NCERT Solutions for Exercise 3.5 Class 10 Maths Chapter 3 - Pair of Linear Equations in two variables

# NCERT Solutions for Exercise 3.5 Class 10 Maths Chapter 3 - Pair of Linear Equations in two variables

Edited By Ramraj Saini | Updated on Nov 13, 2023 03:17 PM IST | #CBSE Class 10th

## NCERT Solutions For Class 10 Maths Chapter 3 Exercise 3.5

NCERT Solutions for class 10 maths ex 3.5 Pair of Linear Equations in two variables is discussed here. These NCERT solutions are created by subject matter expert at Careers360 considering the latest syllabus and pattern of CBSE 2023-24. This ex 3.5 class 10 deals with the cross multiplication method which is one of the methods to solve the pair of linear equations. In exercise 3.5 Class 10 Maths, pair of linear equations can also be solved by many non-graphical methods such as elimination method, substitution method and cross multiplication method but the NCERT book Class 10 Maths chapter 3 exercise 3.5 focuses on solving the equations only by cross multiplication method it is the easiest and straight forward method.

These class 10 maths ex 3.5 solutions are designed as per the students demand covering comprehensive, step by step solutions of every problem. Practice these questions and answers to command the concepts, boost confidence and in depth understanding of concepts. Students can find all exercise together using the link provided below.

##### VMC VIQ Scholarship Test

Register for Vidyamandir Intellect Quest. Get Scholarship and Cash Rewards.

##### Pearson | PTE

Register now for PTE & Unlock 10% OFF : Use promo code: 'C360SPL10'. Limited Period Offer! Trusted by 3,500+ universities globally

## Assess NCERT Solutions for Class 10 Maths chapter 3 exercise 3.5

Pair of Linear Equations in Two Variables Class 10 Chapter 3 Excercise: 3.5

$\\x - 3y -3 = 0\\ 3x - 9y -2 = 0$

Given, two equations,

$\\x - 3y -3 = 0.........(1)\\ 3x - 9y -2 = 0........(2)$

Comparing these equations with $a_1x+b_1y+c_1=0\:and\:a_2x+b_2y+c_2=0$ , we get

$\frac{a_1}{a_2}=\frac{1}{3}$

$\frac{b_1}{b_2}=\frac{-3}{-9}=\frac{1}{3}$

$\frac{c_1}{c_2}=\frac{-3}{-2}=\frac{3}{2}$

As we can see,

$\frac{a_1}{a_2}=\frac{b_1}{b_2}\neq\frac{c_1}{c_2}$

Hence, the pair of equations has no solution.

$\\2x + y = 5 \\ 3x + 2y = 8$

Given, two equations,

$\\2x + y = 5.........(1) \\ 3x + 2y = 8..........(2)$

Comparing these equations with $a_1x+b_1y+c_1=0\:and\:a_2x+b_2y+c_2=0$ , we get

$\frac{a_1}{a_2}=\frac{2}{3}$

$\frac{b_1}{b_2}=\frac{1}{2}$

$\frac{c_1}{c_2}=\frac{5}{8}$

As we can see,

$\frac{a_1}{a_2}\neq\frac{b_1}{b_2}\neq\frac{c_1}{c_2}$

Hence, the pair of equations has exactly one solution.

By Cross multiplication method,

$\frac{x}{b_1c_2-b_2c_1}=\frac{y}{c_1a_2-c_2a_1}=\frac{1}{a_1b_2-a_2b_1}$

$\frac{x}{(1)(-8)-(2)(-5)}=\frac{y}{(-5)(3)-(-8)(2)}=\frac{1}{(2)(2)-(3)(1)}$

$\frac{x}{2}=\frac{y}{1}=\frac{1}{1}$

$x=2,\:and\:y=1$

$\\3x -5 y = 20\\ 6x - 10y = 40$

Comparing these equations with $a_1x+b_1y+c_1=0\:and\:a_2x+b_2y+c_2=0$ , we get

$\frac{a_1}{a_2}=\frac{3}{6}=\frac{1}{2}$

$\frac{b_1}{b_2}=\frac{-5}{-10}=\frac{1}{2}$

$\frac{c_1}{c_2}=\frac{20}{40}=\frac{1}{2}$

As we can see,

$\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$

Hence, the pair of equations has infinitely many solutions.

