculate a₁ / a₂ = 3 / 4
• Calculate b₁ / b₂ = 2 / 8 = 1 / 4
• Calculate c₁ / c₂ = -7 / 11

Comparing the ratios:
a₁ / a₂ = 3 / 4
b₁ / b₂ = 1 / 4
c₁ / c₂ = 7 / 11

Since a₁ / a₂  ≠ b₁ / b₂, the lines are not parallel. Also, since none of the ratios are equal, the lines are not coincident.

Clearly, from the graph, we can see that both lines intersect each other.

Ans: (d)
Sol: Given equations are, y = 2 and y = - 3.

Clearly, from the graph, we can see that both equations are parallel to each other.
So, there will be no solution.

Ans: (b)
Sol: Let age of father be 'x' years and age of son be 'y' years.
According to the question, x = 3y ..(i)
and x + 12 = 2 (y + 12) 
⇒ x - 2y = 12 ..(ii)
From (i) and (ii), we get x = 36, y = 12
∴ x + y = 48 years

Ans: (a)
Sol: Given,17x - 19y = 53 ...(i)
and 19x  - 17y = 55 _(ii)
Multiplying (i) by 19 and (ii) by 17, and by subtracting we get,
323x - 361y -(323x - 289y) = 1007 - 935
⇒ - 72y = 72
⇒ y = - 1
Putting y = - 1 in (i), we get,
17x - 19 (-1) = 53
⇒ 17x = 53 - 19
⇒ 17x = 34
x = 2  
∴ x + y = 2 - 1
= 1

Ans: (b)

Given equations:

x + y − 4 = 0

2x + ky − 3 = 0

The general form of a linear equation is  ax + by + c = 0. So, comparing terms:

For the lines to be parallel (and hence have no solution), we need:

a₁a₂ = b₁b₂ ≠ c₁c₂

So, 1/2 = 1/k

Cross-multiplying gives:

k = 2

Now, let’s check the condition for the

c₁c₂ = -4-3 = 43

Since 1/2 = 1/k when k = 2 but 1/2  ≠ 4/3, the condition for no solution is satisfied.

Thus, the value of $k$k for which the equations have no solution is: 2
So, the correct answer is (b) 2.

Ans: (a)
Sol: (-5, 6) is the solution of x = -5 and y = 6.

Ans: (b)
Sol: For. infinitely many solutions
a₁a₂ = b₁b₂ = c₁c₂

⇒ 3k = 515 = 824

k = 9

Ans: (a)
Sol: 32x + 33y = 34 ...(i)
33x + 32y = 31 ...(ii)
Adding equation (i) and (ii) and subtracting equation (ii) from (i),
we get 65x + 65y = 65 or x + y = 1 ...(iii)
and - x + y = 3 ...(iv)
Adding equation (iii) and (iv),
we get y = 2
Substituting the value of y in equation (iii),
x = -1

Ans: (c)
Sol: If two lines a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 are parallel, then

a₁a₂ = b₁b₂ ≠ c₁c₂
It can only possible between 3x - 2y = 5 and -12x + 8y = 7.

Ans:  5/13
Let the numerator be x and the denominator be y of the fractions. Then, the fraction = x /y.
Given , x + y = 18 - (i)
and
⇒ 3x - y = 2 . . (ii)
Adding (i) and (ii), we get
4x = 20 ⇒ x = 5
Put the value of x in (i), we get
5+ y= 18
⇒ y = 13
∴ The required fraction is 5/13

Ans:  Given, the system of equations

x + 2y = 5
3k + ky = - 15 has no solution.
∴
For K = 6 the given system of equations has no solution.

Ans: (d)
Sol: For Amruta, x + (6 - 2)y = 22
i. e., x + 4y = 22      ...(i)
For Radhika, x + (4 - 2)y = 16 i.e.,x + 2y = 16     ...(ii)
Solving equation (i) and (ii). we get
x = 10 and y = 3
i.e., Fixed charges (x) = 710       ...(iii)
and additional charges per subsequent day
(y) = ₹ 3                ...(iv)
x + 2 y = 16  [From equation (ii)]

Ans: (c)
Sol: For Amruta, x + (6 - 2)y = 22
i. e., x + 4y = 22      ...(i)
For Radhika, x + (4 - 2)y = 16 i.e.,x + 2y = 16     ...(ii)
Solving equation (i) and (ii). we get
x = 10 and y = 3
i.e., Fixed charges (x) = 710       ...(iii)
and additional charges per subsequent day
(y) = ₹ 3                ...(iv)
x + 4 y = 22  [From equation (i)]

