NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.5

NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.5 – Pair of Linear Equations in Two Variables

Exercise 3.5 class 10 - Linear Equations in Two Variables - is one of the most important exercises of chapter 3 as far as the NCERT syllabus is concerned.  This chapter, if you see the textbook, is all about the equation ax + by + c = 0 where a and b both are not equal to zero and a, b and c are real numbers. In NCERT Solutions for Class 10 Maths Chapter 3 exercise 3.5, you get a mixed type of problems based on this equation. To solve these problems, all you have to do is refer to Vedantu’s solution set for 3.5 Maths class 10. It will enable you to get a good grasp of the concept. Eager to get a glimpse of the solved problems in class 10 maths chapter 3 exercise 3.5 solutions? Students can download NCERT Solutions PDF for Ex 3.5 Class 10 Maths at any time. We not only provide Solutions for maths but also NCERT Solutions for Class 10 Science for free only at Vedantu.

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Access NCERT Solutions for Class 10 Maths Chapter 3 – Pair of Linear Equations in Two Variables part-1

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

Exercise 3.5

1. Which of the following pairs of linear equations has unique solution, no solution or infinitely many solutions? In case there is a unique solution, find it by using cross multiplication method.

(i) \[\mathbf{x} - \mathbf{3y} - \mathbf{3}=\mathbf{0};\text{ }\mathbf{3x}- \mathbf{9y} - \mathbf{2}=\mathbf{0}\]

Ans: Calculating \[\frac{{{a}_{1}}}{{{a}_{2}}},\frac{{{b}_{1}}}{{{b}_{2}}}\] and \[\frac{{{c}_{1}}}{{{c}_{2}}}\],

\[\frac{{{a}_{1}}}{{{a}_{2}}}=\frac{1}{3}\]

\[\frac{{{b}_{1}}}{{{b}_{2}}}=\frac{1}{3}\]

\[\frac{{{c}_{1}}}{{{c}_{2}}}=\frac{3}{2}\]

Hence, \[\frac{{{a}_{1}}}{{{a}_{2}}}=\frac{{{b}_{1}}}{{{b}_{2}}}\ne \frac{{{c}_{1}}}{{{c}_{2}}}\].

Therefore there are no solution for the given pair of linear equation as the given lines are parallel to each other and do not intersect.

(ii) \[\mathbf{2x}+\mathbf{y}=\mathbf{5};\text{ }\mathbf{3x}+\mathbf{2y}=\mathbf{8}\]

Ans: Calculating \[\frac{{{a}_{1}}}{{{a}_{2}}},\frac{{{b}_{1}}}{{{b}_{2}}}\] and \[\frac{{{c}_{1}}}{{{c}_{2}}}\],

\[\frac{{{a}_{1}}}{{{a}_{2}}}=\frac{2}{3}\]

\[\frac{{{b}_{1}}}{{{b}_{2}}}=\frac{1}{2}\]

\[\frac{{{c}_{1}}}{{{c}_{2}}}=\frac{-5}{-8}\]

Hence, \[\frac{{{a}_{1}}}{{{a}_{2}}}\ne \frac{{{b}_{1}}}{{{b}_{2}}}\].

Therefore the given pair of linear equations have a unique solution as the given lines intersect each other at a unique point.

Using cross-multiplication method,

\[\frac{x}{{{b}_{1}}{{c}_{2}}-{{b}_{2}}{{c}_{1}}}=\frac{y}{{{c}_{1}}{{a}_{2}}-{{c}_{2}}{{a}_{1}}}=\frac{1}{{{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}}}\]

\[\frac{x}{-8-\left( -10 \right)}=\frac{y}{15+16}=\frac{1}{4-3}\]

\[\frac{x}{2}=\frac{y}{1}=1\]

\[x=2,y=1\]

Therefore, \[x=2\] and \[y=1\].