$\\x - 3y -7 = 0\\ 3x -3y -15 =0$

Given the equations,

$\\x - 3y -7 = 0.........(1)\\ 3x -3y -15 =0........(2)$

Comparing these equations with $a_1x+b_1y+c_1=0\:and\:a_2x+b_2y+c_2=0$ , we get

$\frac{a_1}{a_2}=\frac{1}{3}$

$\frac{b_1}{b_2}=\frac{-3}{-3}=1$

$\frac{c_1}{c_2}=\frac{-7}{-15}=\frac{7}{15}$

As we can see,

$\frac{a_1}{a_2}\neq\frac{b_1}{b_2}\neq\frac{c_1}{c_2}$

Hence, the pair of equations has exactly one solution.

By Cross multiplication method,

$\frac{x}{b_1c_2-b_2c_1}=\frac{y}{c_1a_2-c_2a_1}=\frac{1}{a_1b_2-a_2b_1}$

$\frac{x}{(-3)(-15)-(-3)(-7)}=\frac{y}{(-7)(3)-(-15)(1)}=\frac{1}{(1)(-3)-(3)(-3)}$

$\frac{x}{45-21}=\frac{y}{-21+15}=\frac{1}{-3+9}$

$\frac{x}{24}=\frac{y}{-6}=\frac{1}{6}$

$x=\frac{24}{6}=4,\:and\:y=-1$

Given equations,

$\\2x + 3y = 7 \\(a - b) x + (a + b) y = 3a + b - 2$

As we know, the condition for equations $a_1x+b_1y+c_1=0\:and\:a_2x+b_2y+c_2=0$ to have an infinite solution is

$\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$

So, Comparing these equations with, $a_1x+b_1y+c_1=0\:and\:a_2x+b_2y+c_2=0$ , we get

$\frac{2}{a-b}=\frac{3}{a+b}=\frac{7}{3a+b-2}$

From here we get,

$\frac{2}{a-b}=\frac{3}{a+b}$

$\Rightarrow 2(a+b)=3(a-b)$

$\Rightarrow 2a+2b=3a-3b$

$\Rightarrow a-5b=0.........(1)$

Also,

$\frac{2}{a-b}=\frac{7}{3a+b-2}$

$\Rightarrow 2(3a+b-2)=7(a-b)$

$\Rightarrow 6a+2b-4=7a-7b$

$\Rightarrow a-9b+4=0...........(2)$

Now, Subtracting (2) from (1) we get

$\Rightarrow 4b-4=0$

$\Rightarrow b=1$

Substituting this value in (1)

$\Rightarrow a-5(1)=0$

$\Rightarrow a=5$

Hence, $a=5\:and\:b=1$ .

Given, the equations,

$\\3x + y = 1 \\(2k - 1) x + (k - 1) y = 2k + 1$

As we know, the condition for equations $a_1x+b_1y+c_1=0\:and\:a_2x+b_2y+c_2=0$ to have no solution is

$\frac{a_1}{a_2}=\frac{b_1}{b_2}\neq\frac{c_1}{c_2}$

So, Comparing these equations with, $a_1x+b_1y+c_1=0\:and\:a_2x+b_2y+c_2=0$ , we get

$\frac{3}{2k-1}=\frac{1}{k-1}\neq\frac{1}{2k+1}$

From here we get,

$\frac{3}{2k-1}=\frac{1}{k-1}$

$\Rightarrow 3(k-1)=2k-1$

$\Rightarrow 3k-3=2k-1$

$\Rightarrow 3k-2k=3-1$

$\Rightarrow k=2$

Hence, the value of K is 2.

