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

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: (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
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, 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

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 + 1 = y - 1
⇒ x = y-2                      ...(i)
Also,
⇒ 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;

2k - 1 = 3k - 3
⇒ k = 2
Also,
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: 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

x + 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:

So, 1/2 = 1/k
Cross-multiplying gives:
k = 2
Now, let’s check the condition for the constants:

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

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

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 = 13 - (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.

For any real number except fc = -6
The given equation represent two intersecting lines at unique point.

Ans: (d)
Sol: For no solution;

Hence, option (d) is correct

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

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 o f 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,

 From equations (i) and (ii), we have

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

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.

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 2x 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:

Here,
a₁ = 2, b₁ = 3, c₁ = –7 and 
a₂ = (k – 1), b₂ = (k + 2), c₂ = –3k

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