# NCERT Solutions for Class 10 Maths Chapter 3

NCERT Solutions for class 10 Maths chapter 3 Pair of linear equations in two variables all exercises in Hindi Medium and English medium PDF form (CBSE, Gujrat, MP and UP Board High School updated for CBSE Exam 2020-21) to download free. UP Board students (High School and Intermediate) are now using NCERT Textbooks for most of the subjects. Class 10 Mathematics Books for UP Board is same as the NCERT Books for Class 10 Math in CBSE Board. So, the students of Uttar Pradesh Board can download, UP Board Solutions for class 10 Math Chapter 3 from this page of Tiwari Academy. Solutions are available in Hindi and English Medium.

Kaksha 10 ke adhyay 3 ki sabhi prashnavaliyon ka hal aur samadhan. Graphs are given for all the questions if required. Visit to Discussion Forum to ask your questions. You can reply the questions already asked by other users.

## NCERT Solutions for class 10 Maths chapter 3

 Class: 10 Maths (English and Hindi Medium) Chapter 3: Pair of linear equations in two variables

### 10th Maths Chapter 3 Solutions

In NCERT Solutions Class 10, all exercises are solved in both English as well as in Hindi medium in order to help all type of students for academic session 2020-21. In Maths 10, Exercise 3.1, 3.2, 3.3, 3.4, 3.5, 3.6 and 3.7 solutions, if there is any inconvenient to understand, please inform us, we will try to solve it. All NCERT Solutions 2020-21 are made for the CBSE exam for March, 2021 based on latest CBSE Syllabus 2020-21.

• ### 10th Maths Chapter 3 Exercise 3.7

#### Changes in CBSE Syllabus for Class 10 Maths Chapter 3

CBSE has reduced the syllabus of all subjects in all the classes. The CBSE Syllabus for Class 10 Maths is reduced to 65 percent now. The changes in 10th Maths chapter 3: Linear equations in two variables are given below.

##### The new CBSE Syllabus for 2020-2021 Class 10 Maths Chapter 3

Pair of linear equations in two variables and graphical method of their solution, consistency /inconsistency.
Algebraic conditions for number of solutions. Solution of a pair of linear equations in two variables algebraically – by substitution, by elimination. Simple situational problems. Simple problems on equations reducible to linear equations.

###### Deleted Section from previous Syllabus

Solution of a pair of linear equations in two variables by Cross-multiplication.

#### Class 10 Maths Exercise 3.1 Solutions in English and Hindi Videos

Class 10 Maths Exercise 3.1 Solution in Hindi
Class 10 Maths Chapter 3 Exercise 3.1 Solution

#### Class 10 Maths Exercise 3.2 Solutions in English and Hindi Videos

Class 10 Maths Exercise 3.2 Solution in Hindi
Class 10 Maths Chapter 3 Exercise 3.2 Solution

#### Class 10 Maths Exercise 3.3 Solutions in English and Hindi Videos

Class 10 Maths Exercise 3.3 Solution in Hindi
Class 10 Maths Chapter 3 Exercise 3.3 Solution

#### Class 10 Maths Exercise 3.4 Solutions in English and Hindi Videos

Class 10 Maths Exercise 3.4 Solution in Hindi
Class 10 Maths Chapter 3 Exercise 3.4 Solution

#### Class 10 Maths Exercise 3.5 Solutions in English and Hindi Videos

Class 10 Maths Exercise 3.5 Solution in Hindi
Class 10 Maths Chapter 3 Exercise 3.5 Solution

#### Class 10 Maths Exercise 3.6 Solutions in English and Hindi Videos

Class 10 Maths Exercise 3.6 Solution in Hindi
Class 10 Maths Chapter 3 Exercise 3.6 Solution

#### Class 10 Maths Exercise 3.7 Solutions in English and Hindi Videos

Class 10 Maths Exercise 3.7 Solution in Hindi
Class 10 Maths Chapter 3 Exercise 3.7 Solution