Given the equations

$\\8x + 5y = 9........(1) \\3x + 2y = 4........(2)$

By Substitution Method,

From (1) we have

$y=\frac{9-8x}{5}.........(3)$

Substituting this in (2),

$3x+2\left ( \frac{9-8x}{5} \right )=4$

$\Rightarrow 15x+18-16x=20$

$\Rightarrow -x=20-18$

$\Rightarrow x=-2$

Substituting this in (3)

$y=\frac{9-8x}{5}=\frac{9-8(-2)}{5}=\frac{25}{5}=5$

Hence $x=-2\:and\:y=5$ .

By Cross Multiplication Method

$\frac{x}{b_1c_2-b_2c_1}=\frac{y}{c_1a_2-c_2a_1}=\frac{1}{a_1b_2-a_2b_1}$

$\frac{x}{(5)(-4)-(2)(-9)}=\frac{y}{(3)(-9)-(8)(-4)}=\frac{1}{(8)(2)-(3)(5)}$

$\frac{x}{-20+18}=\frac{y}{32-27}=\frac{1}{16-15}$

$\frac{x}{-2}=\frac{y}{5}=\frac{1}{1}$

$x=-2,\:and\:y=5$

Let the fixed charge be x and the cost of food per day is y,

Now, According to the question

$x+20y=1000.........(1)$

Also

$x+26y=1180.........(2)$

Now subtracting (1) from (2),

$x+26y-x-20y=1180-100$

$\Rightarrow 6y=180$

$\Rightarrow y=30$

Putting this value in (1)

$x+20(30)=1000$

$\Rightarrow x=1000-600$

$\Rightarrow x=400$

Hence, the Fixed charge is Rs 400 and the cost of food per day is Rs 30.

Let numerator of a fraction be x and the denominator is y.

Now, According to the question,

$\frac{x-1}{y}=\frac{1}{3}$

$\Rightarrow 3(x-1)=y$

$\Rightarrow 3x-3=y$

$\Rightarrow 3x-y=3........(1)$

Also,

$\frac{x}{y+8}=\frac{1}{4}$

$\Rightarrow 4x=y+8$

$\Rightarrow 4x-y=8.........(2)$

Now, Subtracting (1) from (2) we get,

$4x-3x=8-3$

$\Rightarrow x=5$

Putting this value in (2) we get,

$4(5)-y=8$

$\Rightarrow y=20-8$

$\Rightarrow y=12$

Hence, the fraction is

$\frac{x}{y}=\frac{5}{12}$ .

Let the number of right answer and wrong answer be x and y respectively

Now, According to the question,

$3x-y=40..........(1)$

And

$\\4x-2y=50\\\Rightarrow 2x-y=25..........(2)$

Now, subtracting (2) from (1) we get,

$x=40-25$

$x=15$

Putting this value in (1)

$3(15)-y=40$

$\Rightarrow y=45-40$

$\Rightarrow y=5$

Hence the total number of question is $x+y=15+5=20.$

Let the speed of the first car is x and the speed of the second car is y.

Let's solve this problem by using relative motion concept,

the relative speed when they are going in the same direction= x - y

the relative speed when they are going in the opposite direction= x + y

The given relative distance between them = 100 km.

Now, As we know,

Relative distance = Relative speed * time .

So, According to the question,

$5\times(x-y)=100$

$\Rightarrow 5x-5y=100$

$\Rightarrow x-y=20.........(1)$

Also,

$1(x+y)=100$

$\Rightarrow x+y=100........(2)$

Now Adding (1) and (2) we get

$2x=120$

$\Rightarrow x=60$

putting this in (1)

$60-y=20$

$\Rightarrow y=60-20$

$\Rightarrow y=40$

Hence the speeds of the cars are 40 km/hour and 60 km/hour.