(iii) \[\mathbf{3x} - \mathbf{5y}=\mathbf{20};\text{ }\mathbf{6x} - \mathbf{10y}=\mathbf{40}\]

Ans: Calculating \[\frac{{{a}_{1}}}{{{a}_{2}}},\frac{{{b}_{1}}}{{{b}_{2}}}\] and \[\frac{{{c}_{1}}}{{{c}_{2}}}\],

\[\frac{{{a}_{1}}}{{{a}_{2}}}=\frac{1}{2}\]

\[\frac{{{b}_{1}}}{{{b}_{2}}}=\frac{1}{2}\]

\[\frac{{{c}_{1}}}{{{c}_{2}}}=\frac{1}{2}\]

Hence, \[\frac{{{a}_{1}}}{{{a}_{2}}}=\frac{{{b}_{1}}}{{{b}_{2}}}=\frac{{{c}_{1}}}{{{c}_{2}}}\].

Therefore there are infinite number of solution for the given pair of linear equation as the given lines are overlapping each other.

(iv) \[\mathbf{x} - \mathbf{3y} - \mathbf{7}=\mathbf{0};\text{ }\mathbf{3x} - \mathbf{3y} - \mathbf{15}=\mathbf{0}\]

Ans: Calculating \[\frac{{{a}_{1}}}{{{a}_{2}}},\frac{{{b}_{1}}}{{{b}_{2}}}\] and \[\frac{{{c}_{1}}}{{{c}_{2}}}\],

\[\frac{{{a}_{1}}}{{{a}_{2}}}=\frac{1}{3}\]

\[\frac{{{b}_{1}}}{{{b}_{2}}}=1\]

\[\frac{{{c}_{1}}}{{{c}_{2}}}=\frac{-7}{-15}\]

Hence, \[\frac{{{a}_{1}}}{{{a}_{2}}}\ne \frac{{{b}_{1}}}{{{b}_{2}}}\].

Therefore the given pair of linear equations have a unique solution as the given lines intersect each other at a unique point.

Using cross-multiplication method,

\[\frac{x}{{{b}_{1}}{{c}_{2}}-{{b}_{2}}{{c}_{1}}}=\frac{y}{{{c}_{1}}{{a}_{2}}-{{c}_{2}}{{a}_{1}}}=\frac{1}{{{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}}}\]

\[\frac{x}{45-\left( 21 \right)}=\frac{y}{-21-\left( -15 \right)}=\frac{1}{-3+\left( 9 \right)}\]

\[\frac{x}{24}=\frac{y}{-6}=\frac{1}{6}\]

\[x=4,y=-1\]

Therefore, \[x=4\] and \[y=-1\].

2. (i) For which values of \[a\] and \[b\] does the following pair of linear equations have an infinite number of solutions?

\[\mathbf{2x}+\mathbf{3y}=\mathbf{7};\text{ }\left( \mathbf{a}\text{ } - \text{ }\mathbf{b} \right)\mathbf{x}+\left( \mathbf{a}\text{ }+\text{ }\mathbf{b} \right)\mathbf{y}=\mathbf{3a}+\mathbf{b} - \mathbf{2}\]

Ans: Calculating \[\frac{{{a}_{1}}}{{{a}_{2}}},\frac{{{b}_{1}}}{{{b}_{2}}}\] and \[\frac{{{c}_{1}}}{{{c}_{2}}}\],

\[\frac{{{a}_{1}}}{{{a}_{2}}}=\frac{2}{a-b}\]

\[\frac{{{b}_{1}}}{{{b}_{2}}}=\frac{3}{a+b}\]

\[\frac{{{c}_{1}}}{{{c}_{2}}}=\frac{-7}{-\left( 3a+b-2 \right)}\]

Condition for infinitely many solutions,

\[\frac{{{a}_{1}}}{{{a}_{2}}}=\frac{{{b}_{1}}}{{{b}_{2}}}=\frac{{{c}_{1}}}{{{c}_{2}}}\]

\[\frac{2}{a-b}=\frac{7}{\left( 3a+b-2 \right)}\]

\[6a+2b-4=7a-7b\]

\[a-9b=-4\]                  …… (i)

\[\frac{2}{a-b}=\frac{3}{a+b}\]

\[2a+2b=3a-3b\]

\[a-5b=0\]                           …… (ii)

Subtracting equation (i) from equation (ii), we get

\[4b=4\]

\[b=1\]                  …… (iii)

Substituting (iii) in equation (ii), we get

\[a-5=0\]

\[a=5\]

Therefore, the given equations will have infinite number of solutions for \[a=5\] and \[b=1\].

