### NCERT Solutions For Class 10 Math Chapter – 3 Exercise – 3.2

- by Dev Bhardwaj

*NCERT Solutions For Class 10 Math Chapter – 3 Linear equation in two variables Exercise – 3.2*

**Q1. From the pair of linear equations in the following problems, and find their solutions graphically.**

**10 students of Class 10 took part in a Mathematics quiz. If the number of girls is 4 more than the number of boys, find the number of boys and girls who took part in the quiz.**

**Solution:-**

** **

Let the number of boys = x

No of girls = y

Given that total number of student is 10 so that

X + y = 10

Subtract y both side we get

X = 10 – y

Putting

X = 10 – 0

X = 10 – 5 = 5

X = 10 – 10 = 0

x | 10 | 5 |

y | 0 | 5 |

Given that if the number of girls is 4 more than the number of boys

So that

Y = x + 4

Putting x = -4 , 0 , 4 and we get

Y = -4 + 4 = 0

Y = 0 + 4 = 4

Y= 4 + 4 = 8

x | -4 | 0 | 4 |

y | 0 | 4 | 8 |

Graphical representation

Therefore , number of boys = 3 and number of girls = 7.

**5 pencils and 7 pens together cost Rs 50 whereas 7 pencils and 5 pens together cost Rs 46 . Find the cost of one pencil and that of one pen.**

**Solution:-**

** **

5x + 7y = 50

5x = 50 – 7y

X = 10 – 7y/5

Putting value of y = 5 , 10 and 15 we get

X = 10 — 7 x 5/5 = 10 – 7 = 3

X = 10 – 7 x 10/5 = 10 – 14 = -4

X = 10 – 7 x 15/5 = 10 – 21 = 11

x | 3 | -4 | -11 |

y | 5 | 10 | 15 |

7x + 5y = 46

5y = 46 – 7x

Y= 9.2 – 1.4x

Putting x = 0.2 and 4 we get

Y = 9.2 – 1.4 x 0 = 9.2

Y = 9.2 – 1.4 x 2 = 6.4

Y = 9.2 – 1.4 (4) = 3.6

x | 0 | 2 | 4 |

y | 9.2 | 6.4 | 3.6 |

Graphical representation

Therefore , cost of one pencil = Rs 3 and cost of cost of one pen = Rs 5.

**Q2. On comparing the ratios a**_{1}**/a**_{2}** , b**_{1}**/b**_{2}** and c**_{1}**/c**_{2}** find out whether the lines representing the following pairs of linear equations intersect at a point, are parallel or coincident.**

**Solution:-**

** **

- 5x – 4y + 8 = 0

7x + 6y – 9 = 0

Therefore , both are intersecting lines at one point.

- 9x + 3y + 12 = 0

18x + 6y + 24 = 0

Therefore , both lines are coincident.

- 6x – 3y + 10 = 0

2x – y + 9 = 0

Therefore , both lines are parallel.

**Q3. On comparing the ratios a**_{1}**/a**_{2}** , b**_{1}**/ and c**_{1}**/c**_{2}** find out whether the following pair of linear equations are consistent , or inconsistent.**

**3x + 2y = 5 ; 2x – 3y = 7****2x – 3y = 8 ; 4x – 6y = 9****3/2x + 5/3y = 7 ; 9x – 10y = 14****5x – 3y = 11 ; -10x + 6y = -22****4/3x + 2y = 8 ; 2x + 3y = 12**

**Solution:-**

** **

- Hence pair of linear equation is consistent.
- Hence the pair of linear equation is inconsistent.
- Hence the pair of linear equation is consistent.
- Hence the pair of linear equation is consistent
- Hence the pair of linear equation is consistent.

**Q4. Which of the following pairs of linear equation are consistent/inconsistent ? If consistent, obtain the solution graphically;**

**X + y = 5 , 2x + 2y = 10****X – y = 8 , 3x – 3y = 16****2x + y – 6 = 0 , 4x – 2y – 4 = 0****2x – 2y – 2 = 0, 4x – 4y – 5 = 0**

**Solution:-**

** **

- X + y = 5 ; 2x + 2y = 10

Hence the pair of linear equation is consistent.

X + y = 5

X = 5 – y

x | 4 | 3 | 2 |

y | 1 | 2 | 3 |

And , 2x + 2y = 10

X = 10 – 2y/2

x | 4 | 3 | 2 |

y | 1 | 2 | 3 |

Graphical representation

From the figure it can be observed that these lines are overlapping each other. Therefore , infinite solutions are possible for the given pair of equations.

- X – y = 8 , 3x – 3y = 16

Hence the pair of linear equation is inconsistent.

- 2x + y – 6 = 0 , 4x – 2y – 4 = 0

Hence the pair of linear equation is consistent.

2x + y – 6 = 0

Y = 6 – 2x

x | 0 | 1 | 2 |

y | 6 | 4 | 2 |

And , 4x – 2y – 4 = 0

Y = 4x – 4/2

x | 1 | 2 | 3 |

y | 0 | 2 | 4 |

Graphical representation

From the figure , it can be observed that these lines are intersecting each other at the only one point (2,2) which is the solution for the given pair of equations.

- 2x – 2y – 2 = 0 , 4x – 4y – 5 = 0

Hence the pair of linear equation is inconsistent

**Q5. Half the perimeter of a rectangular garden , whose length is 4 m more than its width, is 36 m. Find the dimensions of the garden.**

**Solution:-**

** **

Y – x = 4 (i)

Y + x = 36 (ii)

Y – x = 4

Y = x + 4

x | 0 | 8 | 12 |

y | 4 | 12 | 16 |

Y + x = 36

x | 0 | 36 | 16 |

y | 36 | 0 | 20 |

Graphical representation

From the figure , it can be observed that these lines are intersecting each other at only point (16,20) Therefore , the length and width of the given garden is 20 m and 16 m respectively.

**Q6. Given the linear equation 2x + 3y – 8 = 0 write another linear equations in two variables such that the geometrical representation of the pair so formed is:**

**Intersecting lines****Parallel lines****Coincident lines**

**Solution:-**

** **

- Intersecting lines:

For this conditions

A_{1}/a_{2} ≠ b_{1}/b_{2}

- Parallel lines

For this conditions

A_{1}/a_{2} = b_{1}/b_{2} ≠ c_{1} / c_{2}

- Coincident lines

For coincident lines

A_{1}/a_{2} = b_{1}/b_{2} = c_{1}/c_{2}

**Q7. Draw the graph of the equations x – y + 1 = 0 and 3x + 2y – 12 = 0 . Determine the coordinates of the vertices of the triangle formed by these lines and the x – axis , and shade the triangular region.**

**Solution:-**

** **

X – y + 1 = 0

X = y – 1

x | 0 | 1 | 2 |

y | 1 | 2 | 3 |

3x + 2y – 12 = 0

X = 12 – 2y/3

x | 4 | 2 | 0 |

y | 0 | 3 | 6 |

Graphical representation

From the figure it can be observed that these lines are intersecting each other at point (2,3) and x – axis at (1,0) and (4,0) Therefore , the vertices of the triangle are (2,3) (-1,0) and (4,0).

** **

NCERT Solutions For Class 10 Math Chapter – 3 Linear equation in two variables Exercise – 3.2 Q1. From the…

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