148-NCERT-10-3-Pair of Linear Equations in Two Variables - Ex-3.1

NCERT

10th Mathematics

Exercise 3.1

Topic: 3 Pair of Linear Equations in Two Variables

Click here for ⇨ NCERT-10-2-Polynomials - Ex-2.4

EXERCISE 3.1

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 interesting?) Represent this situation algebraically and graphically. 

Explanation:

1) In such cases, always consider their present age.

2) In age-related problems, ago means subtracting those years from both ages.

3) In the case of hence or later any such words will tell us to add those ages in both. 

Solution:

1) Let Aftab's present age be x and his daughter's age be y.

2) 7 years ago their ages were (x - 7) and (y - 7).

3) According to the first relation given in the problem,

(x - 7) = 7(y - 7)

(x - 7) = 7y - 49

x-7+49 = 7y

7y = x+42

y = (x+42)/7  ---------------------- equation 1

4) 3 years from now, their ages will be (x + 3) and (y + 3). 

5) According to the relation given in the problem,

(x + 3) = 3(y + 3)

(x + 3) = 3y + 9

x+3-9 = 3y

3y = x-6

y = (x-6)/3  ---------------------- equation 2

6) Now, we will represent these equations graphically.

a) We will take 2 points for y = (x + 42)/7.

b) We will take 2 points for y = (x - 6)/3.

7) The graphical representation will be as follows.

8) So their present ages are 42 years and 12 years.

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

Explanation:

1) Let x and y be the costs of the bat and balls.

2) Apply the given conditions and frame the equation.

3) We will get two equations from the above two conditions, then solve these equations to get the values of x and y. 

Solution:

1) Let the cost of the bat be Rs x and the cost of the ball be Rs y.

2) As 3 bats and 6 balls cost Rs 3900, we have,

3 x + 6 y = 3900

3 (x + 2 y) = 3900

(x + 2 y) = 1300

2 y = (1300 - x)

   y = (1300 - x)/2 ------------ equation 1

3) As 1 bat and 3 balls cost Rs 1300, we have,

x + 3 y = 1300

      3 y = (1300 - x)

 y = (1300 - x)/3 ------------ equation 2

4) Now, we will represent these equations graphically.

a) We will take 2 points for y = (1300 - x)/2.     

b) We will take 2 points for y = (1300 - x)/3.

5) The graphical representation will be as follows.

6) The point of intersection is (1300,0).

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

Explanation:

1) Here, let x and y be the costs of apples and grapes.

2) Apply the given conditions and frame the equation.

3) We will get two equations from the above two conditions, then solve these equations to get the values of x and y. 

Solution:

1) Let the cost of 1 kg of apple be Rs x and the cost of 1 kg of grapes be Rs y.

2) As 2 kg of apple and 1 kg of grapes cost Rs 160, we have,

2 x + y = 160

 y = 160 - 2x ------------ equation 1

3) As 4 kg of apple and 2 kg of grapes cost Rs 300, we have,

4 x + 2 y = 300

2 x + y = 300/2

2 x + y = 150 

 y = 150 - 2x ------------ equation 2

4) Now, we will represent these equations graphically.

a) We will take 2 points for y = (160 - 2x).

b) We will take 2 points for y = (150 - 2x).

5) The graphical representation will be as follows.

6) Here the lines are parallel, so there will not be any point common.

Click here for ⇨ NCERT-10-3-Pair of Linear Equations in Two Variables - Ex-3.2

ANIL SATPUTE