Ans: (b)
Sol: For Amruta, x + (6 - 2)y = 22
i. e., x + 4y = 22      ...(i)
For Radhika, x + (4 - 2)y = 16 i.e.,x + 2y = 16     ...(ii)
Solving equation (i) and (ii). we get
x = 10 and y = 3
i.e., Fixed charges (x) = 710       ...(iii)
and additional charges per subsequent day
y = ₹ 3                ...(iv)
x = ₹ 10  [From equation (iii)]

Ans: (d)
Sol: For Amruta, x + (6 - 2)y = 22
i. e., x + 4y = 22      ...(i)
For Radhika, x + (4 - 2)y = 16 i.e.,x + 2y = 16     ...(ii)
Solving equation (i) and (ii). we get
x = 10 and y = 3
i.e., Fixed charges (x) = 710       ...(iii)
and additional charges per subsequent day
y = ₹ 3                ...(iv)
y = ₹ 3  [From equation (iv)]

Ans: (c)
For Amruta, x + (6 - 2)y = 22
i. e., x + 4y = 22      ...(i)
For Radhika, x + (4 - 2)y = 16 i.e.,x + 2y = 16     ...(ii)
Solving equation (i) and (ii). we get
x = 10 and y = 3
i.e., Fixed charges (x) = 710       ...(iii)
and additional charges per subsequent day
y = ₹ 3                ...(iv)
Total amount paid for 2 more days by both
= (x + 4 y) + 2 y + (x + 2y ) + 2 y
= 2 x + 10y
= 2 x 10 + 10 x 3
= ₹ 50

Ans: (a) 
Sol: The pair of equations x = a and y = b graphically represent lines which are parallel to the y-axis and x-axis respectively.
The lines will intersect each other at (a, b).

Ans: For any real number except k = -6
kx - 2y = 3 and 3x + y = 5 represent lines intersecting at a unique point.
⇒ k3 ≠ -21
⇒ k ≠ -6
For any real number except k ≠  -6
The given equation represent two intersecting lines at unique point.

Ans: (d)
Sol: For no solution; a₁a₂ = b₁b₂ ≠ c₁c₂

Ans: Solutions of linear equations

2y - x =  8    ..(i)
5y - x = 14    ...(ii)
and  y - 2x =  1    ...(iii)
are given below:

From the graph of lines represented by given equations, we observe that
Lines (i) and (iii) intersect each other at C(2, 5),
Lines (ii) and {iii) intersect each other at B(1, 3) and Lines (i) and (ii) intersect each other at 4(-4, 2).
Coordinates of the vertices of the triangle are A(-4, 2), B(1, 3) and C(2, 5).

Ans: Solution of linear equations

 x + 2y = 6 and 2x - 5y = 12
are given below 

From the graph, the two lines intersect each other at point (6, 0)
∴ x = 6 and y = 0

Ans: Let the required fraction be x/y.

According to question, we have

x - 1y = 13 ... (i)

and

xy + 8 = 14 ... (ii)

From (i), 3x - 3 = y

⇒ 3x - y - 3 = 0 ... (iii)

From (ii), 4x = y +8
so, 4x - y - 8 = 0  ... (iv)
Subtracting (iii) from (iv),
we get x = 5
Substituting the value of x in (iii),
we get y = 12
Thus, the required fraction is 5/12

Ans: Let the present age of son be x years and that of father be y years.

According to question, we have
y = 3x+ 3 ⇒ 3x - y + 3 = 0    (i)
And y + 3 = 2(x + 3) + 10
⇒  y + 3 = 2x + 6 +10
⇒ 2x - y + 13 = 0    (ii)
Subtracting (ii) from (i), we get x = 10
Substituting the value of x in (ii). we get y = 33
So. the present age of the son is 10 years and that of the father is 33 years.