(ii) For which value of \[\mathbf{k}\] will the following pair of linear equations have no solution?\[\mathbf{3x}+\mathbf{y}=\mathbf{1};\text{ }\left( \mathbf{2k}\text{ } - \text{ }\mathbf{1} \right)\mathbf{x}+\left( \mathbf{k}\text{ } - \text{ }\mathbf{1} \right)\mathbf{y}=\mathbf{2k}+\mathbf{1}\].

Ans: Calculating \[\frac{{{a}_{1}}}{{{a}_{2}}},\frac{{{b}_{1}}}{{{b}_{2}}}\] and \[\frac{{{c}_{1}}}{{{c}_{2}}}\],

\[\frac{{{a}_{1}}}{{{a}_{2}}}=\frac{3}{2k-1}\]

\[\frac{{{b}_{1}}}{{{b}_{2}}}=\frac{1}{k-1}\]

\[\frac{{{c}_{1}}}{{{c}_{2}}}=\frac{1}{2k+1}\]

Condition for no solution,

\[\frac{{{a}_{1}}}{{{a}_{2}}}=\frac{{{b}_{1}}}{{{b}_{2}}}\ne \frac{{{c}_{1}}}{{{c}_{2}}}\]

\[\frac{3}{2k-1}=\frac{1}{k-1}\ne \frac{1}{2k+1}\]

\[\frac{3}{2k-1}=\frac{1}{k-1}\]

\[3k-3=2k-1\]

\[k=2\]

Therefore, the given equations will not have any solutions for \[k=2\].

3. Solve the following pair of linear equations by the substitution and cross multiplication methods:

\[\mathbf{8x}+\mathbf{5y}=\mathbf{9};\text{ }\mathbf{3x}+\mathbf{2y}=\mathbf{4}\]

Ans :

Using substitution method,

\[8x+5y=9\]                   …… (i)

\[3x+2y=4\]                  …… (ii)

From equation (ii) we get

\[x=\frac{4-2y}{3}\]      …… (iii)

Substituting (iii) in equation (i), we get

\[8\left( \frac{4-2y}{3} \right)+5y=9\]

\[32-16y+15y=27\]

\[y=5\]              …… (iv)

Substituting (iv) in equation (ii), we obtain

\[3x+10=4\]

\[x=-2\]

Therefore, \[x=-2\] and \[y=5\].

Using cross multiplication method:

\[8x+5y=9\]

\[3x+2y=4\]

\[\frac{x}{-20-\left( -18 \right)}=\frac{y}{-27-\left( -32 \right)}=\frac{1}{16-15}\]

\[\frac{x}{-2}=\frac{y}{5}=1\]

\[x=-2,y=5\]

Therefore, \[x=-2\] and \[y=5\].

4. Form the pair of linear equations in the following problems and find their solutions (if they exist) by any algebraic method:

(i) A part of monthly hostel charges is fixed and the remaining depends on the number of days one has taken food in the mess. When a student \[\mathbf{A}\] takes food for \[\mathbf{20}\] days she has to pay \[\mathbf{Rs}\text{ }\mathbf{1000}\] as hostel charges whereas a student \[\mathbf{B}\], who takes food for \[\mathbf{26}\] days, pays \[\mathbf{Rs}\text{ }\mathbf{1180}\] as hostel charges. Find the fixed charges and the cost of food per day.

Ans: Assuming that the fixed charge is \[Rs\text{ }x\] and the charge for food per day is \[Rs\text{ y}\].

