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