Important Questions Class 10 Maths Chapter 3 Linear Equations In Two Variables

Important questions of Chapter 3 Linear Equations In Two Variables with solutions for class 10 Maths are available for students preparing for the CBSE board exam 2022-2023. By practising these questions, students will score good marks, covering the major topics included in the NCERT syllabus. These questions are created by subject experts after thorough research and are based on the latest exam pattern. Students can expect a few questions from these important questions in the exam. So, they are advised to practice them well to score high marks in the exam.

Check: Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables MCQs

Chapter 3 linear equations In two variables deals with finding the solutions for pairs of such equations with two variables present. These solutions can also be represented in graphs. Let us solve the questions which are important for the final exams of the 10th standard. Students can practice the additional questions provided at the end of the page for a better understanding of the concept and score maximum marks in the board exam.

Also Check:

• Important 2 Marks Questions for CBSE 10th Maths
• Important 3 Marks Questions for CBSE 10th Maths
• Important 4 Marks Questions for CBSE 10th Maths

Class 10 Maths Chapter 3 Important Questions With Answers

Q.1: The cost of 2 kg of apples and 1kg of grapes on a day was found to be Rs.160. After a month, the cost of 4 kg of apples and 2 kg of grapes is Rs.300. Represent the situation algebraically.

Solution:

Let the cost of 1 kg of apples be ‘Rs. x’.
And, let the cost of 1 kg of grapes be ‘Rs. y’.
According to the question, the algebraic representation is
2x + y = 160
And 4x + 2y = 300

For, 2x + y = 160 or y = 160 − 2x, the solution table is;

For 4x + 2y = 300 or y = (300 – 4x)/ 2, the solution table is;

Note: Students can also represent these two equations graphically, by using the given points of x-coordinate and y-coordinate.

Q.2: Half the perimeter of a rectangular garden, whose length is 4 m more than its width, is 36 m. Find the dimensions of the garden.

Solution:

Given, half the perimeter of a rectangular garden = 36 m

so, 2(l + b)/2 = 36

(l + b) = 36 ……….(1)
Given, the length is 4 m more than its width.

Let width = x

And length = x + 4

Substituting this in eq(1), we get;

x + x + 4 = 36
2x + 4 = 36
2x = 32
x = 16

Therefore, the width is 16 m and the length is 16 + 4 = 20 m.

Q.3: On comparing the ratios a₁/a₂, b₁/b₂, and c₁/c₂, find out whether the following pair of linear equations are consistent, or inconsistent.

(i) 3x + 2y = 5 ; 2x – 3y = 7

(ii) 2x – 3y = 8 ; 4x – 6y = 9

Solution:

(i) Given : 3x + 2y = 5 or 3x + 2y – 5 = 0
and 2x – 3y = 7 or 2x – 3y – 7 = 0

Comparing the above equations with a₁x + b₁y + c₁=0
And a₂x + b₂y + c₂ = 0
We get,
a₁ = 3, b₁ = 2, c₁ = -5
a₂ = 2, b₂ = -3, c₂ = -7

a_1/a₂ = 3/2, b₁/b₂ = 2/-3, c₁/c₂ = -5/-7 = 5/7
Since, a₁/a₂≠b₁/b₂ the lines intersect each other at a point and have only one possible solution.

Hence, the equations are consistent.

(ii) Given 2x – 3y = 8 and 4x – 6y = 9
Therefore,
a₁ = 2, b₁ = -3, c₁ = -8
a₂ = 4, b₂ = -6, c₂ = -9

a₁/a₂ = 2/4 = 1/2, b₁/b₂ = -3/-6 = 1/2, c₁/c₂ = -8/-9 = 8/9

Since, a₁/a₂=b₁/b₂≠c₁/c₂

Therefore, the lines are parallel to each other and they have no possible solution. Hence, the equations are inconsistent.

Q.4: Solve the following pair of linear equations by the substitution method.