Writing the algebraic representation using the information given in the question:

\[x+20y=1000\]             …… (i)

\[x+26y=1180\]    …… (ii)

Subtracting equation (i) from equation (ii), we obtain

\[6y=180\]

\[y=30\]     …… (iii)

Substituting (iii) in equation (i), we obtain

\[x+20\left( 30 \right)=1000\]

\[x=400\]              …… (iv)

Therefore, \[x=400\] and \[y=30\].

Hence, fixed charges are \[Rs\text{ 400}\] and charges per day are \[Rs\text{ 30}\].


(ii) A fraction becomes \[\frac{1}{3}\] when \[\mathbf{1}\] is subtracted from the numerator and it becomes \[\frac{1}{4}\] when \[\mathbf{8}\] is added to its denominator. Find the fraction.

Ans: Assuming that the fraction is \[\frac{x}{y}\].

Writing the algebraic representation using the information given in the question:

\[\frac{x-1}{y}=\frac{1}{3}\]

\[3x-y=3\]             …… (i)

\[\frac{x}{y+8}=\frac{1}{4}\]

\[4x-y=8\]    …… (ii)

Subtracting equation (i) from equation (ii), we obtain

\[x=5\]     …… (iii)

Substituting (iii) in equation (i), we obtain

\[15-y=3\]

\[y=12\]              …… (iv)

Therefore, \[x=5\] and \[y=12\].

Hence, the fraction is \[\frac{5}{12}\].

(iii) Yash scored \[\mathbf{40}\] marks in a test, getting \[\mathbf{3}\] marks for each right answer and losing \[\mathbf{1}\] mark for each wrong answer. Had \[\mathbf{4}\] marks been awarded for each correct answer and \[\mathbf{2}\] marks been deducted for each incorrect answer, then Yash would have scored \[\mathbf{50}\] marks. How many questions were there in the test?

Ans: Say the number of right answers be \[\text{ }x\] and the number of wrong answers be \[\text{ y}\].

Writing the algebraic representation using the information given in the question:

\[3x-y=40\]             …… (i)

\[2x-y=25\]    …… (ii)

Subtracting equation (ii) from equation (i), we obtain

\[x=15\]     …… (iii)

Substituting (iii) in equation (ii), we obtain

\[y=5\]              …… (iv)

Therefore, \[x=15\] and \[y=5\].

Hence, number of right answers = \[\text{15}\] and number of wrong answers = \[\text{ 5}\].

Therefore, the total number of questions = \[\text{20}\]

(iv) Places A and B are \[\mathbf{100}\text{ }\mathbf{km}\] apart on a highway. One car starts from \[\mathbf{A}\] and another from \[\mathbf{B}\] at the same time. If the cars travel in the same direction at different speeds, they meet in \[\mathbf{5}\] hours. If they travel towards each other, they meet in \[\mathbf{1}\] hour. What are the speeds of the two cars?

Ans: Assuming that the speed of first car is \[u\text{ }km/h\] and the speed of second car is \[\text{v }km/h\].

Relative speed when both cars are travelling in same direction \[\text{=}\left( u-v \right)\text{ }km/h\]

Relative speed when both cars are travelling in opposite direction \[\text{=}\left( u+v \right)\text{ }km/h\]

Writing the algebraic representation using the information given in the question:

\[5\left( u-v \right)=100\]

\[u-v=20\]             …… (i)

\[1\left( u+v \right)=100\]    …… (ii)

Adding equation (i) and (ii), we get

\[2u=120\]

\[u=60\]     …… (iii)

Substituting (iii) in equation (ii), we obtain

\[v=40\]

Hence, speed of one car is \[\text{60 }km/h\] and speed of the other car is \[\text{40 }km/h\].

(v) The area of a rectangle gets reduced by \[\mathbf{9}\] square units, if its length is reduced by \[\mathbf{5}\] units and breadth is increased by \[\mathbf{3}\] units. If we increase the length by \[\mathbf{3}\] units and the breadth by \[\mathbf{2}\] units, the area increases by \[\mathbf{67}\] square units. Find the dimensions of the rectangle.

