Complete solutions of Ex 3.1 Class 10 Maths Chapter 3 with additional questions and answers from new NCERT syllabus textbook Class 10 Maths.
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
Ex 3.1 Class 10 Maths 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.
You can also Download NCERT Solutions for class 10 Maths in Hindi to help you to revise complete Syllabus and score more marks in your examinations.
Ex 3.1 Class 10 Maths 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.
Ex 3.1 Class 10 Maths 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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