Ncert Solutions For Class 10 Maths Ex 3.1

Ncert Solutions For Class 10 Maths Chapter 3 Ex 3.1

Question 1:

Astitva tells his daughter, “Seven years ago I was seven times as old as you were then and also three years from now, I shall be three times as old as you will be.”  Represent this situation algebraically and graphically.

Solution:

Let the present age of Astitva be ‘x’.

And, the present age of his daughter be ‘y’.

Seven years ago,

Age of Astitva = x-7

Age of his daughter =y-7

According to the question,

x7=7(y7) x7=7y49

x7y=42         ………………………(i)

 

Three years from now,

Age of Astitva will be ‘x+3’

Age of his daughter will be ‘y+3’

According to the question,

x+3=3(y+3) x+3=3y+9

x3y=6       …………………(ii)

 

Subtracting equation (i) from equation (ii) we have

(x3y)(x7y)=6(42) 3y+7y=6+42 4y=48 y=12

 

The algebraic equation is represented by

x7y=42 x3y=6

For x7y=42

x=42+7y

 

The solution table is

 

 

X -7 0 7
Y 5 6 7

 

For  x3y=6   or     x=6+3y

The solution table is

 

X 6 4 0
Y 0 -1 -2

 

The graphical representation is-

1

 

Question 2:

A cricket team coach purchases 3 bats and 3 balls for Rs 3900. Later on, he buys 1 bat and 2 more balls of the same kind for Rs 1200. Represent the situation algebraically and geometrically.

Solution:

Let us assume that the cost of a bat be ‘Rs x’

And,the cost of a ball be ‘Rs y’

According to the question, the algebraic representation is

 

3x+6y=3900 x+2y=1300

For 3x+6y=3900

x=39006y3

 

The solution table is

 

x 300 100 -100
y 500 600 700

 

For,

x+2y=1300 x=13002y

 

The solution table is

 

x 300 100 -100
y 500 600 700

 

The graphical representation is as follows.

2

 

Question 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 given situation algebraically and geometrically.

Solution:

Let the cost of 1 kg of apples be ‘Rs x’

And, cost of 1 kg of grapes be ‘Rs y’

According to the question, the algebraic representation is

2x+y=160 4x+2y=300

For,  2x+y=160

y=1602x

 

The solution table is

 

X 50 60 70
Y 60 40 20

 

For 4x+2y=300,

y=3004x2

 

The solution table is

 

x 70 80 75
y 10 -10 0

The graphical representation is as follows.

3

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