#### Previous Years Questions – 1 Mark or 2 Marks

1. Find whether the lines representing the following pair of linear equations intersect at a point, are parallel or coincident: 3x + y = 7 & 6x + 2y = 8. [CBSE 2016]
2. Find the value of k for which the system of equations 3𝑥 − 4𝑦 = 7; 𝑘𝑥 + 3𝑦 − 5 = 0 has no solution. [CBSE 2014]
3. A father is three times as old as his son. After five years, his age will be two and a half times as old as his son. Represent this situation algebraically only. [CBSE 2013]
4. For which value of p does the pair of equations given below have a unique solutions? 4x + py + 8 = 0; 2x + 2y + 2 = 0. [CBSE 2010, 2011, 2013]
5. For what value of k, the following system of linear equations has no solutions? 3x + y = 1; (2k – 1)x + (k – 1)y =2k + 1. [CBSE 2010, 2011, 2012]

##### Previous Years Questions – 3 Marks
1. Solve for x and y: 11/x – 1/y = 10 & 9/x – 4/y = 5. [CBSE 2016]
2. Solve using cross multiplication method: 5x + 4y – 4 = 0 & x – 12y – 20 = 0. [CBSE 2016]
3. A man has certain notes of denomination ₹ 20 and ₹ 5 which amount to ₹ 380. If the number of notes of each kind are interchanged, they amount to ₹ 60 less than before. Find the number of notes of each denomination. [CBSE 2015]
4. Find the value of ‘k’ for which the following system of equations represents a pair of coincident lines: 𝑥 + 2𝑦 = 3; (𝑘 − 1)𝑥 + (𝑘 + 1)𝑦 = 𝑘 + 3. [CBSE 2014]
5. Check graphically, whether the pair of equations x + 3y = 6 & 2x – 3y = 12 is consistent. If so, then solve them graphically. [CBSE 2013]
6. The path of a train A is given by the equation x + 2y – 4 = 0 and path of another train B is given by the equation 2x + 4y – 12 = 0. Represent this situation graphically and find whether the two trains meet each other at some place. [CBSE 2013]
7. Form a pair of linear equations in two variables from the data given and solve it graphically: Tina went to a book shop to get some story books and textbooks. When her friends asked her how many of each she had bought, she answered – ‘The number of textbooks is two more than twice the number of story books bought. Also, the number of textbooks is four less than four times the number of story books bought. Help her friends to find the number of textbooks and story books she had bought. [CBSE 2013]
8. Determine graphically, the coordinates of the vertices of a triangle whose sides are graphs of the equations 2x – 3y + 6 = 0, 2x + 3y – 18 = 0 and y – 2 = 0. Also, find the area of this triangle. [CBSE 2010, 2011]
###### Previous Years Questions – 4 Marks
1. For Uttarakhand flood victims’ two sections A and B of class X contributed ₹ 1500. If the contribution of X A was ₹ 100 less than that of X B, find graphically the amounts contributed by both the sections. [CBSE 2016]
2. Three lines 3x + 5y = 15, 6x – 5y = 30 and x = 0 are enclosing a beautiful triangular park. Find the points of intersection of the lines graphically and the area of the park if all measurements are in km. [CBSE 2016]
3. Some people collected money to be donated in two Old Age Homes. A part of the donation is fixed and remaining depends on the number of old people in the home. If they donated ₹ 14500 in the home having 60 people and ₹ 19500 with 85 people, find the fixed part of donation and the amount donated for each people. What is the inspiration behind this? [CBSE 2016]
4. While teaching about the Indian National flag, teacher asked the students that how many lines are there in Blue colour wheel? One student replies that it is 8 times the number of colours in the flag. While other says that the sum of the number colours in the flag and number of lines in the wheel of the flag is 27. Convert the statements given by the students into linear equation of two variables. Find the number of lines in the wheel. [CBSE 2015]
5. Determine the value of k for which the following system of linear equations has infinite number of solutions: (k – 3)x + 3y = k & kx + ky = 12. [CBSE 2015]
6. Draw the graph of the following pair of linear equation: x + 3y = 6 & 2x – 3y = 12. Find the ratio of the areas of the two triangles formed by first line, x = 0, y = 0 and second line, x = 0, y = 0. [CBSE 2015]
7. Places A and B are 200km apart on a high way. One car starts from A and another from B at the same time. If the cars travel in the same directions at different speeds, they meet in 10 hours. Find the speeds of the two cars. [CBSE 2014]
8. Show graphically that the system of equations𝑥 + 2𝑦 = 4 and 7𝑥 + 4𝑦 = 18 is consistent with a unique solution (2, 1). [CBSE 2014]

