NCERT Solutions for Class 10 Maths Chapter 3 PDF Download
NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables contains all the questions with detailed solutions. Students are advised to solve these questions for better understanding of the concepts in Chapter 3.
Ananya Gupta19 Mar, 2024
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NCERT Solutions for Class 10 Maths Chapter 3: NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables are designed to help students in comprehending the solution methods for problems within this topic.
Mathematics is a subject that necessitates ample practice, and students preparing for 10th-grade board exams can utilize NCERT Solutions Class 10 as a reference. These solutions for the Chapter Pair of Linear Equations in Two Variables provide step-by-step explanations for all the mathematical problems found in the NCERT textbook. A linear equation in two variables x and y can be expressed in the form ax + by + c = 0, where a, b, and c are real numbers, and neither a nor b is zero.NCERT Solutions for Class 10 Maths Chapter 3 PDF
You can access the PDF link for NCERT Solutions for Class 10 Maths Chapter 3 by clicking the provided link. These solutions provide detailed explanations and step-by-step guidance to help you understand the concepts covered in this chapter. Whether you're preparing for exams or simply seeking to enhance your understanding of pair of linear equations in two variables, these NCERT solutions serve as a valuable resource. With clear explanations and solved examples, you can strengthen your grasp of the subject and tackle mathematical problems with confidence.NCERT Solutions for Class 10 Maths Chapter 3 PDF
NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables
These solutions provide a comprehensive guide for students preparing for their board exams, aiding in thorough comprehension of the topic.NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.1
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.Answer:
Let the present age of Aftab and his daughter be x and y respectively. Seven years ago, Age of Aftab = x – 7 and Age of his daughter = y – 7 According to the given condition,
Thus, the given conditions can be algebraically represented as:
x – 7y = –42
And x – 3y = 6
The graphical representation is as follows:
2. 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 graphically.
Answer:<.
Let the cost of a bat and a ball be Rs x and Rs y respectively. The given conditions can be algebraically represented as: 3x + 6y = 3900 x + 2y = 1300
Three solutions of this equation can be written in a table as follows:
| x | 3900 | 1300 | -1300 |
| y | -1300 | 0 | 1300 |
Three solutions of this equation can be written in a table as follows:
| x | 3900 | 1300 | -1300 |
| y | -1300 | 0 | 1300 |
3. 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.
Answer:
Let cost of 1 kg of apples = Rs x and let cost of 1 kg of grapes= Rs y According to given conditions, we have 2x + y = 160… (1) 4x + 2y = 300 ⇒ 2x + y = 150… (2) So, we have equations (1) and (2), 2x + y = 160 and 2x + y = 150 which represent given situation algebraically. For equation 2x + y = 160, we have following points which lie on the line
We plot the points for both of the equations and it is the graphical representation of the given situation.
NCERT Solutions for Class 10 Maths Chapter 3 Exercise 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 |
| 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:
| x | 3 | -4 | -11 |
| y | 5 | 10 | 15 |
Three solutions of this equation can be written in a table as follows:
| x | 0 | 2 | 4 |
| y | 9.2 | 6.4 | 3.6 |
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
Hence,
Therefore, these linear equations will intersect each other at one point and have only one possible solution.
And,pair of linear euations is consistent
(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
And,pair of linear euations is consistent
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 = 0Answer:
(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).