Ans: Assuming length of the rectangle be \[x\] and the breadth be \[\text{y}\].

Writing the algebraic representation using the information given in the question:

\[\left( x-5 \right)\left( y+3 \right)=xy-9\]

\[3x-5y-6=0\]             …… (i)

\[\left( x+3 \right)\left( y+2 \right)=xy+67\]    …… (ii)

Using cross multiplication method:

\[\frac{x}{305-\left( -18 \right)}=\frac{y}{-12-\left( -183 \right)}=\frac{1}{9-\left( -10 \right)}\]

\[\frac{x}{323}=\frac{y}{171}=\frac{1}{19}\]

\[x=17,y=9\]

Hence, the rectangle has length of \[17\text{ }units\] and breadth of \[\text{9 }units\].

Class 10 Chapter 3 Exercise 3.5 Solution: A Brief Overview

Exercise 3.5 class 10 Maths consists of various questions related to Linear Equations in Two Variables. In this section, you will get familiar with the few questions related to the above-mentioned topic. Rigorous practising of class 10 Maths ex. 3.5 will definitely help you to understand the entire chapter. This will also increase your problem-solving ability along with coping up with the latest CBSE question pattern. Ex. 3.5 class 10 Maths mainly deals with the following types of questions:

Question 1: Which of the following pairs of linear equations consists of a unique solution, without solution or infinitely numerous solutions? Derive the solution in case of a unique solution applying a cross multiplication method.

i) x – 3y – 3= 0 and 3x – 9y - 2 = 0 ii) 2x+ y = 5 and 3x + 2y = 8 iii) x – 3y – 7= 0 and 3x – 3y -15 = 0 iv) 3x – 5y = 20 and 6x – 10y = 40.

Solution:

i) Given, x -3y -3 = 0 and 3x -9y – 2= 0.

a1/a2 = 1/3, b1/b2 = -3/-9= 1/3, c1/c2 = -3/-2 = 3/2

(a1/a2) = (b1/b2) ≠ (c1/c2)

There is no solution for the above equation as the said pair of lines are parallel to each other and there will not be any intersection between them.

Here are a few more solved problems from exercise 3.5 Maths class 10.

ii) Given, 2x + y = 5 and 3x + 2y = 8

a1/a2 = 2/3, b1/b2 = ½ and c1/c2 = -5/-8

(a1/a2) ≠ (b1/b2)

These equations should have a unique solution by applying cross multiplication technique as there is an intersection between them at a unique point.

x/(b1c2 – c1b2) = y/(c1a2 – c2a1) = 1/ (a1b2 – a2b1)

x/(-8-(-10)) = y/(15+16) = 1/(4-3) 

x/2= y/1 = 1

Hence, x = 2 and y = 1

iii) Given, x - 3y - 7 = 0 and 3x - 3y - 15 = 0

(a1/a2) = 1/3

(b1/b2) = -3/-3 = 1

(c1/c2) = -7/-15

(a1/a2) ≠ (b1/b2) 

There will definitely be a unique solution as the set of lines are intersecting each other at a unique point.

Applying cross multiplication method we get,

x/(45-21) = y /(-21+ 15) = 1/(-3 + 9)

x/24 = y/-6 = 1/6

x/24 = 1/6 and y/-6 = 1/6

Hence, x = 4 and y = 1

iv) Given, 3x - 5y = 20 and 6x – 10y = 40

(a1/a2) = 3/6 = ½

(b1/b2) = -5/-10 = ½

(c1/c2) = 20/40 = ½

(a1/a2) = (b1/b2) = (c1/c2)

There will be an infinite number of solutions as the given pair of lines is overlapping each other.

Why Should You Choose Vedantu?

If you are looking for the solutions of Ex. 3.5 class 10 Maths Vedantu will be the best option for you. There are various reasons in support of the above statement.

  • Vedantu provides the exercise 3.5 class 10 Maths NCERT solutions which are entirely exam-oriented and based on the latest CBSE guidelines.