Ans: Given lines are 2x + 3y = 2 and x - 2y = 8
 2x + 3y = 2

and x - 2y = 8

∴ We will plot the points (1, 0), (-2, 2) and (4, - 2 ) and join them to get the graph of 2x + 3y = 2 and we will plot the points (0, -4), (8, 0) and (2, -3) and join them to get the graph of x - 2y = 8

Ans: Let the original uniform speed of the train be x km/hr and the total length of journey be l km. Then, scheduled time taken by the train to cover a distance of l km = l/x hours
Now,

lx + 6 = lx - 4

⇒  lx - lx + 6 = 4

⇒  x + 6 - xx(x + 6) = 4

⇒  6lx(x + 6) = 4

⇒  l = 2x(x + 6)3 ... (i)

Also,

lx - 6 = lx + 6

⇒  lx - 6 - lx = 6

⇒  x - x + 6(x - 6)x = 6

⇒  6l(x - 6)x = 6

⇒ l = x(x - 6) ... (ii)

From equations (i) and (ii), we have

2x(x + 6)3 = x(x - 6)

⇒ 2x + 12 = 3x - 18

⇒ x = 30

Putting the value of x in eq. (ii), we get
l = 30(30 – 6)
= 30 × 24
= 720
Hence, the length of the journey is 720 km.

Ans: Solutions of linear equation
x - y + 1 = 0    ...(i)
and 3x + 2y - 12 = 0    ...(ii)
are given below:

From the graph, the two lines intersect each other at the point (2, 3)
∴ x = 2, y = 3.

Ans: Let the larger angle be x° and the smaller angle be y°. We know that the sum of two supplementary pairs of angles is always 180°.

We have x° + y° = 180°    (i)
and x° - y° = 18°    (ii) [Given]
By (1), we have x° = 180° - y°    _(iii)
Put the value of x° in (ii), we get
180° - y° - y° = 18°
⇒ 162° = 2y°
⇒ y = 81
From (3), we have x° = 180° - 81° = 99°
The angles are 99° and 81°

Ans: Given, pair of linear equations:

3x - 5y = 4,  ....... (i)
2y+ 7 = 9x  
9x - 2y = 7 ........ (ii)
Multiply (i) by 3 and subtract from (ii), as

⇒ 9x - 2y - 9x + 15y = -5 → 13y = -5

⇒ y = -513

Put y = -513 in (i), we get

3x - 5 -513 = 4

⇒ 3x + 2513 = 4

⇒ 3x = 4 - 2513

⇒ x = 2713 × 3 = 913

Hence, x  = 9/13 and y = -5/13

Ans: Let the ages of two children be x and y respectively.

Father's present age = 3(x +y)
After 5 years, sum of ages of children = x + 5 + y + 5
= x + y + 10
and age of father = 3(x + y) + 5
According to the question,
3(x + y) + 5 = 2(x + y+ 10)
3x + 3y + 5 = 2x + 2y + 20
⇒ x + y = 15
Hence, present age of father = 3(x + y)
= 3 x 15 = 45 years

Ans: Let the fraction be x/y

Then, according to question.

x - 2y = 13 and xy - 1 = 12

⇒ 3x - 6 = y and 2x = y - 1

⇒ 3x - y - 6 = 0 ... (i) and 2x - y + 1 = 0 ... (ii)

Subtracting (ii) from (i), we get x - 7 = 0
So, x = 7
From (i) ,3(7) - y - 6 = 0
⇒ 21 - 6 = y
⇒ y = 15
Therefore required fraction is 7/15

Ans:  The given pair of linear equations is
x + 2y = 5
3x + ky= -15
Since the system of equations has a unique solution

∴ For all values of k except k = 6, the given pair of linear equations will have a unique solution.

Ans: Given pair of equations are

x - 4y + p = 0    (i)
and 2x + y - q - 2 = 0    (ii)
It is given that x = 3 and y = 1 is the solution of (i) and (ii)
∴ 3 - 4 x 1+ p = 0
⇒ p = 1
and 2 x 3 + 1 - q - 2 = 0
⇒ q = 5  
∴ q = 5p

Ans: The given system of linear equations are:      
2x + 3y = 7 
(k – 1)x + (k + 2)y = 3k 
For infinitely many solutions:

a₁a₂ = b₁b₂ = c₁c₂

Here,

a₁ = 2, b₁ = 3, c₁ = -7

a₂ = (k - 1), b₂ = (k + 2), c₂ = -3k

⇒ 2k - 1 = 3k + 2 = -7-3k

⇒ 2(k + 2) = 3(k - 1); 3(3k) = 7(k + 2)

⇒ 2k – 3k = –3 – 4; 9k – 7k = 14 
⇒  –k = –7; 2k = 14 
⇒  k = 7; k = 7 Hence, the value of k is 7