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.” Represent this situation algebraically and graphically.

**Solution:** Let us assume Aftab’s current age = x and his daughter’s current age = y

Seven years ago: Aftab’s age `= x - 7` and daughter’s age `= y - 7`

As per question;

`x – 7 = 7(y – 7)`

Or, `x – 7 = 7y – 49`

Or, `x = 7y – 49 + 7`

Or, `x = 7y – 42`

Or, `7y – x – 42 = 0` ……(1)

This equation gives following values for x and y

x | - 35 | - 28 | - 21 | - 14 | - 7 | 0 | 7 | 14 | 21 | 28 | 35 | 42 | 49 |

y | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 |

Three years from now:

Aftab’s age `= x + 3` and daughter’s age `= y + 3`

As per question;

`x + 3 = 3(y + 3)`

Or, `x + 3 = 3y + 9`

Or, `x = 3y + 9 – 3`

Or, `x = 3y + 6`

Or, `3y – x + 6 = 0` ……..(2)

This equation gives following values for x and y:

x | 9 | 12 | 15 | 18 | 21 | 24 | 27 | 30 | 33 | 36 | 39 | 42 | 45 |

y | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 |

The following graph is plotted for the given pair of linear equations:

Daughter’s age = 12 years and Aftab’s age = 42 years

Question 2: The coach of a cricket team buys 3 bats and 6 balls for Rs. 3900. Later, she buys another bat and 3 more balls of the same kind for Rs. 1300. Represent this situation algebraically and geometrically.

**Solution:** Let us assume that price of one bat = x and price of one ball = y. Following equations can be written as per the question:

`3x + 6y = 3900`

Or, `x + 2y = 1300` ……..(1)

This equation will give following values for x and y

x | 1500 | 1400 | 1300 | 1200 | 1100 |

y | - 100 | - 50 | 0 | 50 | 100 |

`x + 3y = 1300` ……….(2)

This equation will give following values for x and y

x | 1600 | 1450 | 1300 | 1150 | 1000 |

y | - 100 | - 50 | 0 | 50 | 100 |

The following graph is plotted for the given pair of linear equations.

Price of one bat = Rs. 1300 and Price of one ball = zero

Question 3: The cost of 2 kg apples and 1 kg grapes was found to be Rs. 160. After a month, the cost of 4 kg apples and 2 kg grapes is Rs. 300. Represent this situation algebraically and graphically.

**Solution:** Let us assume that cost of 1 kg apple = x and cost of 1 kg grapes = y. Following equations can be written as per the question.

`2x + y = 160` ……..(1)

This equation gives following values for x and y

x | 70 | 60 | 50 | 40 |

y | 20 | 40 | 60 | 80 |

`4x + 2y = 300`

Or, `2x + y = 150` ……..(2)

This equation will give following values for x and y:

x | 65 | 55 | 45 | 35 |

y | 20 | 40 | 60 | 80 |

The following graph is plotted for the given pair of linear equations.

Since we get parallel lines so there will be no solution for this pair of linear equations.

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