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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