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

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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.

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

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.

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

Q1(i) Which of the following pairs of linear equations has a unique solution, no solution, or infinitely many solutions? In case there is a unique solution, find it by using a cross multiplication method.

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

Answer:

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.

Q1(ii) Which of the following pairs of linear equations has a unique solution, no solution, or infinitely many solutions? In case there is a unique solution, find it by using a cross multiplication method.

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

Answer:

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,

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

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

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

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

Q1(iii) Which of the following pairs of linear equations has a unique solution, no solution, or infinitely many solutions? In case there is a unique solution, find it by using a cross multiplication method.

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

Answer:

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.

Q1(iv) Which of the following pairs of linear equations has a unique solution, no solution, or infinitely many solutions? In case there is a unique solution, find it by using a cross multiplication method.

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

Answer:

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,

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

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

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

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

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

Q2 (i) For which values of a and b does the following pair of linear equations have an infinite number of solutions? $\\2x + 3y = 7 \\(a - b) x + (a + b) y = 3a + b - 2$

Answer:

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

Q2 (ii) For which value of k will the following pair of linear equations have no solution? $\\3x + y = 1 \\(2k - 1) x + (k - 1) y = 2k + 1$

Answer:

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.

Q3 Solve the following pair of linear equations by the substitution and cross-multiplication methods :
$\\8x + 5y = 9 \\3x + 2y = 4$

Answer:

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

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

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

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

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

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

Q4(i) Form the pair of linear equations in the following problems and find their solutions (if they exist) by any algebraic method : 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 A takes food for 20 days she has to pay Rs 1000 as hostel charges whereas a student B, who takes food for 26 days, pays Rs 1180 as hostel charges. Find the fixed charges and the cost of food per day.

Answer:

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.

Q4(ii) Form the pair of linear equations in the following problems and find their solutions (if they exist) by any algebraic method : A fraction becomes $\frac{1}{3}$ when 1 is subtracted from the numerator and it becomes $\frac{1}{4}$ when 8 is added to its denominator. Find the fraction.

Answer:

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

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

Q4(iii) Form the pair of linear equations in the following problems and find their solutions (if they exist) by any algebraic method : Yash scored 40 marks in a test, getting 3 marks for each right answer and losing 1 mark for each wrong answer. Had 4 marks been awarded for each correct answer and 2 marks been deducted for each incorrect answer, then Yash would have scored 50 marks. How many questions were there in the test?

Answer:

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 $x+y=15+5=20.$

Q4(iv) Form the pair of linear equations in the following problems and find their solutions (if they exist) by any algebraic method : Places A and B are 100 km apart on a highway. One car starts from A and another from B at the same time. If the cars travel in the same direction at different speeds, they meet in 5 hours. If they travel towards each other, they meet in 1 hour. What are the speeds of the two cars?

Answer:

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.

Q4(v) Form the pair of linear equations in the following problems and find their solutions (if they exist) by any algebraic method : (v) The area of a rectangle gets reduced by 9 square units if its length is reduced by 5 units and breadth is increased by 3 units. If we increase the length by 3 units and the breadth by 2 units, the area increases by 67 square units. Find the dimensions of the rectangle.

Answer:

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.

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-

NCERT Solutions of Class 10 Subject Wise

Subject Wise NCERT Exemplar Solutions

Frequently Asked Questions (FAQs)

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 ?

Graphical method
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 ?

Substitution method
Elimination method
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.

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show the previous year exams ?

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as an icse student in class 10 is above 80 percent good enough to get into commerce in 11th cbse

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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.

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Option 1) $11.2\, L\, H_{2(g)}$ at STP is produced for every mole $HCL_{(aq)}$ consumed	Option 2) $6L\, HCl_{(aq)}$ is consumed for ever $3L\, H_{2(g)}$ produced
Option 3) $33.6 L\, H_{2(g)}$ is produced regardless of temperature and pressure for every mole Al that reacts	Option 4) $67.2\, L\, H_{2(g)}$ at STP is produced for every mole Al that reacts .

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

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.

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

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

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

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

NCERT Solutions For Class 10 Maths Chapter 3 Exercise 3.5

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