-
Q1 Aftab tells his daughter, “Seven years ago, I was seven times as old as you were then. Also, three years from now, I shall be three times as old as you will be.” (Isn’t this interesting?) Represent this situation algebraically and graphically.
Ans: Let the present age of Aftab and his father be x years and y years respectively.
According to question,
7 years ago, we have
x – 7 = 7 (y - 7)
Or, x – 7 = 7y – 49
Or, x – 7y = - 42 …………… (1)
3 years from now, we have
(x + 3) = 3 (y + 3)
Or, x + 3 = 3y + 9
Or, x – 3y = 6 ……………. (2)
Graphical Representation
From equation (1), x – 7y = -42
Table value of x and y
x:
0
-42
-35
y:
6
0
1
From equation (2), x – 3y = 6
Table value of x and y
x:
0
9
6
y:
-2
1
0
Plotting the tables on the graph:
Q2 The coach of a cricket team buys 3 bats and 6 balls for ` 3900. Later, she buys another bat and 3 more balls of the same kind for ` 1300. Represent this situation algebraically and geometrically.
Ans: Let the cost of one bat and one ball be x Rupees and y Rupees respectively.
According to first condition,
3x + 6y = ₨ 3900 …………….(1)
According to second condition,
x + 3y = ₨ 1300 …………….(2)
Graphical Representation
From equation (1), 3x + 6y = 3900
Table value of x and y
x:
100
300
700
y:
600
500
300
From equation (2), x + 3y = 1300
Table value of x and y
x:
100
700
400
y:
400
200
300
1 Unit = 100
Q3 The cost of 2 kg of apples and 1kg of grapes on a day was found to be ` 160. After a month, the cost of 4 kg of apples and 2 kg of grapes is ` 300. Represent the situation algebraically and geometrically.
Ans: Let the cost of one kg apple be x ₨ and 1 kg grapes be y ₨.
According to first condition,
2x + y = 160 ₨ …………..(1)
According to second condition,
4x + 2y = 300
2x + y = 150 ……………….(2)
Graphical Representation
Table for equation (1), 2x + y = 160
Table value of x and y
x:
40
60
80
y:
60
40
0
Table for equation (2), 2x + y = 150
Table value of x and y
x:
40
60
20
y:
70
30
110
Pair of Linear Equations in Two Variables Question Answers: NCERT Class 10 Mathematics
Welcome to the Chapter 3 - Pair of Linear Equations in Two Variables, Class 10 Mathematics - NCERT Solutions page. Here, we provide detailed question answers for Chapter 3 - Pair of Linear Equations in Two Variables.The page is designed to help students gain a thorough understanding of the concepts related to natural resources, their classification, and sustainable development.
Our solutions explain each answer in a simple and comprehensive way, making it easier for students to grasp key topics Graphical and algebraic solutions, consistency/inconsistency, substitution and elimination methods and excel in their exams. By going through these Pair of Linear Equations in Two Variables question answers, you can strengthen your foundation and improve your performance in Class 10 Mathematics. Whether you're revising or preparing for tests, this chapter-wise guide will serve as an invaluable resource.
Exercise 1 ( Page No. : 44 )
Exercise 2 ( Page No. : 50 )
-
Q5 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.Ans: 
Q6 Given the linear equation 2x + 3y – 8 = 0, write another linear equation in two variables such that the geometrical representation of the pair so formed is:
(i) intersecting lines (ii) parallel lines (iii) coincident linesAns: 
Q7 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.
Ans: Given equations,
x + y + 1 = 0 …………………. (1)
3x + 2y + 2 = 0 ………………….. (2)
From the equation (1), we get
x
0
1
2
y
1
2
3
From the equation (2), we get
x
4
3
0
y
0
3
6
The coordinates of vertices of triangle formed by these lines and the x- axis are (-1, 0), (4, 0), (2, 3).
Exercise 3 ( Page No. : 53 )
-
Q1 Solve the following pair of linear equations by the substitution method.
