NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.2 PDF

NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.2 (Pair of Linear Equations in Two Variables)

Here are the NCERT Solutions for Class 10 Maths Chapter 3 Ex 3.2: 1. 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 Rs 50, whereas 7 pencils and 5 pens together cost Rs. 46. Find the cost of one pencil and that of one pen.

Answer:

(i) Let number of boys = x Let number of girls = y According to given conditions, we have x + y = 10 And, x = 10 - y putting y=0,5,10,we get, X=10-0=10 X=10-5=5, X=10-10=0

x	10	5	0
y	0	5	10

Number of girls is 4 more than number of boys ........Given, so, Y=x+4 putting x=-4,0,4 we get, Y=-4+4=0 Y=0+4 Y=4+4=8

x	-4	0	4
y	0	4	8

We plot the points for both of the equations to find the solution. (ii) Let the cost of one pencil=Rs.X and Let the cost of one pen=Rs.Y According to the given conditions, we have: =5x + 7y = 50 =5x=50-7y =x=10-7/5y Three solutions of this equation can be written in a table as follows: Three solutions of this equation can be written in a table as follows: The graphical representation is as follows:

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2. 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: (i) 5x − 4y + 8 = 0 (ii)9x + 3y + 12 = 0 7x + 6y – 9 = 018x + 6y + 24 = 0 (iii) 6x − 3y + 10 = 0 2x – y + 9 = 0

Answer:

(i) 5x − 4y + 8 = 0, 7x + 6y – 9 = 0 Comparing equation 5x − 4y + 8 = 0 with a ₁ x + b ₁ y + c ₁ = 0and 7x + 6y – 9 = 0 with a ₂ x + b ₂ y + c ₂ = 0, We get, Hence, we find that, (ii) 9x + 3y + 12 = 0, 18x + 6y + 24 = 0 Comparing equation 9x + 3y + 12 = 0 with a ₁ x + b ₁ y + c ₁ = 0and 7x + 6y – 9 = 0 with a ₂ x + b ₂ y + c ₂ = 0, We get, Hence We find that, Hence, lines are coincident. (iii) 6x − 3y + 10 = 0, 2x – y + 9 = 0 Comparing equation 6x − 3y + 10 = 0 with a ₁ x + b ₁ y + c ₁ = 0and 7x + 6y – 9 = 0 with a ₂ x + b ₂ y + c ₂ = 0, We get, Hence We find that, Hence, lines are parallel to each other. 3. On comparing the ratios a ₁ /a ₂ ,b ₁ /b ₂ and c ₁ /c ₂ , find out whether the following pair of linear equations are consistent, or inconsistent. (i) 3x + 2y = 5, 2x − 3y = 8 (ii) 2x − 3y = 7, 4x − 6y = 9 (iii) 3x/2 + 5y/3 = 7, 9x − 10y = 14 (iv) 5x − 3y = 11, −10x + 6y = −22

Answer:

(i) 3x + 2y = 5, 2x − 3y = 7 Comparing equation 3x + 2y = 5 with a ₁ x + b ₁ y + c ₁ = 0and 7x + 6y – 9 = 0 with a ₂ x + b ₂ y + c ₂ = 0, We get, Hence, Therefore these linear equations will intersect at one point only and have only one possible solution. And,pair of linear euations is consistent (ii) 2x − 3y = 8, 4x − 6y = 9 Comparing equation 2x − 3y = 8 with a ₁ x + b ₁ y + c ₁ = 0and 7x + 6y – 9 = 0 with a ₂ x + b ₂ y + c ₂ = 0, We get, Hence, Therefore these linear equations are parallel to each other and have no possible solution.in And, pair of linear euations is inconsistent (iii) 9x − 10y = 14 We get, Hence, Therefore, these linear equations will intersect each other at one point and have only one possible solution. (iv) 5x − 3y = 11, −10x + 6y = −22 Comparing equation 5x − 3y = 11 with a ₁ x + b ₁ y + c ₁ = 0and 7x + 6y – 9 = 0 with a ₂ x + b ₂ y + c ₂ = 0, We get, Hence, Therefore these pair of lines have infinite number of solutions

4. Which of the following pairs of linear equations are consistent/inconsistent? If consistent, obtain the solution graphically:

4. (i) x + y = 5, 2x + 2y = 10 (ii) x – y = 8, 3x − 3y = 16 (iii) 2x + y = 6, 4x − 2y = 4 (iv) 2x − 2y – 2 = 0, 4x − 4y – 5 = 0

Answer:

(i) x + y = 5, 2x + 2y = 10 We get, Hence, (ii) x – y = 8, 3x − 3y = 16   We get, Hence, Therefore, these linear equations are intersecting each other at one point and thus have only one possible solution. Hence,the pair of linear equations is inconsistent. (iii) 2x + y = 6, 4x − 2y = 4 We get, Hence, Therefore, these linear equations are intersecting each other at one point and thus have only one possible solution. Hence the pair of linear equations is consistent (iv) 2x − 2y – 2 = 0, 4x − 4y – 5 = 0 We get, Hence, Therefore, these linear equations are parallel to each other and have no possible solution, Hence,the pair of linear equations is inconsistent. 5. Half the perimeter of a rectangle garden, whose length is 4 m more than its width, is 36 m. Find the dimensions of the garden.

Answer:

Let width of rectangular garden = x metres and length=y So, Hence, the graphic representation is as follows. 6. Given the linear equation (2x + 3y – 8 = 0), write another linear equation in two variables such that the geometrical representation of the pair so formed is: (i) Intersecting lines (ii) Parallel lines (iii) Coincident lines

Answer:

(i) Let the second line be equal to a ₂ x + b ₂ y + c ₂ = 0, Intersecting Lines:For this Condition, The Second line such that it is intersecting the given line is 2x+4y-6=0 As, (ii) Let the second line be equal to a ₂ x + b ₂ y + c ₂ = 0, parallel Lines: For this Condition, Hence,the second line can be 4x+6y-8=0 As, (iii) Let the second line be equal to a ₂ x + b ₂ y + c ₂ = 0, Coincident lines: For coincident lines, Hence,the second line can be 6x+9y-24=0 As, 7. Draw the graphs 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.

Answer:

For equation x – y + 1 = 0, we have following points which lie on the line For equation 3x + 2y – 12 = 0, we have following points which lie on the line We can see from the graphs that points of intersection of the lines with the x–axis are (–1, 0), (2, 3) and (4, 0).

Benefits of Solving NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.2

• Enhanced Understanding : The step-by-step solutions help students better understand the methods of solving pairs of linear equations, like the Substitution Method and Elimination Method . This strengthens conceptual clarity.
• Improved Problem-Solving Skills : By practicing these exercises, students can improve their ability to solve complex problems, building strong analytical and logical skills.
• Increased Accuracy : The solutions guide students on how to approach problems methodically, minimizing errors and improving accuracy in solving linear equations.
• Better Exam Preparation : Regular practice with NCERT solutions enhances preparation for exams, as students become familiar with the types of questions and efficient problem-solving techniques.
• Time Management : With consistent practice, students learn how to solve problems more quickly and efficiently, an important skill during exams.
• Confidence Boost : Solving a variety of problems in the exercise helps students build confidence in their ability to handle similar questions in exams.