(i) x + y = 14
x – y = 4

(ii) 3x – y = 3
9x – 3y = 9

Solution:

(i) Given,
x + y = 14 and x – y = 4 are the two equations.
From 1st equation, we get,
x = 14 – y
Now, put the value of x in second equation to get,
(14 – y) – y = 4
14 – 2y = 4
2y = 10
Or y = 5
By the value of y, we can now find the value of x;
∵ x = 14 – y
∴ x = 14 – 5
Or x = 9
Hence, x = 9 and y = 5.

(ii) Given,
3x – y = 3 and 9x – 3y = 9 are the two equations.
From 1st equation, we get,
x = (3 + y)/3
Now, substitute the value of x in the given second equation to get,
9[(3 + y)/3] – 3y = 9

⇒ 3(3+y) – 3y = 9
⇒ 9 + 3y – 3y = 9
⇒ 9 = 9
Therefore, y has infinite values and since, x = (3 + y)/3, so x also has infinite values.

Q.5: Solve 2x + 3y = 11 and 2x – 4y = – 24 and hence find the value of ‘m’ for which y = mx + 3.

Solution:

2x + 3y = 11…………………………..(i)
2x – 4y = -24………………………… (ii)
From equation (i), we get;
x = (11 – 3y)/2 ……….…………………………..(iii)

Putting the value of x in equation (ii), we get

2[(11 – 3y)/2] – 4y = −24
11 – 3y – 4y = -24
-7y = -35
y = 5……………………………………..(iv)
Putting the value of y in equation (iii), we get;

x = (11 – 15)/2 = -4/2 = −2
Hence, x = -2, y = 5
Also,
y = mx + 3
5 = -2m +3
-2m = 2
m = -1
Therefore, the value of m is -1.

Q.6: The coach of a cricket team buys 7 bats and 6 balls for Rs.3800. Later, she buys 3 bats and 5 balls for Rs.1750. Find the cost of each bat and each ball.
Solution:
Let the cost of a bat be x and the cost of a ball be y.
According to the question,
7x + 6y = 3800 ………………. (i)
3x + 5y = 1750 ………………. (ii)
From (i), we get;
y = (3800 – 7x)/6 …………………… (iii)
Substituting (iii) in (ii). we get,
3x + 5[(3800 – 7x)/6] = 1750
⇒3x + (9500/3) – (35x/6) = 1750
3x – (35x/6) = 1750 – (9500/3)

(18x – 35x)/6 = (5250 – 9500)/3
⇒-17x/6 = -4250/3
⇒-17x = -8500
x = 500
Putting the value of x in (iii), we get;

y = (3800 – 7 × 500)/6 = 300/6 = 50
Hence, the cost of a bat is Rs 500 and the cost of a ball is Rs 50.

Q.7: A fraction becomes 9/11 if 2 is added to both the numerator and the denominator. If, 3 is added to both the numerator and the denominator it becomes 5/6. Find the fraction.

Solution:
Let the fraction be x/y.

According to the question,
(x + 2)/(y + 2) = 9/11

11x + 22 = 9y + 18

11x – 9y = -4 …………….. (1)
(x + 3)/(y + 3) = 5/6

6x + 18 = 5y +15
6x – 5y = -3 ………………. (2)
From (1), we get

x = (-4 + 9y)/11 …………….. (3)

Substituting the value of x in (2), we get
6[(-4 + 9y)/11] – 5y = -3

-24 + 54y – 55y = -33
-y = -9
y = 9 ………………… (4)
Substituting the value of y in (3), we get
x = (-4 + 81)/11 = 77/11 = 7

Hence, the fraction is 7/9.