Ans: (i) x + y = 14 …………….(1)
x – y = 4 …………….(2)
From the equation (1), we get
x = 14 - y …………….(3)
Putting the value of x in equation (2), we get
(14 - y) – y = 4
14 – y – y = 4
- 2y = - 10
Putting the value of y in equation (3),
x = 14 – 5
x = 9
Hence, x = 9 and y = 5
(iii) 3x - y = 3 …………….(1)
9x – 3y = 9…………….(2)
From the equation (1), we get
Putting the value of y in equation (2), we get
9x – 3 (3x - 3) = 9
9x – 9x + 9 = 9
9 = 9, which is true.
Therefore, pair of linear equation has infinite many solutions.
Q2 Solve 2x + 3y = 11 and 2x – 4y = – 24 and hence find the value of ‘m’ for which y = mx + 3.
Ans: 
Q3 Form the pair of linear equations for the following problems and find their solution by substitution method.
(i) The difference between two numbers is 26 and one number is three times the other. Find them.
(ii) The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.
(iii) 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.(iv) The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is Rs. 105 and for a journey of 15 km, the charge paid is Rs. 155. What are the fixed charges and the charge per km? How much does a person have to pay for travelling a distance of 25 km?
(v) A fraction becomes
, , if 2 is added to both the numerator and the denominator. If, 3 is added to both the numerator and the denominator it becomes 
Find the fraction.(vi) 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?
Ans: (i) Let the numbers be x and y, such that x > y
Therefore, according to question
x - y = 26 …………….(1)
x = 3y…………….(2)
Putting the value of x from equation (2) to equation (1), we get
3y – y = 26
2y = 26
y = 13
Putting the value in equation (2), we get
x = 3 x 13
x = 39
Hence, the numbers are 39 and 13.
(ii) Let one be xâ—¦ and other be yâ—¦ such that (xâ—¦ > yâ—¦)
Therefore, according to question
xâ—¦ + yâ—¦ = 180â—¦…………….(1) (Supplementary angles)
xâ—¦ = 18 + yâ—¦…………….(2)
Putting the value of x from equation (2) to equation (1), we get
18 + yâ—¦ + yâ—¦ = 180â—¦
2yâ—¦ = 162â—¦
Putting the value of y in equation (2), we get
xâ—¦ = 18 + 81
x = 99
(iv) Let the fixed charge be = ₨ x
Let the charge for 1 km distance be = ₨ y
According to first condition,
x + 10y = ₨ 105
x = 105 – 10y …………….(1)
According to second condition,
x + 15y = 155 ………………(2)
Putting the value of x in equation (2), we get
105 – 10y + 15y = 155
5y = 50
y = 10
Putting the value of y in equation (2), we get
x + 15 x 10 = 155
x = 5
Hence, the fixed charge for taxi is ₨ 5 and, the charge for one km distance is ₨ 50.
Charge for 25 km distance
= 25 x 10 + 5
= ₨ 255
(vi) Let the age of Jacob be = x years
Let the age of Jacob’s father be = y years
After 5 years,
Jacob’s age x + 5 years
Son’s age y + 5 years
According to question,
x + 5 = 3 (y + 5)
x + 5 = 3y + 15
x = 3y + 10………………(1)
Five years ago,
(x- 5) = 7 (y - 5)
x – 5 = 7y – 35
x – 7y = -30……………….(2)
Putting the value of x in equation (2), we get
3y + 10 - 7y = -30
-4y = -40
y = 10
Putting the value of y in equation (1), we get
x = 3 (10) + 10
x = 40
Hence, the present age of Jacob is 40 years and the age of his son is 10 years.
Key Features of NCERT Class 10 Mathematics Chapter 'Pair of Linear Equations in Two Variables' question answers :
- All chapter question answers with detailed explanations.
- Simple language for easy comprehension.
- Aligned with the latest NCERT guidelines.