Related Links -
NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.3
1. Solve the following pair of linear equations by the substitution method.(i) x + y = 14
x – y = 4
(ii) s – t = 3
s/3 + t/2 = 6
(iii) 3x – y = 3
9x − 3y = 9
(iv) 0.2x + 0.3y = 1.3
0.4x + 0.5y = 2.3
(v) √2x+ √3y = 0
√3x – √8y = 0
(vi) 3x/2 - 5y/3 = -2
x/3 + y/2 = 13/6
Answer:
(i) x + y = 14 …(1)
x – y = 4 … (2) x = 4 + y from equation (2) Putting this in equation (1), we get 4 + y + y = 14 ⇒ 2y = 10⇒ y = 5 Putting value of y in equation (1), we get x + 5 = 14 ⇒ x = 14 – 5 = 9 Therefore, x = 9 and y = 5(ii) s – t = 3 … (1)
…(2)
Using equation (1), we can say that s = 3 + t
Putting this in equation (2), we get
⇒ 2t + 6 + 3t =36
⇒ 5t + 6 = 36
⇒ 5t = 30⇒ t = 6
Putting value of t in equation (1), we get
s – 6 = 3⇒ s = 3 + 6 = 9
Therefore, t = 6 and s = 9
(iii) 3x – y = 3 … (i)
9x − 3y = 9 … (ii) From equation (i),we get, y =3x − 3 … (iii) putting value of y from equation(iii) to equation(ii) 9x-3(3x-3)=9 =9x-9x+9=9 =9=9 This is always true,and pair of these equations have infinite possible solutions. Therefore one possible solutions is x=1 and y=0(iv) 0.2x + 0.3y = 1.3 … (1)
0.4x + 0.5y = 2.3 … (2) Using equation (1), we can say that 0.2x = 1.3 − 0.3y ⇒ x =
Putting this in equation (2), we get
0.4x(6.5-1.5y)+ 0.5y = 2.3
⇒ 2.6 − 0.6y + 0.5y = 2.3
⇒ −0.1y = −0.3 ⇒ y = 3
Putting value of y in (1), we get
0.2x + 0.3 (3) = 1.3
⇒ 0.2x + 0.9 = 1.3
⇒ 0.2x = 0.4 ⇒ x = 2
Therefore, x = 2 and y = 3
(v)
……….(1)
……….(2)
Using equation (1), we can say that
x =
Putting this in equation (2), we get
⇒
⇒
⇒ y = 0
Putting value of y in (1), we get x = 0
Therefore, x = 0 and y = 0
(vi)
… (1)
… (2)
Using equation (2), we can say that
⇒ x =
Putting this in equation (1), we get
⇒
⇒
⇒
⇒
⇒ y = 3
Putting value of y in equation (2), we get
⇒
⇒
⇒ x = 2
Therefore, x = 2 and y = 3
2. Solve 2x + 3y = 11 and 2x − 4y = −24 and hence find the value of ‘m’ for which
y = mx + 3.
Answer:
2x + 3y = 11 … (1) 2x − 4y = −24 … (2) Using equation (2), we can say that 2x = −24 + 4y ⇒ x = −12 + 2y Putting this in equation (1), we get 2 (−12 + 2y) + 3y = 11 ⇒ −24 + 4y + 3y = 11 ⇒ 7y = 35 ⇒ y = 5 Putting value of y in equation (1), we get 2x + 3 (5) = 11 ⇒ 2x + 15 = 11 ⇒ 2x = 11 – 15 = −4⇒ x = −2 Therefore, x = −2 and y = 5 Putting values of x and y in y = mx + 3, we get 5 = m (−2) + 3 ⇒ 5 = −2m + 3 ⇒ −2m = 2 ⇒ m = −1NCERT Solutions for Class 10 Maths Chapter 3 FAQs
How can understanding Pair of Linear Equations in Two Variables benefit students?
Understanding this concept helps students solve problems involving two variables in various real-life situations such as time-distance, cost-profit, and mixture-related problems.
How can students use NCERT Solutions for Pair of Linear Equations effectively?
Students can use NCERT Solutions to practice different types of problems, clarify doubts, and improve their problem-solving skills, which are essential for scoring well in exams.
Are NCERT Solutions for Pair of Linear Equations in Two Variables comprehensive?
Yes, NCERT Solutions provide detailed explanations and step-by-step solutions to all the questions in the textbook, ensuring that students grasp the concepts thoroughly.
What are the key concepts covered in the Pair of Linear Equations in Two Variables chapter of Class 10 Maths?
This chapter covers topics such as introduction to linear equations, graphical representation of linear equations, algebraic methods to solve linear equations, and practical applications of linear equations in real-life scenarios.
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