NCERT Solutions For Class 10 Math Chapter – 3 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 you will be” Represent this situation algebraically and graphically.
Solution:-
A.T.Q
(x-7) = 7y – 49
X = 7y – 42
Putting y = 5 , 6 and 7 we get
X = 7 x 5 – 42 = 35 – 42 = -7
X = 7 x 6 – 42 = 42 – 42 = 0
X = 7 x 7 – 42 = 49 – 42 = 7
| x | -7 | 0 | 7 |
| y | 5 | 6 | 7 |
A.T.Q
(x + 3) = 3(y + 3)
X = 3y + 6
Putting , y = -2 ,-1 and 0 , we get
X = 6
| x | 0 | 3 | 6 |
| y | -2 | -1 | 0 |
Algebraic representation
From equation (I) and (ii)
X – 7y = -42
X – 3y = 6
Graphical representation.
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.
Solution:-
A.T.Q
3x + 6y = 3900 (i)
Dividing equation by 3 , we get
X + 2y = 1300
Subtracting 2y both side we get
X = 1300 – 2y
Putting y = -1300 , 0 and 1300 we get
x = 1300 – 2(-1300)
= 1300 + 2600
= 3900
X = 1300 – 2(0)
= 1300 – 0
= 1300
| x | 3900 | 1300 | -1300 |
| y | -1300 | 0 | 1300 |
X + 2y = 1300
X = 1300 – 2y
X = 1300 – 2(-1300)
= 1300 + 2600
= 3900
X = 1300 – 2(0)
= 1300 – 2(1300)
= -1300
| x | 3900 | 1300 | -1300 |
| y | -1300 | 0 | 1300 |
Algebraic representation
3x + 6y = 3900
X + 2y = 1300
Graphical representation,
Q3. 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:-
2x + y = 160 (i)
2x = 160 – y
X = (160 – y)/2
X = 0
| x | 80 | 40 | 0 |
| y | 0 | 80 | 160 |
4x + 2y = 300 (ii)
Dividing by 2 we get
2x + y = 150
Y = 150 – 2x
Y = 150 – 2 x 0
Y = 150
Y = 150 – 2 x 50 = 50
Y = 150 – 2 x (100) = – 50
| x | 0 | 50 | 100 |
| y | 150 | 50 | -50 |
Algebraic representation,
2x + y = 160 (i)
4x + 2y = 300 (ii)
Graphical representation,
