The taxi fare in a city is as follows: For the first kilometre, the fare is 8 and for the
subsequent distance it is 5 per km. Taking the distance covered as x km and total
fare as y .write a linear equation for this information, and draw its graph.
Answers
Let Total fare of the taxi = y
And let Total distance covered = x
It is given that - Fair for the subsequent distance after 1st kilometer = Rs 5
Fair for 1st kilometer = Rs 8
Total taxi fare = y
Fare for the 1st km = Rs. Remaining distance = (x – 1) km
∴ Fare for (x – 1) km = Rs. 5 × (x – 1) km
Total taxi fare = Rs. 8 + Rs. 5(x – 1)
∴ According to the condition,
y = 8 + 5(x – 1)
y = 8 + 5x – 5
y = 5x + 3
Graph: We have y = 5x + 3
When × = 0, y = 5(0) + 3
⇒ y = 3
When x = –1, y = 5(–1) + 3
⇒ y = –2
Solution:
Given,
Total distance covered = x
Total fare = y
Fare for the first kilometer = 8 per km
Fare after the first 1km = 5 per km
If x is the total distance, then the distance after one km = (x-1)km
i.e., Fare after the first km = 5(x-1)
According to the question,
The total fare = Fare of first km+ fare after the first km
y = 8+5(x-1)
y = 8+5(x-1)
y = 8+5x – 5
y = 5x+3
Solving the equation,
When x = 0,
y = 5x+3
y = 5×0+3
y = 3
When y = 0,
y = 5x+3
o = 5x+3
5x = -3
x = -3/5
The points to be plotted are (0, 3) and (-3/5, 0).