Jharkhand Board JAC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.1 Textbook Exercise Questions and Answers.

## JAC Board Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Exercise 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 be x years and the present age of his daughter be y years. Then, seven years ago, the age of Aftab was x – 7 years and the age of his daughter was y – 7 years.

So, from the given data.

x – 7 = 7(y – 7)

∴ x – 7 = 7y – 49

x – 7y + 42 = 0 …….. (1)

Similarly, three years from now, the age of Aftab will be x + 3 years and the age of his daughter will be y + 3 year.

So, according to the given data,

x + 3 = 3(y + 3)

∴ x + 3 = 3y + 9

∴ x – 3y – 6 = 0 ….. (2)

Thus, the equations x – 7y + 42 = 0 and x – 3y – 6 = 0 represent the given situation algebraically.

To represent the given situation graphically. we draw the graphs of both the equations.

x – 7y + 42 = 0

∴ y = \(\frac{42+x}{7}\)

x | 0 | 35 |

y | 6 | 11 |

x – 3y – 6 = 0

∴ y = \(\frac{x-6}{3}\)

x | 0 | 30 |

y | -2 | 8 |

The above graph represents the situation graphically.

We observe that the lines intersect at point (42, 12).

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 one bat be ₹ x and the cost of one ball be ₹ y.

Then, the total cost of 3 bats is ₹ 3x and that of 6 balls is ₹ 6y. From the data, the total cost is ₹ 3900.

∴ 3x + 6y = 3900

∴ x + 2y = 1300

Similarly, the cost of 1 bat is ₹ x and the total cost of 3 balls is ₹ 3y. From the data, the total cost is ₹ 1300.

∴ x + 3y = 1300

Thus, the equations x + 2y = 1300 and x + 3y = 1300 represent the given situation algebraically.

To represent the given situation geometrically. we draw the graphs of both the equations.

x + 2y = 1300

x | 100 | 1300 |

y | 600 | 0 |

x + 3y = 1300

x | 100 | 1300 |

y | 400 | 0 |

The above graph represents the situation geometrically.

We observe that the lines intersect at point (1300, 0).

Question 3.

The cost of 2 kg of apples and 1 kg of grapes on a day was found to be 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 be ₹ x and the cost of 1 kg of grapes be ₹ y.

Then, from the given data, 2x + y = 160 and 4x + 2y = 300.

Thus, the equations 2x + y = 160 and 4x + 2y = 300 represent the given situation algebraically.

To represent the given situation geometrically. we draw the graphs of both the equations.

2x + y = 160

x | 0 | 80 |

y | 160 | 0 |

4x + 2y = 300

x | 0 | 75 |

y | 150 | 0 |

The above graph represents the situation geometrically.

We observe that the lines are parallel.