The efficiency of a car is the measure of how far can it travel in one liter of gas.
Thus the efficiency can be calculated as,
![\eta=\frac{\text{distance traveled}}{volume\text{ of the gas}}](https://img.qammunity.org/2023/formulas/physics/college/610t6czhopluk9od32i9wy9qnjkjvj4uca.png)
a. Given, the car traveled for d=80 km in V=5.0 L of gas.
Thus the efficiency in this case is,
![\begin{gathered} \eta_a=(80)/(5) \\ =16\text{ km/L} \end{gathered}](https://img.qammunity.org/2023/formulas/physics/college/paeqgywvb7guoqi98h92f5cawu2oemabub.png)
b.
Given, the car traveled for 90 km using 6.0 L of gas
Thus the efficiency in this case is,
![\begin{gathered} \eta_b=(90)/(6) \\ =15\text{ km/L} \end{gathered}](https://img.qammunity.org/2023/formulas/physics/college/riloicjwugyjd33zqd1b2tdp6okinasymk.png)
c.
Given, the car travels for 50 km on 4.0 L of gas.
Thus the efficiency of the car, in this case, is given by,
![\begin{gathered} \eta_c=(50)/(4) \\ =12.5\text{ km/L} \end{gathered}](https://img.qammunity.org/2023/formulas/physics/college/dmywjbq93coacs1wvr9a299hq982cr9b56.png)
d.
Given,
The car travels for a distance of 70 km using 5.0 L of gas.
Therefore, the efficiency of the car, in this case, is given by,
![\begin{gathered} \eta_d=(70)/(5) \\ =14\text{ km/L} \end{gathered}](https://img.qammunity.org/2023/formulas/physics/college/8inwmfahicrjly7712ymo1t0x6rwvkdwze.png)
Thus on comparing, the car in the option a achieved the greatest efficiency.
Therefore, the correct answer is option a.