### NCERT Solutions For Class 10 Maths Chapter – 3 Exercise – 3.1

- by Dev Bhardwaj

*NCERT Solutions For Class 10 Math Chapter – 3 Linear equation in two variables Exercise – 3.1*

* *

**Q1. 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 you will be” Represent this situation algebraically and graphically.**

**Solution:-**

** **

A.T.Q

(x-7) = 7y – 49

X = 7y – 42

Putting y = 5 , 6 and 7 we get

X = 7 x 5 – 42 = 35 – 42 = -7

X = 7 x 6 – 42 = 42 – 42 = 0

X = 7

x | -7 | 0 | 7 |

y | 5 | 6 | 7 |

A.T.Q

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

X = 3y + 6

Putting , y = -2 ,-1 and 0 , we get

X = 6

x | 0 | 3 | 6 |

y | -2 | -1 | 0 |

** **

Algebraic representation

From equation (I) and (ii)

X – 7y = -42

X – 3y = 6

Graphical representation.

*NCERT Solutions For Class 10 Math Chapter – 3 Linear equation in two variables Exercise – 3.1*

* *

**Q1. 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 you will be” Represent this situation algebraically and graphically.**

**Solution:-**

** **

A.T.Q

(x-7) = 7y – 49

X = 7y – 42

Putting y = 5 , 6 and 7 we get

X = 7 x 5 – 42 = 35 – 42 = -7

X = 7 x 6 – 42 = 42 – 42 = 0

X = 7 x 7 – 42 = 49 – 42 = 7

x | -7 | 0 | 7 |

y | 5 | 6 | 7 |

A.T.Q

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

X = 3y + 6

Putting , y = -2 ,-1 and 0 , we get

X = 6

x | 0 | 3 | 6 |

y | -2 | -1 | 0 |

** **

Algebraic representation

From equation (I) and (ii)

X – 7y = -42

X – 3y = 6

Graphical representation.

**Q2. 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:-**

** **

A.T.Q

3x + 6y = 3900 (i)

Dividing equation by 3 , we get

X + 2y = 1300

Subtracting 2y both side we get

X = 1300 – 2y

Putting y = -1300 , 0 and 1300 we get

x = 1300 – 2(-1300)

= 1300 + 2600

= 3900

X = 1300 – 2(0)

= 1300 – 0

= 1300

x | 3900 | 1300 | -1300 |

y | -1300 | 0 | 1300 |

X + 2y = 1300

X = 1300 – 2y

X = 1300 – 2(-1300)

= 1300 + 2600

= 3900

X = 1300 – 2(0)

= 1300 – 2(1300)

= -1300

x | 3900 | 1300 | -1300 |

y | -1300 | 0 | 1300 |

Algebraic representation

3x + 6y = 3900

X + 2y = 1300

Graphical representation,

**Q3. The cost of 2 kg of apples and 1 kg of grapes on a day was found to be Rs 160 After a month , the cost of 4 kg of apples and 2 kg of grapes is Rs 300. Represent the situation algebraically and geometrically.**

**Solution:-**

** **

** **

2x + y = 160 (i)

2x = 160 – y

X = (160 – y)/2

X = 0

x | 80 | 40 | 0 |

y | 0 | 80 | 160 |

4x + 2y = 300 (ii)

Dividing by 2 we get

2x + y = 150

Y = 150 – 2x

Y = 150 – 2 x 0

Y = 150

Y = 150 – 2 x 50 = 50

Y = 150 – 2 x (100) = – 50

x | 0 | 50 | 100 |

y | 150 | 50 | -50 |

Algebraic representation,

2x + y = 160 (i)

4x + 2y = 300 (ii)

Graphical representation,

**Q2. 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:-**

** **

A.T.Q

3x + 6y = 3900 (i)

Dividing equation by 3 , we get

X + 2y = 1300

Subtracting 2y both side we get

X = 1300 – 2y

Putting y = -1300 , 0 and 1300 we get

x = 1300 – 2(-1300)

= 1300 + 2600

= 3900

X = 1300 – 2(0)

= 1300 – 0

= 1300

x | 3900 | 1300 | -1300 |

y | -1300 | 0 | 1300 |

X + 2y = 1300

X = 1300 – 2y

X = 1300 – 2(-1300)

= 1300 + 2600

= 3900

X = 1300 – 2(0)

= 1300 – 2(1300)

= -1300

x | 3900 | 1300 | -1300 |

y | -1300 | 0 | 1300 |

Algebraic representation

3x + 6y = 3900

X + 2y = 1300

Graphical representation,

**Q3. The cost of 2 kg of apples and 1 kg of grapes on a day was found to be Rs 160 After a month , the cost of 4 kg of apples and 2 kg of grapes is Rs 300. Represent the situation algebraically and geometrically.**

**Solution:-**

** **

** **

2x + y = 160 (i)

2x = 160 – y

X = (160 – y)/2

X = 0

x | 80 | 40 | 0 |

y | 0 | 80 | 160 |

4x + 2y = 300 (ii)

Dividing by 2 we get

2x + y = 150

Y = 150 – 2x

Y = 150 – 2 x 0

Y = 150

Y = 150 – 2 x 50 = 50

Y = 150 – 2 x (100) = – 50

x | 0 | 50 | 100 |

y | 150 | 50 | -50 |

Algebraic representation,

2x + y = 160 (i)

4x + 2y = 300 (ii)

Graphical representation,

NCERT Solutions For Class 10 Math Chapter – 3 Linear equation in two variables Exercise – 3.1 Q1. Aftab tells…

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