- Perfect for exam preparation and revision.
Popular Questions of Class 10 Mathematics
- Q:-
Is the following situation possible? If so, determine their present ages.
The sum of the ages of two friends is 20 years. Four years ago, the product of their ages in years was 48. - Q:-
A cottage industry produces a certain number of pottery articles in a day. It was observed on a particular day that the cost of production of each article (in rupees) was 3 more than twice the number of articles produced on that day. If the total cost of production on that day was Rs 90, find the number of articles produced and the cost of each article.
- Q:-
Find two consecutive positive integers, sum of whose squares is 365.
- Q:-
Find the nature of the roots of the following quadratic equations. If the real roots exist, find them:
(i) 2x2 – 3x + 5 = 0
(iii) 2x2– 6x + 3 = 0 - Q:-
Find two numbers whose sum is 27 and product is 182.
- Q:-
Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.
(i) x2 – 2x – 8 (ii) 4s2 – 4s + 1 (iii) 6x2 – 3 – 7x (iv) 4u2 + 8u (v) t2 – 15 (vi) 3x2 – x – 4
- Q:-
Refer to Example 13. (i) Complete the following table:
(ii) A student argues that ‘there are 11 possible outcomes 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 and 12. Therefore, each of them has a probability
Do you agree with this argument? Justify your answer.
- Q:-
Two poles of equal heights are standing opposite each other on either side of the road, which is 80 m wide. From a point between them on the road, the angles of elevation of the top of the poles are 60° and 30°, respectively. Find the height of the poles and the distances of the point from the poles.
- Q:-
Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.
- Q:-
The graphs of y = p(x) are given in Fig. 2.10 below, for some polynomials p(x). Find the number of zeroes of p(x), in each case.
Recently Viewed Questions of Class 10 Mathematics
- Q:-
A cottage industry produces a certain number of pottery articles in a day. It was observed on a particular day that the cost of production of each article (in rupees) was 3 more than twice the number of articles produced on that day. If the total cost of production on that day was Rs 90, find the number of articles produced and the cost of each article.
- Q:-
Two poles of equal heights are standing opposite each other on either side of the road, which is 80 m wide. From a point between them on the road, the angles of elevation of the top of the poles are 60° and 30°, respectively. Find the height of the poles and the distances of the point from the poles.
- Q:-
Complete the following statements:
(i) Probability of an event E + Probability of the event ‘not E’ = .
(ii) The probability of an event that cannot happen is . Such an event is called .
(iii) The probability of an event that is certain to happen is . Such an event is called .
(iv) The sum of the probabilities of all the elementary events of an experiment is .(v) The probability of an event is greater than or equal to and less than or equal to .
- Q:-
A game of chance consists of spinning an arrow which comes to rest pointing at one of the numbers 1, 2, 3, 4, 5, 6, 7, 8 (see Fig. 15.5 ), and these are equally likely outcomes. What is the probability that it will point at
(i) 8 ?
(ii) an odd number?
(iii) a number greater than 2?
(iv) a number less than 9?
- Q:-
Find the nature of the roots of the following quadratic equations. If the real roots exist, find them:
(i) 2x2 – 3x + 5 = 0
(iii) 2x2– 6x + 3 = 0 - Q:-
A jar contains 24 marbles, some are green and others are blue. If a marble is drawn at random from the jar, the probability that it is green is 2/3.
Find the number of blue balls in the jar.
- Q:-
Find two consecutive positive integers, sum of whose squares is 365.
- Q:-
Find two numbers whose sum is 27 and product is 182.
- Q:-
The cost of 2 kg of apples and 1kg of grapes on a day was found to be ` 160. After a month, the cost of 4 kg of apples and 2 kg of grapes is ` 300. Represent the situation algebraically and geometrically.
- Q:-
Write down the decimal expansions of those rational numbers in Question 1 above which have terminating decimal expansions.
- All Chapters Of Class 10 Mathematics
- All Subjects Of Class 10