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,
[tex]\eta=\frac{\text{distance traveled}}{volume\text{ of the gas}}[/tex]a. Given, the car traveled for d=80 km in V=5.0 L of gas.
Thus the efficiency in this case is,
[tex]\begin{gathered} \eta_a=\frac{80}{5} \\ =16\text{ km/L} \end{gathered}[/tex]b.
Given, the car traveled for 90 km using 6.0 L of gas
Thus the efficiency in this case is,
[tex]\begin{gathered} \eta_b=\frac{90}{6} \\ =15\text{ km/L} \end{gathered}[/tex]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,
[tex]\begin{gathered} \eta_c=\frac{50}{4} \\ =12.5\text{ km/L} \end{gathered}[/tex]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,
[tex]\begin{gathered} \eta_d=\frac{70}{5} \\ =14\text{ km/L} \end{gathered}[/tex]Thus on comparing, the car in the option a achieved the greatest efficiency.
Therefore, the correct answer is option a.