We the team of subject wise have provided detailed solution of class 10 ncert exercise 3.1 so please read this post completely if you have any doubt please comment it down also follow our you-tube channel.
Chapter 3-Pairs of linear equation in 1 variable Exercise Ex.3.1
here it is solution of question 1
Chapter 3 - Pairs of Linear Equations in Two Variables Exercise Ex. 3.1
Solution 1
Let the present age of Aftab and his daughter be x and y respectively.
Seven years ago,
Age of Aftab = x - 7
Age of his daughter = y - 7
According to the given condition,
Three years hence,
Age of Aftab = x + 3
Age of his daughter = y + 3
According to the given condition,
Thus, the given conditions can be algebraically represented as:
x - 7y = -42
x - 3y = 6
Three solutions of this equation can be written in a table as follows:
Three solutions of this equation can be written in a table as follows:
The graphical representatino is as follows:
Concept insight: In order to represent a given situation mathematically, first see what we need to find out in the problem. Here, Aftab and his daughter's present age needs to be found so, so the ages will be represented by variables x and y. The problem talks about their ages seven years ago and three years from now. Here, the words 'seven years ago' means we have to subtract 7 from their present ages, and 'three years from now' or 'three years hence' means we have to add 3 to their present ages. Remember in order to represent the algebraic equations graphically the solution set of equations must be taken as whole numbers only for the accuracy. Graph of the two linear equations will be represented by a straight line.
Seven years ago,
Age of Aftab = x - 7
Age of his daughter = y - 7
According to the given condition,
Three years hence,
Age of Aftab = x + 3
Age of his daughter = y + 3
According to the given condition,
Thus, the given conditions can be algebraically represented as:
x - 7y = -42
x - 3y = 6
Three solutions of this equation can be written in a table as follows:
| x | -7 | 0 | 7 |
| y | 5 | 6 | 7 |
Three solutions of this equation can be written in a table as follows:
| x | 6 | 3 | 0 |
| y | 0 | -1 | -2 |
The graphical representatino is as follows:
Concept insight: In order to represent a given situation mathematically, first see what we need to find out in the problem. Here, Aftab and his daughter's present age needs to be found so, so the ages will be represented by variables x and y. The problem talks about their ages seven years ago and three years from now. Here, the words 'seven years ago' means we have to subtract 7 from their present ages, and 'three years from now' or 'three years hence' means we have to add 3 to their present ages. Remember in order to represent the algebraic equations graphically the solution set of equations must be taken as whole numbers only for the accuracy. Graph of the two linear equations will be represented by a straight line.
Chapter 2-Pairs of linear equation in 1 variable Exercise Ex.3.1
here it is solution of question 2
Solution 2
Let the cost of a bat and a ball be Rs x and Rs y respectively.
The given conditions can be algebraically represented as:
3x + 6y = 3900
x + 2y = 1300 .... (i)
x + 3y = 1300 .... (ii)
From (i), we get
x = 1300 - 2y
Three solutions of this equation can be written in a table as follows:
From (ii), we get
x = 1300 - 3y
Three solutions of this equation can be written in a table as follows:
The graphical representation is as follows:
Concept insight: Cost of bats and balls needs to be found so the cost of a ball and bat will be taken as the variables. Applying the conditions of total cost of bats and balls algebraic equations will be obtained. Then, in order to represent the obtained equations graphically take atleast three ordered pairs on both the equations in order to avoid any computational errors.
The given conditions can be algebraically represented as:
3x + 6y = 3900
x + 2y = 1300 .... (i)x + 3y = 1300 .... (ii)
From (i), we get
x = 1300 - 2y
Three solutions of this equation can be written in a table as follows:
| x | 300 | 100 | -100 |
| y | 500 | 600 | 700 |
x = 1300 - 3y
Three solutions of this equation can be written in a table as follows:
| x | 100 | 400 | -200 |
| y | 400 | 300 | 500 |
The graphical representation is as follows:
Concept insight: Cost of bats and balls needs to be found so the cost of a ball and bat will be taken as the variables. Applying the conditions of total cost of bats and balls algebraic equations will be obtained. Then, in order to represent the obtained equations graphically take atleast three ordered pairs on both the equations in order to avoid any computational errors.
Chapter 2-Pairs of linear equation in 1 variable Exercise Ex.3.1
here it is solution of question 3
Solution 3
Let the cost of 1 kg of apples and 1 kg grapes be Rs. x and Rs. y.
The given condtions can be algebraically represented as:
Three solutions of this equation can be written in a table as follows:
Three solutions of this equation can be written in a table as follows:
The graphical representation is as follows:
Concept insight: cost of apples and grapes needs to be found so the cost of 1 kg apples and 1 kg grapes will be taken as the variables. From the given conditions of collective cost of apples and grapes, a pair of linear equations in two variables will be obtained. Then, in order to represent the obtained equations graphically, take the values of variables as whole numbers only. Since these values are large so take the suitable scale like 1 cm = 20.
The given condtions can be algebraically represented as:
Three solutions of this equation can be written in a table as follows:
x |
50 | 60 | 70 |
| y | 60 | 40 | 20 |
Three solutions of this equation can be written in a table as follows:
x |
70 | 80 | 75 |
| y | 10 | -10 | 0 |
The graphical representation is as follows:
Concept insight: cost of apples and grapes needs to be found so the cost of 1 kg apples and 1 kg grapes will be taken as the variables. From the given conditions of collective cost of apples and grapes, a pair of linear equations in two variables will be obtained. Then, in order to represent the obtained equations graphically, take the values of variables as whole numbers only. Since these values are large so take the suitable scale like 1 cm = 20.
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if you not read the previous blog please read it. it is the solution to exercise 1.1 AND EXERCISE 2.3
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https://www.youtube.com/watch?v=Wjsw6GuJnz0&list=PLcMqfTNwNJuQ6ut9yqZwWaHiaPRgJIJ57&index=2
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https://youtu.be/JvGpUNF5dhA
Class 10 physics s chand-
https://amzn.to/2ygIG5a
Interesting question video-https://mydomainscan.com/ZsbiLmG
Maths problem-
https://mydomainscan.com/wxSr5VQg
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https://mydomainscan.com/wxSr5VQg
best books-
https://amzn.to/3fThv1f
https://amzn.to/2LEbPKA
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