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 Pair of Linear Equations in Two Variables 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
A:

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:

A:

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

A:

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

      Exercise 3

A:

(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.

A:


A:

(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 x◦ and other be y◦ such that (x◦ > y◦)

Therefore, according to question          

                          x◦ + y◦ = 180◦…………….(1) (Supplementary angles)

x◦ = 18 + y◦…………….(2)

Putting the value of x from equation (2) to equation (1), we get                                             

18 + y◦ + y◦ = 180◦

2y◦ = 162◦

Putting the value of y in equation (2), we get

                      x◦ = 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.


Exercise 2


A:

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).


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Exam Preparation Tips for Pair of Linear Equations in Two Variables

The Pair of Linear Equations in Two Variables is an important chapter of 10 Mathematics. This chapter’s important topics like Pair of Linear Equations in Two Variables are often featured in board exams. Practicing the question answers from this chapter will help you rank high in your board exams.

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