NCERT Solutions Maths Class 10 Chapter 3

NCERT Solutions Maths Class 10 Chapter 3

 

NCERT Solutions Maths Class 10 Exercise 3.1

 

1. 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.” Represent this situation algebraically and graphically.

Solution: Let us suppose Aftab’s current age = x years and his daughter’s current age = y years.

Seven years ago, Aftab’s age = (x – 7) years and daughter’s age = (y – 7) years

As per question,

x – 7 = 7(y – 7)    
Or, x – 7 = 7y – 49   
Or, x = 7y – 49 + 7      
Or, x = 7y – 42      …… (1)

This equation gives the following values of x and y.

x

7

0

7

y

5

6

7

Now, draw a line joining points (7, 5), (0, 6) and (7, 7).

Three years from now,

Aftab’s age = (x + 3) years and his daughter’s age = (y + 3) years

As per question,

x + 3 = 3(y + 3)       
Or, x + 3 = 3y + 9      
Or, x = 3y + 9 – 3                
Or, x = 3y + 6      …… (2)

This equation gives the following values of x and y.

x

3

0

3

y

1

2

3

Now, draw a line joining points (3, 1), (0, 2) and (3, 3).

The following graph is plotted for the given pair of linear equations.

 


Since both the straight lines meet at a point (42, 12), therefore, x = 42 and y = 12.

Hence, Aftab’s current age is 42 years and his daughter’s current age is 12 years.


2. 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.

Solution: Let us suppose that price of one bat = Rs x and price of one ball = Rs y.

According to first condition,

3x + 6y = 3900         
x + 2y = 1300       .……..(1)

This equation will give the following values for x and y:

x

1400

1300

1200

1100

y

-50

0

50

100

 

According to second condition,

x + 3y = 1300        ……….(2)

This equation will give the following values for x and y:

x

1450

1300

1150

1000

y

-50

0

50

100

 The following graph is plotted for the given pair of linear equations.

 


Since both the straight lines meet at a point (1300, 0), therefore, x = 1300 and y = 0.

Thus, price of one bat is Rs 1300 and price of one ball is Rs 0.

It means a ball is free with a bat.

 

3. The cost of 2 kg of apples and 1 kg of grapes on 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.

Solution: Let us suppose that the cost of 1 kg of apple = Rs x and the cost of 1 kg of grapes = Rs y.

According to the first condition,

2x + y = 160          ……..(1)

This equation gives the following values for x and y:

x

70

60

50

40

y

20

40

60

80

 

According to the second condition,

4x + 2y = 300       
2x + y = 150       ……..(2)

This equation gives the following values for x and y:

x

65

55

45

35

y

20

40

60

80

 The following graph is plotted for the given pair of linear equations.

 


Since we get parallel lines so there will be no solution for this pair of linear equations.

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