Let $l$be the length of the rectangle and $b$ be the width,

Now, According to the question,

$(l-5)(b+3)=lb-9$

$\Rightarrow lb+3l-5b-15=lb-9$

$\Rightarrow 3l-5b-6=0..........(1)$

Also,

$(l+3)(b+2)=lb+67$

$\Rightarrow lb+2l+3b+6=lb+67$

$\Rightarrow 2l+3b-61=0..........(2)$

By Cross multiplication method,

$\frac{x}{b_1c_2-b_2c_1}=\frac{y}{c_1a_2-c_2a_1}=\frac{1}{a_1b_2-a_2b_1}$

$\frac{x}{(-5)(-61)-(3)(-6)}=\frac{y}{(-6)(2)-(-61)(3)}=\frac{1}{(3)(3)-(2)(-5)}$

$\frac{x}{305+18}=\frac{y}{-12+183}=\frac{1}{9+10}$

$\frac{x}{323}=\frac{y}{171}=\frac{1}{19}$

$x=17,\:and\:y=9$

Hence the length and width of the rectangle are 17 units and 9 units respectively.

## More About NCERT Solutions for Class 10 Maths Exercise 3.5

Class 10 Maths chapter 3 exercise 3.5: The questions in exercise 3.5 Class 10 Maths, broadly consist of four types of questions. In question one we have to identify whether the pair of equations has a unique solution, no solution, or infinitely many solutions. In question two we have to find the extra variable a and b they can be found with the help of the condition which is applied to the question of equation three is a direct question in which we have to solve both substitution method and cross multiplication method. in question four there are word problems based on real-life application. Also students can get access of Pair of linear equations in two variables notes to revise all the concepts discussed in this chapter.

## Benefits of NCERT Solutions for Class 10 Maths Exercise 3.5

• Chapter 3 exercise 3.5 in Class 10 Maths includes a wide range of problems regarding solving a pair of linear equations using the cross multiplication approach.
• NCERT Class 10 Maths chapter 3 exercise 3.5, will be helpful in solving NCERT Class 11 chapter 6- linear inequalities
• Exercise 3.5 in Class 10 Maths involves calculating a pair of linear equations using a non-graphical approach, namely the cross multiplication method, which is a key idea in the chapter.

Also see-

##### JEE Main Important Mathematics Formulas

As per latest 2024 syllabus. Maths formulas, equations, & theorems of class 11 & 12th chapters

## Subject Wise NCERT Exemplar Solutions

1. How many ways are there to solve a pair of linear equations with two variables according to NCERT solutions for Class 10 Maths 1 exercise 3.5 ?
1. Graphical method

2. Algebraic method

2. How many types of algebraic methods are there to solve a pair of linear equations with two variables according to NCERT solutions for Class 10 Maths 1 exercise 3.5 ?

According to NCERT solutions for Class 10 Maths 1 exercise 3.5 there are three ways to solve a pair of linear equations with a single variable by the algebraic method.

3. Name all the types of algebraic methods to solve a pair of linear equations with two variables according to NCERT solutions for Class 10 Maths 1 exercise 3.5 ?
1. Substitution method

2. Elimination method

3. Cross-multiplication Method

4. Sometimes we can get the statement with no variable. What does that mean ?

Statement with no variable means all the values for that variable is true we can say this when the variable itself get eliminated from the equation leaving the same constant at both sides

5. What is the type of linear equation when we get a statement with no variable?

The pair of linear equations can be said to have an infinite number of solutions.

6. When we have defined the value for both the variables, what is the type of linear equation ?

We might remark that the solutions to the pair of linear equations are inconsistent.

7. What is the total number of solved examples based on the cross multiplication method prior to Exercise 3.5 Class 10 Maths?

There are mainly 3 questions that are solved before the Class 10 Maths chapter 3 exercise 3.5

8. How many questions are there in the Exercise 3.5 Class 10 Maths?

There are 4  questions and question 1 has 4 subparts, question 2 has 2 subparts and question 4  has  5  subparts.

