# MATH IA- Filling up the petrol tank ARWA and BAO

## Filling up the Petrol Tank

In this world money is one of the most important things, and hence we try to save as much money as possible. Even drivers try to save as much money as possible. Some places sell fuel for lower prices than others. Is it more economical to travel a larger distance to get cheaper fuel or is it more economic to travel a shorter distance but to buy relatively expensive fuel? Let’s investigate!

There are two people Arwa and Bao, who share the same driving route r of 20km.

Let’s assume:

- The two are neighbors and work at the same place.
- They drive to work every day.
- They buy fuel in the morning, before going to work, every second Monday, as they are busy people.
- They used their car only to go to work and they work only from Monday to Friday.
- They are cautious people and so they always have at least 5litres petrol as reserve. If it goes below that they will go and refuel at that instant.
- They have enough petrol to reach the gas station initially. After reaching the gas station they buy exactly the fuel required to drive to work the next ten days and to drive to and from the station (if the gas station is not on the route).

Since Arwa and Bao work only Monday to Friday and use their car only to go to work, we can say that they fuel up every 10 working days (number of working days after which, they refuel is w working days).

The fuel efficiency f1 of Arwa’s Toyota Etios is 18km/l where as the fuel efficiency f2 of Bao’s i10 is 20km/l. (Fuel Efficiency of a car is f km/l in general). The capacity of fuel tank t of both cars is equal to 35 liters.

Therefore number of kilometers Arwa travels after which he has to refuel

=(t-5) ×f1

=30l ×18km/l

=540km (5l is for reserve)

Therefore number of kilometers Bao travels after which he has to refuel

=(t-5) ×f2

=600km (5l is for reserve)

Arwa fills up her tank at her station along her normal route for US$ p1 per liter. On the other hand Bao drives an extra d kilometers out of his normal route to fill up his vehicle tank for US$ p2 per liter, where p1 is less than p2.

Note: The diagram is not drawn to scale

A diagram has been used so that it is easier to see the route taken by Arwa and Bao.

Since Arwa and Bao are being compared, they both should be able to get to d. Since Arwa can travel less distance before having to refuel, his maximum distance of traveling before having to fuel up will determine the maximum value of d.

∴2× distance of detour+2× distance to work × number of working days after which refueling is done = distance Arwa travels after which he has to refuel.

2dMAX+2rw=600km

d MAX + rw =300km

d MAX=300km-rw

d MAX= 300km- (20km)(10)

d MAX=100km

If d=10km, p1=1.00$ and p2=0.98$ Let’s try to find the cost to Arwa and Bao each time they go to refuel.

Cost to Arwa= [(number of working days)(distance driven per day)/fuel

Efficiency](cost of fuel)

= [(w)(2r)/f1](p1)

= (10days× 2×20kmday-1 /18km liter-1)(1.00$/per liter)

≈ 22.22$ (2 decimal places)

Cost to Bao = [(number of kilometers driven)/(fuel efficiency) ] × (cost of fuel)

=(2 × distance of detour+10 working days × number of

kilometers driven to work each day)(p2)/(f2)