**NCERT Solutions Class 10 Maths Chapter 3 Pair Of Linear Equations In Two Variables** – Here are all the NCERT solutions for Class 10 Maths Chapter 3. This solution contains questions, answers, images, explanations of the complete Chapter 3 titled Linear Equations In Two Variables of Maths taught in Class 10. If you are a student of Class 10 who is using NCERT Textbook to study Maths, then you must come across Chapter 3 Pair Of Linear Equations In Two Variables. After you have studied lesson, you must be looking for answers of its questions. Here you can get complete NCERT Solutions for Class 10 Maths Chapter 3 Pair Of Linear Equations In Two Variables in one place.

## NCERT Solutions Class 10 Maths Chapter 3 Pair Of Linear Equations In Two Variables

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Class | 10 |

Subject | Maths |

Book | Mathematics |

Chapter Number | 3 |

Chapter Name |
Pair Of Linear Equations In Two Variables |

### NCERT Solutions Class 10 Maths chapter 3 Pair Of Linear Equations In Two Variables

Class 10, Maths chapter 3, Pair Of Linear Equations In Two Variables solutions are given below in PDF format. You can view them online or download PDF file for future use.

### Pair Of Linear Equations In Two Variables Download

**NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables**

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### Question & Answer

Q.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.

Ans :Let the present age of Aftab be x. And, present age of his daughter = y Seven years ago, Age of Aftab = x - 7 Age of his daughter = y - 7 According to the question, ( x - 7 ) = 7( y - 7 ) x - 7 = 7y - 49 x - 7y = - 42 ………… (1) Three years hence, Age of Aftab = x + 3 Age of his daughter = y + 3 According to the question, ( x + 3 ) = 3( y + 3 ) x + 3 = 3y + 9 x - 3y = 6 ………… (2) Therefore, the algebraic representation is x - 7y = - 42 x - 3y = 6 For, x - 7y = - 42 x = - 42 + 7y The solution table is For x - 3y = 6 x = 6 + 3y The solution table is The graphical representation is as follows:

Q.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.

Ans :Let the cost of a bat be Rs x. And, cost of a ball Rs y According to the question, the algebraic representation is 3x + 6y = 3900 x + 2y - 1300 For 3x + 6y = 3900, \( x=\frac{3900-6 y}{3}\) The solution table is For x + 2y = 1300 x = 1300 - 2y The solution table is The graphical representation is as follows:

Q.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.

Ans :Let the cost of 1 kg of apples be Rs x. And, cost of 1 kg of grapes = Rs y According to the question, the algebraic representation is 2x + y= 160 4x + 2y = 300 For 2x + y = 160, y = 160 - 2x The solution table is For 4x + 2y = 300 \( y=\frac{300-4 x}{2} \) The solution table is The graphical representation is as follows:

Q.4:Form the pair of linear equations in the following problems, and find their solutions graphically. (i) 10 students of Class X 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. (ii) 5 pencils and 7 pens together cost ₹ 50, whereas 7 pencils and 5 pens together cost ₹ 46. Find the cost of one pencil and that of one pen.

Ans :(i) Let the number of girls be x and the number of boys be y. According to the question, the algebraic representation is x + y = 10 x - y = 4 For x + y = 10 x = 10 - y For x - y = 4, x = 4 + y Hence, the graphic representation is as follows: From the figure, it can be observed that these lines intersect each other at point (7, 3). Therefore, the number of girls and boys n the class a e 7 and 3 respectively. (ii) Let the cost of 1 pencil be Rs x and the cost of 1 pen be Rs y. According to the question, the algebraic representation is 5x + 7y - 50 7x + 5y = 46 For 5x + 7y = 50, \( x=\frac{50-7 y}{5} \) 7x + 5y = 46 \( x=\frac{46-5 y}{7} \) Hence, the graphic representation is as follows: From the figure, it can be observed that these lines intersect each other at point (3, 5). Therefore, the cost of a pencil and a pen are Rs 3 and Rs 5 respectively.

Q.5:On comparing the ratios \( \frac{a_{1}}{a_{2}}, \frac{b_{1}}{b_{2}} \text { and } \frac{c_{1}}{c_{2}} \) find out whether the lines representing the following pairs of linear equations intersect at a point, are parallel or coincident: (i) 5x – 4y + 8 = 0 7x + 6y – 9 = 0 (ii) 9x + 3y + 12 = 0 18x + 6y + 24 = 0 (iii) 6x – 3y + 10 = 0 2x – y + 9 = 0

Ans :(i) 5x - 4y + 8 = 0 7x + 6y - 9 = 0 Comparing these equations with \( a_{1} x+b_{1} y+c_{1}=0 \) and \( a_{2} x+b_{2} y+c_{2}=0 \), we obtain \( a_{1}=5, \quad b_{1}=-4, \quad c_{1}=8 \) \( a_{2}=7, \quad b_{2}=6, \quad c_{2}=-9 \) \( \frac{a_{1}}{a_{2}}=\frac{5}{7} \) \( \frac{b_{1}}{b_{2}}=\frac{-4}{6}=\frac{-2}{3} \) \( \operatorname{since}^{\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}} \) Hence, the lines representing the given pair of equations have a unique solution and the pair of lines intersects at exactly one point. (ii) 9x + 3y + 12 = 0 18x + 6y + 24 = 0 Comparing these equations with \( a_{1} x+b_{1} y+c_{1}=0 \) and \( a_{2} x+b_{2} y+c_{2}=0, \), we obtain \( a_{1}=9, \quad b_{1}=3, \quad c_{1}=12 \) \( a_{2}=18, \quad b_{2}=6, \quad c_{2}=24 \) \( \frac{a_{1}}{a_{2}}=\frac{9}{18}=\frac{1}{2} \) \( \frac{b_{1}}{b_{2}}=\frac{3}{6}=\frac{1}{2} \) \( \frac{c_{1}}{c_{2}}=\frac{12}{24}=\frac{1}{2} \) Since, \( \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}} \) Hence, the lines representing the given pair of equations are coincident and there are infinite possible solutions for the given pair of equations (iii) 6x - 3y + 10 = 0 2x - y + 9 = 0 Comparing these equations with \( a_{1} x+b_{1} y+c_{1}=0 \) and \( a_{2} x+b_{2} y+c_{2}=0 \), we obtain \( a_{1}=6, \quad b_{1}=-3, \quad c_{1}=10 \) \( a_{2}=2, \quad b_{2}=-1, \quad c_{2}=9 \) \( \frac{a_{1}}{a_{2}}=\frac{6}{2}=\frac{3}{1} \) \( \frac{b_{1}}{b_{2}}=\frac{-3}{-1}=\frac{3}{1} \) \( \frac{c_{1}}{c_{2}}=\frac{10}{9} \) Since \( \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}} \) Hence, the lines representing the given pair of equations are parallel to each other and hence, these lines will never intersect each other at any point or there is no possible solution for the given pair of equations.

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