NCERT Solutions for Class 10 Maths Exercise 3.3 of Chapter 3 Pair of Linear Equations in Two Variable - Future Study Point

NCERT Solutions for Class 10 Maths Exercise 3.3 of Chapter 3 Pair of Linear Equations in Two Variable

class 10 maths exercise 3 linear equation in two variables

NCERT Solutions for Class 10 Maths Exercise 3.3 of Chapter 3 Pair of Linear Equations in Two Variable

NCERT Solutions for Class 10 Maths Exercise 3.3 of Chapter 3 Pair of Linear Equations in Two Variables are created for boosting the preparation of class 10 maths students for the exams and in doing their homework and class 10 maths assignments on chapter 3 exercise 3.3.NCERT Solutions for Class 10 Maths Exercise 3.3 of Chapter 3 Pair of Linear Equations in Two Variables are solved by a step-by-step method. These NCERT Solutions for Class 10 Maths Exercise 3.3 of Chapter 3 Pair of Linear Equations in Two Variables are the most important study material for clearing the concept of linear equations in two variables.

class 10 maths exercise 3 linear equation in two variables

See the video for Class 10 Maths Exercise 3.2 Linear Equations in Two Variable

NCERT Solutions for Class 10 Maths Exercise 3.3 of Chapter 3 Linear Equations in Two Variable

Q1.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

Ans. The given equations are x +y = 14….(i) and x -y = 4……(ii)

Taking equation (i)

x +y =14

Solving it for y

y = 14 -x

Putting the value of y in equation (ii)

x -(14 -x)= 4

x -14 +x = 4

2x = 4 +14 =18

x = 18/2 = 9

Putting the value of x =9 in equation (i)

y = 14 -9 = 5

Hence the solutions of the given pair of linear equations are x =9 and y =5

(ii)The given equations are s -t = 3……(i) and (s/3) +(t/2) =6 ….(ii)

Taking equation (i)

s -t = 3

Solving it for s

s = 3+t

Putting the value of s in equation (ii)

(s/3) +(t/2) =6

(3+t)/3 +(t/2) =6

(6+2t +3t)/6 =6

6+5t =36

5t =36-6 =30

t = 30/5 = 6

Putting the value of t =6 in equation (i)

s = 3+6 = 9

Hence the solutions of the given pair of linear equations are x =9 and y =5

Q2.Solve 2x + 3y = 11 and 2x – 4y = -24 and hence find the value of ‘m’ for which y = mx + 3.

Ans. The given equations are 2x + 3y = 11 ….(i) and 2x – 4y = -24……(ii)

Taking equation (i)

2x + 3y = 11

Solving it for y

y = (11 -2x )/3

Putting the value of y in equation (ii)

2x – 4(11 -2x)/3 = – 24

(6x -44 +8x)/3 = -24

6x -44+8x = -72

14x = -72 +44 =-28

x = -28/14= -2

Putting the value of x =-2 in equation (i)

2×-2+ 3y = 11

-4 + 3y =11

3y = 11 +4 =15

y = 15/3 = 5

Hence the solutions of the given pair of linear equations are x =-2 and y =5

Putting the value of x =-2 and y =5 in the given relation y =mx +3

5 = -2m +3

2m = 3 -5 = -2

m = -1

Hence the value of m is -1

Q3.Form the pair of linear equations for the following problems and find their solution by substitution method.

(i) The difference between two numbers is 26 and one number is three times the other.Find them.

Ans.Let the two numbers are x and y

According to the first condition

The difference between the numbers is =26

x – y =26…..(i)

According to the second condition

One number = 3× Other number

x = 3y……(ii)

Putting the value of x from equation (ii) to in equation (i)

3y – y =26

2y = 26⇒ y = 13

Putting the value of y  in equation (i)

x – 13 = 26

x = 26+ 13 = 39

Hence the numbers are 13 and 39

(ii) Larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.

Ans. Let the smaller angle is x and larger angle is y

According to first condition two angles are complimentary to each other

x + y = 180°……(i)

According to second condition one angle is smaller by 18 degree than the larger one

x = y -18°…….(ii)

Substituting the value of x from (ii) equation to equation (i)

y -18° + y = 180°

2y = 180° +18° = 198°

y = 99°

Putting the value of y in equation (i) in the equation

x + 99°= 180°

x = 180°- 99° = 81°

Hence the solutions of the given pair of linear equations is x = 81°and y = 99°

(iii)The coach of a cricket team buys 7 bats and 6 balls for Rs.3800. Later, she buys 3 bats and 5 balls for Rs.1750. Find the cost of each bat and each ball.

Ans. Let the cost of one bat is x and the cost of one ball is y

According to first condition

7x + 6y = 3800 …..(i)

According to second condition

3x + 5y = 1750 …..(ii)

Taking equation (i)

7x + 6y = 3800

Solving it for y

y = (3800 -7x )/6

Putting the value of y in equation (ii)

3x + 5(3800 -7x )/6 =1750

(18x + 19000-35x)/6 = 1750

18x +19000 -35x = 10500

-17x = 10500 -19000 = -8500

x = 500

Putting the value of x in equation (i)

7×500 +6y = 3800

6y = 3800 -3500 =300

y = 50

Hence the cost of one bat is Rs 500 and the cost of one ball is Rs 50

Ans(iv)The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is Rs 105 and for a journey of 15 km, the charge paid is Rs 155. What are the fixed charges and the charge per km? How much does a person have to pay for travelling a distance of 25 km?

Ans. Let the fixed charge of the taxi is x and the charge of per km is y

According to the first condition

x + 10y =105…..(i)

According to the second condition

x + 15y =155…..(ii)

Taking equation (i)

x + 10y =105

Solving it for y

y = (105 -x )/10

Putting the value of y in equation (ii)

x + 15 (105 -x )/10=155

(10x +1575-15x )/10 = 155

-5x +1575 = 1550

-5x = 1550 -1575 =-25

x = 5

Putting the value of x in equation (i)

5 + 10y =105

10y = 105 – 5 = 100

y = 10

Hence the value of the fixed charge is Rs 5 and the per km charge of the taxi is Rs 10

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

Exercise 3.2

Exercise 3.3

Exercise 3.4

Exercise 3.5

Exercise 3.6

Exercise 3.7

Class 10 maths NCERT solutions of important questions of chapter 3 Pair of Linear Equations

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