In this article, we will share MP Board Class 10th Maths Book Solutions Pair of Linear Equations in Two Variables Exercise 3.1 Pdf solution file.

## MP Board Class 10th Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Exercise 3.1 Solution

**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:**

At present: Let Aftab’s age = x years

His daughter’s age = y years

Seven years ago : Aftab’s age = (x – 7) years

His daughter’s age = (y – 7)years

According to the condition,

[Aftab’s age] = 7[His daughter’s age]

⇒ [x – 7] = 7[y – 7] = x – 7 = 7y – 49

⇒ x – 7y – 7 + 49 = 0 ⇒ x – 7y + 42 = 0 …. (1)

After three years : Aftab’s age = (x + 3) years

His daughter’s age = (y + 3) years

According to the condition,

[Aftab’s age] = 3[His daughter’s age]

⇒ [x + 3] = 3[y + 3]

⇒ x + 3 = 3y + 9 ⇒ x — 3y + 3 — 9 = 0

⇒ x — 3y – 6 = 0 …. (2)

Graphical representation of equation (1) and (2): From equation (1), we have :

### MP Board Class 10th Maths Exercise 3.1 Solution

**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 a bat = ₹ x

and the cost of a ball = ₹ y

Cost of 3 bats = ₹ 3x

and cost of 6 balls = ₹ 6y

Again, cost of 1 bat = ₹ x

and cost of 3 balls = ₹ 3y

Algebraic representation:

Cost of 3 bats + Cost of 6 balls = ₹ 3900

⇒ 3x + 6y = 3900 ⇒ x + 2y = 1300 …. (1)

Also, cost of 1 bat + cost of 3 balls = ₹ 1300

⇒ x + 3y = 1300 …. (2)

Thus, (1) and (2) are the algebraic representations of the given situation.

Geometrical representation:

We have for equation (1),

We can also see from the obtained graph that the straight lines representing the two equations intersect at (1300, 0).

10th Maths Exercise 3.1 Solutions

**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 the cost of 1 kg of apples = ₹ x

And the cost of 1 kg of grapes = ₹ y

Algebraic representation:

2x + y = 160 … (1)

and 4x + 2y = 300

⇒ 2x + y = 150 … (2)

Geometrical representation:

We have, for equation (1),

The straight lines l_{1} and l_{2} are the geometrical representations of the equations (1) and (2) respectively. The lines are parallel.

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