Q.8 Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method:

(i) Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu?
Solution:
Let us assume, the present age of Nuri be x.
And the present age of Sonu is y.
According to the given condition, we can write as;
x – 5 = 3(y – 5)
x – 3y = -10…………………………………..(1)
Now,
x + 10 = 2(y +10)
x – 2y = 10…………………………………….(2)
Subtract eq. 1 from 2, to get,
y = 20 ………………………………………….(3)

Substituting the value of y in eq.1, we get,
x – 3(20) = -10
x – 60 = -10
x = 50
Therefore,
The age of Nuri is 50 years
The age of Sonu is 20 years.

(ii) A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid Rs.27 for a book kept for seven days, while Susy paid Rs.21 for the book she kept for five days. Find the fixed charge and the charge for each extra day.
Solution:

Let the fixed charge for the first three days be Rs. A and the charge for each day extra be Rs. B.
According to the information given,
A + 4B = 27 …………………………………….…………………………. (i)
A + 2B = 21 ……………………………………………………………….. (ii)
When equation (ii) is subtracted from equation (i) we get,
2B = 6
B = 3 …………………………………………………………………………(iii)
Substituting B = 3 in equation (i) we get,
A + 12 = 27
A = 15
Hence, the fixed charge is Rs. 15.
And the Additional charge per day is Rs. 3.

Q.9: Solve the following pair of linear equations by the substitution and cross-multiplication methods:
8x + 5y = 9
3x + 2y = 4

Solution:

8x + 5y = 9 …………………..(1)
3x + 2y = 4 ……………….….(2)
From equation (2) we get;

x = (4 – 2y) / 3 ……………………. (3)
Substituting this value in equation 1, we get
8[(4 – 2y)/3] + 5y = 9
32 – 16y + 15y = 27
-y = -5
y = 5 ……………………………….(4)
Substituting this value in equation (2), we get
3x + 10 = 4

3x = -6
x = -2
Thus, x = -2 and y = 5.

Now, Using Cross Multiplication method:
8x +5y – 9 = 0
3x + 2y – 4 = 0

x/(-20 + 18) = y/(-27 + 32 ) = 1/(16 – 15)
-x/2 = y/5 = 1/1
∴ x = -2 and y =5.

Q.10: Formulate the following problems as a pair of equations, and hence find their solutions:

(i) Ritu can row downstream 20 km in 2 hours, and upstream 4 km in 2 hours. Find her speed of rowing in still water and the speed of the current.

Solution:

(i) Let us consider,
Speed of boat is still water = x km/hr
Speed of current = y km/hr
Now, speed of Ritu, during,
Downstream = x + y km/hr
Upstream = x – y km/hr
As per the question given,
2(x + y) = 20
Or x + y = 10……………………….(1)
And, 2(x – y) = 4
Or x – y = 2………………………(2)
Adding both the eq.1 and 2, we get,
2x = 12
x = 6
Putting the value of x in eq.1, we get,
y = 4
Therefore,

Speed of Ritu is still water = 6 km/hr
Speed of current = 4 km/hr

Q.11: Solve the equations x + 2y – 4 = 0 and 2x + 4y – 12 = 0 graphically.

Solution:

Given,

x + 2y – 4 = 0….(i)

2x + 4y – 12 = 0….(ii)

From (i),

x + 2y = 4

2y = 4 – x

y = (4 – x)/2

From (ii),

2x + 4y = 12

x + 2y = 6

2y = 6 – x

y = (6 – x)/2

Plotting the points on the graph, we get;

Here, the lines represent the given pair of linear equations are parallel.

Thus, there is no solution to the given pair of linear equations.

Q.12: Find the value(s) of k so that the pair of equations x + 2y = 5 and 3x + ky + 15 = 0 has a unique solution.

Solution:

Given,

x + 2y = 5

3x + ky + 15 = 0

Also, given that the pair of equations has a unique solution.

Comparing the given equations with standard form,

a₁ = 1, b₁ = 2, c₁ = -5

a₂ = 3, b₂ = k, c₂ = 15

Condition for unique solution is:

a₁/a₂ ≠ b₁/b₂

1/3 ≠ 2/k

k ≠ (2)(3)

k ≠ 6

Thus, for all real values of k except 6, the given pair of equations has a unique solution.

