NCERT Solutions for Class 10 Maths Exercise 3.1 of Chapter 3 Linear Equations in Two Variables - Future Study Point

NCERT Solutions for Class 10 Maths Exercise 3.1 of Chapter 3 Linear Equations in Two Variables

class 10 maths ex 3.1

NCERT Solutions for Class 10 Maths Exercise 3.1 of Chapter 3 Linear Equations in Two Variables

NCERT Solutions for Class 10 Maths Exercise 3.1 of Chapter 3 Linear Equations in Two Variables will give you an idea of solving two linear equations in two variables. In the questions of the exercise of chapter 3 -Linear Equation In two variables two conditions of two variables would be given and you are required to build two relations between the two variables in the form of two equations and then solve them for two or more than 2  values and draw the graphs wherever both graphs intersect each other, the corresponding values of x and y of the intersecting point is the solution of the linear equation.

class 10 maths ex 3.1

NCERT Solutions for Class 10 Maths Exercise 3.1 of Chapter 3 Linear Equations in Two Variables

Q1.Aftab tells his daughter,”seven years ago I was seven times as old as you were then. Also three years from now,I shall be three times as old as you will be.”(Isn’t this be interesting?)Represent this situation algebraically and graphically.

Ans. Let the age of Aftab is x and of his daughter is y

According to first condition, seven years ago the age of Aftab = x – 7 and of his daughter =y-7,further it is given that

Seven years back the age of Afteb = 7× Seven years back his daughter’s age

x – 7 = 7(y-7)

x -7 = 7y -49

x – 7y = -49 + 7

x – 7y = -42……(i)

According to second condition,after 3 years the age of Aftab is = x + 3 and of his daughter’s age = y +3,furher it is given that

After 3 years the age of Aftab = 3× His daughter’s age after 3 years

x + 3 = 3(y + 3)

x + 3 = 3y + 9

x – 3y = 6……(ii)

Solutions of the equation x – 7y = -42

            x       0          7       -7
             y       6         7       5

Solutions of the equation x – 3y = 6

          x       0       6      -3
        y       -2      0      -3

Graphical representation of both equations

Q1 exercise 3.1 class 10 maths

 

Q2.The coach of a cricket team buys 3 bats and 6 balls for ₹ 3900. Later, she buys another bat and 3 more balls of the same kind for ₹1300. Represent this situation algebraically and geometrically.

Ans. Let the cost of 1 bat is x and of 1 ball is y

According to fist condition

3 × cost of 1 bat + 6× cost of 1 ball = Rs 3900

3x + 6y = 3900…..(i)

According to second condition

Cost of 1 bat+ 3× cost of 1 ball = Rs 1300

x + 3y = 1300….(ii)

Solutions of equation (i) 3x + 6y = 3900

       x        100       -100        300
       y        600       700        500

Solutions of equation (ii) x + 3y = 1300

           x        100      400     -200
         y       400     300      500

Graphical representation of both equations

Q2. exercise 3.1 class 10 maths

 

Q3.The cost of 2 kg of apples and 1 kg of grapes on a day was found to be ₹160.  After a month, the cost of 4 kg of apples and 2 kg of grapes is ₹300. Represent the situation algebraically and geometrically.

Ans. Let the cost of 1 kg apple is = x and the cost of 1 kg grapes is= y

According to first condition of the question

2×cost of 1 kg apple +cost of 1 kg grapes = Rs 160

2x + y = 160…..(i)

According to second condition of the question

4×cost of 1 kg apple +2×cost of 1 kg grapes = Rs 300

4x + 2y = 300…..(ii)

Solutions of the first equation 2x + y = 160

      x       80         60      40
     y       0         40      80

 

Solutions of the second equation 4x + 2y = 300

           x       60      40     20
          y      30       70    110

Graphical representation of both equations

 

Q3 exercise 3.1 class 10 maths

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

Exercise 3.2

Exercise 3.3

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