###### 10th Maths Chapter 3 Questions for Practice
• Solve for x and y: 99x + 101y = 1499; 101x + 99y = 1501. [CBSE 2010, 2011, 2012, 2013, 2014]
• The age of father is equal to sum of ages of his 4 children. After 30 years, sum of the ages of the children will be twice the age of the father. Find the age of the father. [CBSE 2013]
• A person can row a boat 8 km upstream and 24 km downstream in 4 hours. He can row 12 km downstream and 12 km upstream in 4 hours. Find the speed of rowing in still water and the speed of the current. [CBSE 2013]
• Solve for x and y: 37x + 43y = 123; 43x + 37y = 117. [CBSE 2010, 2011, 2012]
• Draw the graph of the equations: x – y + 1 = 0 & 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. Also calculate the area of the triangle so formed. [CBSE 2011]
• The sum of a 2 digit number and number obtained by reversing the order of digits is 99. If the digits of the number differ by 3, find the number. [CBSE 2011]
• Check graphically whether the pair of linear equations 4x – y – 8 = 0 and 2x – 3y + 6 = 0 is consistent. Also determine the vertices of the triangle form by these lines and x – axis. [CBSE 2006, 2011]
• The sum of the digits of a two digit number is 9. Nine times this number is twice the number obtained by reversing the digits. Find the number. [CBSE 2010]
• A leading library has a fixed charge for the first three days and an additional charge for each day thereafter. Sarita paid ₹ 27 for a book kept for seven days. While Susy paid ₹ 21 for the book she kept for five days. Find the fixed charge and the charge for each extra day. [CBSE 2010]
• Solve the following system of linear equations by elimination method: 6(ax + by) = 2a + 2b and 6(bx – ay) = 3b – 2a. [CBSE 2006, 2004]

###### Historical Facts !

History about Linear Equations with two variables.

• Around 4000 years ago, Babylon knew how to solve a simple linear equation with two variables.
• Around 200 BC, the Chinese published that “Nine Chapters of the Mathematical Art,” they displayed the ability to solve a system of equations in three variables (Perotti).
• Evidence from about 300 BC indicates that the Egyptians also knew how to solve problems involving a system of two equations in two unknown quantities, including quadratic equations.
• Euler brought to light the idea that a system of equations doesn’t necessarily have to have a solution (Perotti). He recognized the need for conditions to be placed upon unknown variables in order to find a solution.
• With the turn into the 19th century Gauss introduced a procedure to be used for solving a system of linear equations.
• Cayley, Euler, Sylvester, and others changed linear systems into the use of matrices to represent them. Gauss brought his theory to solve systems of equations proving to be the most effective basis for solving unknowns.