  • The class 10 Maths ch 3 ex. 3.5 is designed by the pool of experienced teachers of Vedantu.

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  • The questions of Maths class 10 chapter 3 exercise 3.5 are architected from moderate to difficult problems so that students can get familiar with the exam pattern gradually.

Not only NCERT class 10 maths chapter 3 exercise 3.5, but all the study materials of Vedantu are going to be equally helpful for you in your journey towards the CBSE board exam.

FAQs (Frequently Asked Questions)

1. What is the Procedure of Accessing Class 10 Chapter 3 Exercise 3.5 Solutions?

If you want to access the class 10th maths chapter 3 exercise 3.5 you have to follow the following steps:

a) Go to the official site of Vedantu and hit the search bar.

b) Type the words NCERT solutions class 10 maths chapter 3 exercise 3.5.

c) You will be redirected to the page containing NCERT solutions for class 10 maths chapter 3.5.

d) You can either practise Maths class 10 ex 3.5 online or download it for future reference.

e) Solve the ex 3.5 class 10 maths on your own first and then, refer to the solutions.

Ex 3.5 class 10 maths solutions are not the only study materials. You get all the chapter-wise solutions available on Vedantu’s website.

2. How Helpful are Maths PDFs like Class 10 Maths Ex 3.5 Solutions?

Class 10 Maths ch 3 ex 3.5 solutions are just one of the many study materials that are considered as the Bible of CBSE Mathematics. The questions covered in class 10 Maths Chapter 3 ex 3.5 and are completely designed as per the latest CBSE guidelines. If you practise the NCERT class 10 Maths 3.5 rigorously you will definitely score high marks in this portion of CBSE Maths paper.

3. What are the different processes of solving a pair of linear equations in two variables?


Any pair of linear equations with two variables can be solved by the following processes:

Graphical process: A graph of the pair of linear equations with two variables can be presented with two lines on the graphical plane.

Algebraic processes: There are three methods to find the solution of any given pair of linear equations:

  • Substitution

  • Elimination 

  • Cross-multiplication

4. What is the underlying concept of Chapter 3 of Class 10 Maths?

Students learn about linear equations with two variables in Chapter 3 of Class 10 Maths. Any equation may be written as  ax + by + c = 0.

When a, b and c have to be real numbers, with a and b being non zero, it is referred to as a linear equation with two variables, i.e,  x and y. The solution to these kinds of equations is two values, one to replace x and another to replace y.  Students will learn that all solutions to the equation are points on the representing line.

5. What are the topics and sub-topics included in Chapter 3 of Class 10 Maths?

Chapter 3 of Class 10 Maths is important to understand future concepts easily. The following are the topics and sub-topics of this Chapter: 

  • Pair of Linear Equations with Two Variables

  • Solving a Pair of Linear Equations with Graphical method

  • Substitution 

  • Elimination 

  • Cross-Multiplication 

  • Reducing equations to a Pair of Linear Equations with Two Variables

6. What are the most important formulas that come in Chapter 3 of Class 10 Maths?

If

a1 + b1y + c1 = 0  

a2 + b2y + c2 = 0,

the kind of roots or the solutions can be determined by:

If a1/a2 ≠ b1/b2 

  • Unique solution

  • Intersecting lines

If a1/a2 ≠ b1/b2 ≠ c1/c2

  • No solution

  • Parallel lines

If a1/a2 = b1/b2 = c1/c2

  • Infinite solutions

  • Coincident lines

7. Where can I get the NCERT Solution for Chapter 3 of Class 10 Maths?

Chapter 3 “Pair of Linear Equations with Two Variables” of Class 10 Maths is a topic that is important for 10th boards as well as future classes. This implies that it is extremely crucial to understand the concepts thoroughly with the help of NCERT Solutions. The NCERT Solutions by Vedantu are your best choice because they have been carefully curated by the subject matter experts who have kept in mind the marking scheme as well as the primary objective of concept clearing. These solutions are available on the Vedantu website and on the Vedantu app at free of cost.

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