## Upcoming School Exams

#### National Means Cum-Merit Scholarship

Application Date:05 August,2024 - 20 September,2024

#### National Rural Talent Scholarship Examination

Application Date:05 September,2024 - 20 September,2024

#### National Institute of Open Schooling 10th examination

Admit Card Date:13 September,2024 - 07 October,2024

#### National Institute of Open Schooling 12th Examination

Admit Card Date:13 September,2024 - 07 October,2024

#### Nagaland Board of Secondary Education 12th Examination

Application Date:17 September,2024 - 30 September,2024

Edx
1113 courses
Coursera
804 courses
Udemy
394 courses
Futurelearn
222 courses
IBM
85 courses

## Explore Top Universities Across Globe

University of Essex, Colchester
Wivenhoe Park Colchester CO4 3SQ
University College London, London
Gower Street, London, WC1E 6BT
The University of Edinburgh, Edinburgh
Old College, South Bridge, Edinburgh, Post Code EH8 9YL
University of Bristol, Bristol
Beacon House, Queens Road, Bristol, BS8 1QU
University of Nottingham, Nottingham
University Park, Nottingham NG7 2RD

### Questions related to CBSE Class 10th

Have a question related to CBSE Class 10th ?

Hello

The Sadhu Ashram in Aligarh is located in Chhalesar . The ashram is open every day of the week, except for Thursdays . On Mondays, Wednesdays, and Saturdays, it's open from 8:00 a.m. to 7:30 p.m., while on Tuesdays and Fridays, it's open from 7:30 a.m. to 7:30 p.m. and 7:30 a.m. to 6:00 a.m., respectively . Sundays have varying hours from 7:00 a.m. to 8:30 p.m. . You can find it at Chhalesar, Aligarh - 202127 .

Yes, scoring above 80% in ICSE Class 10 exams typically meets the requirements to get into the Commerce stream in Class 11th under the CBSE board . Admission criteria can vary between schools, so it is advisable to check the specific requirements of the intended CBSE school. Generally, a good academic record with a score above 80% in ICSE 10th result is considered strong for such transitions.

hello Zaid,

Yes, you can apply for 12th grade as a private candidate .You will need to follow the registration process and fulfill the eligibility criteria set by CBSE for private candidates.If you haven't given the 11th grade exam ,you would be able to appear for the 12th exam directly without having passed 11th grade. you will need to give certain tests in the school you are getting addmission to prove your eligibilty.

best of luck!

According to cbse norms candidates who have completed class 10th, class 11th, have a gap year or have failed class 12th can appear for admission in 12th class.for admission in cbse board you need to clear your 11th class first and you must have studied from CBSE board or any other recognized and equivalent board/school.

You are not eligible for cbse board but you can still do 12th from nios which allow candidates to take admission in 12th class as a private student without completing 11th.

A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

 Option 1) Option 2) Option 3) Option 4)

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

 Option 1) 2.45×10−3 kg Option 2)  6.45×10−3 kg Option 3)  9.89×10−3 kg Option 4) 12.89×10−3 kg

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

 Option 1) Option 2) Option 3) Option 4)

A particle is projected at 600   to the horizontal with a kinetic energy . The kinetic energy at the highest point

 Option 1) Option 2) Option 3) Option 4)

In the reaction,

 Option 1)   at STP  is produced for every mole   consumed Option 2)   is consumed for ever      produced Option 3) is produced regardless of temperature and pressure for every mole Al that reacts Option 4) at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, will contain 0.25 mole of oxygen atoms?

 Option 1) 0.02 Option 2) 3.125 × 10-2 Option 3) 1.25 × 10-2 Option 4) 2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

 Option 1) decrease twice Option 2) increase two fold Option 3) remain unchanged Option 4) be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

 Option 1) Molality Option 2) Weight fraction of solute Option 3) Fraction of solute present in water Option 4) Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

 Option 1) twice that in 60 g carbon Option 2) 6.023 × 1022 Option 3) half that in 8 g He Option 4) 558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

 Option 1) less than 3 Option 2) more than 3 but less than 6 Option 3) more than 6 but less than 9 Option 4) more than 9