### Important Questions on Class 10 Maths Chapter 3

रितु धारा के अनुकूल 2 धंटे में 20 km तैर सकती है और धारा के प्रतिकूल 2 घंटे में 4 km तैर सकती है। उसकी स्थिर जल में तैरने की चाल तथा धारा की चाल ज्ञात कीजिए।
माना, रितु की स्थिर जल में तैरने की चाल = x km/h
माना, धारा की चाल = y km/h
धारा के अनुकूल, चाल = x + y km/h,
दूरी = 20 km, समय = 2 धंटे
प्रश्नानुसार, चाल = दूरी/ समय
x + y = 20/2
⇒ y = 10 – x … (1)
धारा के प्रतिकूल, चाल = x – y km/h,
दूरी = 4 km, समय = 2 धंटे
x – y = 4/2
⇒ x – y = 2
समीकरण (1) से y का मान रखने पर
x – (10 – x) = 2
⇒ 2x = 12
⇒ x = 6
समीकरण (1) में x का मान रखने पर
y= 10 – 6 = 4
अतः, रितु की स्थिर जल में तैरने की चाल 6 km/h तथा धारा की चाल 4 km/h है।
Solve the following pair of linear equation by the substitution method: x + y =14 ; x – y = 4.
x + y = 14 … (1)
x – y = 4 … (2)
From the equation (1), we get
y = 14 – x … (3)
Putting the value of y in equation (2),
we get
x – (14 – x) = 4
⇒ 2x = 18
⇒ x = 9
Putting the value of x in equation (3),
we get
y = 14 – 9 = 5
Hence, x = 9 and y = 5.
Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacob’s age was seven times that of his son. What are their present ages?
Let the age of Jacob = x years
Let the age of son = y years
After 5 years,
Jacob’s = x + 5 years
Son’s age = y + 5 years
According to question,
x + 5 = 3(y + 5)
⇒ x + 5 = 3y + 15
⇒ x = 3y + 10 … (1)
5 years ago,
Jacob’s age = x – 5 years
Son’s age = y – 5 years
According to question,
x – 5 = 7(y – 5)
⇒ x – 5 = 7y – 35
⇒ x – 7y = -30 … (2)
Putting the value of x in equation (2),
we get
3y + 10 – 7y = -30
⇒ – 4y = – 40
⇒ y = 10
Putting the value of y in equation (1),
we get
x = 3(10) + 10 = 40
Hence, the age of Jacob is 40 years and the age of his son is 10 years.
The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.
Let the larger angle = x
Let the smaller angle = y
According to question,
x = y + 18 … (1)
Both angles are supplementary, therefore
x + y = 180 … (2)
Putting the value of x in equation (2), we get
y + 18 + y = 180
⇒ 2y = 162
⇒ y = 81
Putting the value of y in equation (1), we get
x = 81 + 18 = 99
Hence, one angle is 81° and the other one is 99°.
पाँच वर्ष पूर्व नूरी की आयु सोनू की तीन गुनी थी। दस वर्ष पश्चात, नूरी की आयु सोनू की आयु की दो गुनी हो जाएगी। नूरी और सोनू की आयु कितनी है।
माना, नूरी की वर्तमान आयु = x वर्ष
माना, सोनू की वर्तमान आयु = y वर्ष
पाँच वर्ष पूर्व
नूरी की आयु = x – 5 वर्ष
सोनू की आयु = y – 5 वर्ष
प्रश्नानुसार,
x – 5 = 3(y – 5)
⇒ x – 5 = 3y – 15
⇒ x – 3y = -10 … (1)
दस वर्ष पश्चात
नूरी की आयु = x + 10 वर्ष
सोनू की आयु = y + 10 वर्ष
प्रश्नानुसार,
x + 10 = 2(y + 10)
⇒ x + 10 = 2y + 20
⇒ x – 2y = 10 … (2)
समीकरण (1) से समीकरण (2) को घटाने पर
– y = -20
⇒ y = 20
y का मान समीकरण (1) में रखने पर
x- 3(20) = -10
⇒ x = 50
अतः, नूरी की वर्तमान आयु 50 वर्ष तथा सोनू की वर्तमान आयु 20 वर्ष है।
One says, “Give me a hundred, friend! I shall then become twice as rich as you”. The other replies, “If you give me ten, I shall be six times as rich as you”. Tell me what is the amount of their (respective) capital?
Let the amount with first friend = ₹x
Let the amount with second friend = ₹y
According to first condition,
x + 100 = 2(y – 100)
⇒ x = 2y – 300 … (1)
According to second condition,
y + 10 = 6(x – 10)
y – 6x = -70
Putting the value of x from equation (1), we get
y – 6(2y – 300) = -70
⇒ y – 12y + 1800 = -70
⇒ -11y = -1870
⇒ y = 170
Putting the value of y in equation (1), we get
x = 2(170) – 300 = 40
Hence, one friend have ₹170 and the other have ₹40.
The students of a class are made to stand in rows. If 3 students are extra in a row, there would be 1 row less. If 3 students are less in a row, there would be 2 rows more. Find the number of students in the class.
Let the number of rows = x
Let the number of students in each row = y
Therefore, the total number of students = xy
According to first condition,
If 3 students are extra in a row, there would be 1 row less.
(x – 1)(y + 3) = xy
⇒ xy + 3x – y – 3 = xy
⇒ y = 3x – 3 … (1)
According to second condition,
If 3 students are less in a row, there would be 2 rows more.
(x + 2)(y – 3) = xy
⇒ xy – 3x + 2y – 6 = xy
⇒ -3x + 2y = 6
Putting the value of y from equation (1), we get
-3x + 2(3x – 3) = 6
⇒ -3x + 6x – 6 = 6
⇒ 3x = 12
⇒ x = 4
Putting the value of x in equation (1), we get
y = 3(4) – 3 = 9
The total number of students in class
= xy = (4)(9) = 36
Hence, there are 36 students in the class.              