NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.1 are part of NCERT Solutions for Class 10 Maths. Here we have given NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.1.
- Pair of Linear Equations in Two Variables Class 10 Ex 3.1
- Pair of Linear Equations in Two Variables Class 10 Ex 3.2
- Pair of Linear Equations in Two Variables Class 10 Ex 3.3
- Pair of Linear Equations in Two Variables Class 10 Ex 3.4
- Pair of Linear Equations in Two Variables Class 10 Ex 3.5
- Pair of Linear Equations in Two Variables Class 10 Ex 3.6
- Pair of Linear Equations in Two Variables Class 10 Ex 3.7
| Board | CBSE |
| Textbook | NCERT |
| Class | Class 10 |
| Subject | Maths |
| Chapter | Chapter 3 |
| Chapter Name | Pair of Linear Equations in Two Variables |
| Exercise | Ex 3.1 |
| Number of Questions Solved | 3 |
| Category | NCERT Solutions |
NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.1
Question 1.
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.
Solution:
Let the present age of Aftab’s daughter be x years and that of Aftab be y years.
Then, seven years ago:
Aftab’s age was (y – 7) years
and his daughter’s age was (x – 7) years.
According to the first condition:
y – 7 = 7(x – 7)
⇒ y – 7 = 7x – 49
⇒ 7x – y = 42 …(i)
Five years later:
Aftab’s age will be (y + 5) years,
and his daughter’s age will be (x + 5) years
According to the second condition:
y + 5 = 3(x + 5)
⇒ y + 5 = 3x + 15
⇒ 3x – y = -10 …(ii)
For graphical representation:
From equation (i), we have:
y = 7x – 42
When x = 11, then y = 7 x 11 – 42 = 35
When x = 12, then y = 7 x 12 – 42 = 42
When x = 13, then y = 7 x 13 – 42 = 49
Thus, we have the following table of points:
From equation (ii), we have:
y = 3x + 10
When x = 9, then y = 3 x 9 + 10 = 37
When x = 10, then y = 3 x 10 + 10 = 40
When x = 13, then y = 3 x 13 + 10 = 49
Thus, we have the following table of points:
Plotting the points of each table, we obtain the required graph of two intersecting lines.
Question 2.
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.
Solution:
Let the cost of 1 bat be ₹ x and the cost of 1 ball be ₹ y.
Then according to the question, we have:
3x + 6y = 3900 …(i)
x + 3y = 1300 …(ii)
For geometrical representation:
Thus, we have the following table of points:
Plotting the points of each table, we obtain the required graph of two intersecting lines.
Question 3.
The cost of 2 kg of apples and 1 kg 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.
Solution:
Let cost of 1 kg apples be ₹ x and the cost of 1 kg grapes be ₹y.
Then according to the question, we have:
2x + y = 160 …(i)
4x + 2y = 300 …(ii)
For geometrical representation:
From equation (i), we have: y = -2x + 160
When x = 50, then y = -2 x 50 + 160 = 60
When x = 20, then y = -2 x 20 + 160 = 120
When x = 45, then y = -2 x 45 + 160 = 70
Thus, we have the following table of points:
Thus, we have the following table of points:
Plotting the points of each table on a graph paper, we obtain the required graph of two parallel lines.
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