NCERT Solutions for class 10 Maths Chapter 3 Exercise 3.2 in Hindi and English Medium revised and updated for 2023-24 session. The solutions of each question is modified as per the new rationalised NCERT books issued for academic year 2023-24 exams.

Class: 10 | Mathematics |

Chapter 3: | Exercise 3.2 |

Content: | NCERT Textbook Solutions and Extra Questions |

Content Type: | Text, Images, PDF and Videos |

Session: | 2023-24 |

Medium Type: | English and Hindi Medium |

### NCERT Solutions for class 10 Maths Chapter 3 Exercise 3.2

Class 10 Maths Ex. 3.2 solution pair of linear equations in two variables in Hindi Medium and English Medium. 10th Class Maths CBSE Solutions are in PDF format and Videos format. You can view all the answers explained in Video Format free, which are updated for new academic session 2023-24.

Uttar Pradesh Board of Education, Prayagraj also implemented NCERT Books for their High School Students. So, these 10th Maths solutions are helpful for UP Board students also. They can download UP Board Solutions for Class 10 Maths Exercise 3.2 in Hindi Medium.

NCERT Solutions Online and Offline Apps 2023-24 with PDF Solutions have been updated according to latest CBSE Curriculum 2023-24 for CBSE Board, Gujrat Board, MP Board and UP Board schools whoever using NCERT Books 2023-24.

### 10 Maths Chapter 3 Exercise 3.2 Solutions

NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.2 in Hindi and English medium are given below to use it online or download in PDF form free. If you want to download, the link is given at the same page. NCERT Solutions Offline Apps 2023-24 are updated for current academic session 2023-24 to use offline.

#### Class 10 Chapter 3 Exercise 3.2 Solution in Hindi Medium Video

#### Important Question based on 10th Maths Chapter 3

- Form a pair of linear equations for: The sum of the numerator and denominator of the fraction is 3 less than twice the denominator. If the numerator and denominator both are decreased by 1, the numerator becomes half the denominator. [Answer: x-y=-3, 2x-y=1]
- A jeweler has bars of 18 carat gold and 12 carat gold. How much of each must be melted together to obtain a bar of 16 carat gold, weighing 120g? It is given that pure gold is 24 carat. [Answer: 80g of 18 carat gold and 40g of 12 carat gold]
- For what value of p the pair of linear equations (p + 2)x – (2p + 1)y = 3(2p – 1) and 2x – 3y = 7 has a unique solution. [Answer: p≠4]
- The sum of the prices of a Sofa and table is ₹2340 and the difference of their prices is ₹140. 5. Find the price of each. [Answer: Sofa = ₹1240, table = ₹1100]

60 copies of Mathematics books and 40 copies of English books cost ₹2220. Also, 40 copies of Mathematics books and 50 copies of English books cost ₹1900. Find the cost of one Mathematics book and one English book separately. [Answer: Mathematics book = ₹25, English book = ₹18]

##### Important Question on Linear Equations for practice

- Solve for x and y: 3x + 2y = 11 and 2x + 3y = 4. Also find p if p = 8x + 5y. [Answer: x=5, y=-2, p=30]
- If three times, the greater of the two numbers is divided by the lesser, we get the quotient 6 and remainder 6. If 5 times the lesser number is divided by the greater, we get the quotient 2 and the remainder 3. Find the two numbers. [Answer: 16, 7]
- Solve the pair of linear equations by substitution method x – 7y + 42 = 0 and x – 3y – 6 = 0. [Answer: 42, 12]
- Ram is walking along the line joining (1, 4) and (0, 6). Rahim is walking along the line Joining (3, 4) and (1, 0). Represent on graph and find the point where both of them cross each other. [Answer: (2, 2)]

### Important Questions 10th Maths Exercise 3.2

### How many solutions are there for the equations 2x + 3y = 9 and 4x + 6y = 18?

If we plot these equations graphically, we observe that every point on the line is a common solution to both the equations. So, the equations 2x + 3y = 9 and 4x + 6y = 18 have infinitely many solutions.

### What is meant by inconsistent pair of Linear Equations?

A pair of linear equations which has no solution, is called an inconsistent pair of linear equations.

### What do you understand by consistent pair of linear equations?

A pair of linear equations in two variables, which has a solution, is called a consistent pair of linear equations.

### Which set of linear equations are dependent pair of linear equations?

A pair of linear equations which are equivalent has infinitely many distinct common solutions. Such a pair is called a dependent pair of linear equations in two variables.

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### How many questions are there in exercise 3.2 of class 10th Maths?

There are 3 questions in exercise 3.2 of class 10th mathematics chapter 3 (Pair of linear equations in two variables). Q1 contains 6 parts and Q3 contains 6 word problems. Q1 (iv, v, vi), 2, 3 are important and Q2 is unique.

### How many examples are based on exercise 3.2 of class 10th mathematics?

Only 4 examples (examples 7, 8, 9 and 10) are based on exercise 3.2 (chapter 3 Pair of linear equations in two variables) of class 10th mathematics. Examples 9 and 10 are very important.

### What will students learn in exercise 3.2 of 10th mathematics?

In exercise 3.2 (chapter 3 Pair of linear equations in two variables) of class 10th mathematics students will learn how to solve pairs of linear equations in two variables by SUBSTITUTION METHOD.

### Which method is used in exercise 3.2 of 10th Maths to solve the questions.

Substitution Method (step-wise):

a_1x + b_1y + c_1 = 0

a_2x + b_2y + c_2 = 0

**Step 1**: Find the value of one variable, say y in terms of the other variable, i.e., x from either equation, whichever is convenient.**Step 2**: Substitute this value of y in the other equation, and reduce it to an equation in one variable, i.e., in terms of x, which can be solved. Sometimes, you can get statements with no variable. If this statement is true, you can conclude that the pair of linear equations has infinitely many solutions. If the statement is false, then the pair of linear equations is inconsistent.**Step 3**: Substitute the value of x (or y) obtained in Step 2 in the equation used in Step 1 to obtain the value of the other variable.