Q.13: Determine graphically the coordinates of vertices of a triangle, the equation of whose sides are given by 2y – x = 8, 5y – x = 14 and y – 2x = 1.

Solution:

Given,

2y – x = 8….(i)

5y – x = 14….(ii)

y – 2x = 1….(iii)

From (i),

2y = x + 8

y = (x + 8)/2

From (ii),

5y = x + 14

y = (x + 14)/5

From (iii),

y = 2x + 1

Let us plot all these points on the graph.

From the graph, we can write the coordinates of vertices of triangle formed are:

P(-4, 2), Q(1, 3), and R(2, 5)

Q.14: Use elimination method to find all possible solutions of the following pair of linear equation:

2x + 3y = 8 

4x + 6y = 7

Solution:

Given,

2x + 3y = 8….(i)

4x + 6y = 7….(ii)

Multiply Equation (1) by 2 and Equation (2) by 1 to make the coefficients of x equal.

4x + 6y = 16….(iii)

4x + 6y = 7….(iv)

Subtracting (iv) from (iii),

4x + 6y – 4x – 6y = 16 – 7

0 = 9, it is not possible

Therefore, the pair of equations has no solution.

Q.15: Solve the following pairs of equations by reducing them to a pair of linear equations:

1/2x + 1/3y = 2

1/3x + 1/2y = 13/6

Solution: 

Given,

1/2x + 1/3y = 2

1/3x + 1/2y = 13/6

Let us assume 1/x = m and 1/y = n , then the equations will change as follows.

m/2 + n/3 = 2

⇒ 3m+2n-12 = 0….(1)

m/3 + n/2 = 13/6

⇒ 2m+3n-13 = 0….(2)

Now, using cross-multiplication method, we get,

m/(-26-(-36) ) = n/(-24-(-39)) = 1/(9-4)

m/10 = n/15 = 1/5

m/10 = 1/5 and n/15 = 1/5

So, m = 2 and n = 3

1/x = 2 and 1/y = 3

x = 1/2 and y = 1/3

Pair of Linear Equations in Two Variables Questions for Practice

1. A motorboat whose speed is 18 km/hr in still water takes 1 hr more to go 24 km upstream than to return downstream to the same spot. Find the speed of the stream.
2. For which values of a and b do the following pair of linear equations have infinite solutions?
   2x + 3y = 7
   (a – b) x + (a + b) y = 3a + b – 2
3. Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacob’s age was seven times that of his son. What are their present ages?
4. Draw the graphs of the equations x – y + 1 = 0 and 3x + 2y – 12 = 0. Determine the coordinates of the vertices of the triangle formed by these lines and the x-axis, and shade the triangular region.
5. Two lines are represented by the equations x + 2y – 4 = 0 and 2x + 4y – 12 = 0. Represent this situation geometrically.
6. Solve the following pair of equations by substitution method:
   7x – 15y = 2
   x + 2y = 3
7. The ratio of incomes of two persons is 9 : 7 and the ratio of their expenditures is 4 : 3. If each of them manages to save Rs. 2000 per month, find their monthly incomes.
8. Solve the following pair of linear equations by the substitution and cross-multiplication methods :
   8x + 5y = 9
   3x + 2y = 4
9. Solve the following pairs of equations by reducing them to a pair of linear equations:
   6x + 3y = 6xy
   2x + 4y = 5xy
10. The students of a class are made to stand in rows. If 3 students are extra in a row, there would be 1 row less. If 3 students are less in a row, there would be 2 rows more. Find the number of students in the class.

Students must have found these important questions of Class 10 Maths Chapter 3 Linear Equations In Two Variables helpful for the exam preparation. Keep practising and stay tuned to BYJU’S for the latest update on CBSE/ICSE/State Board/Competitive exams.