The first step to having a good solution in the present question, building an equation that gives us the number of Gallons (G) in the function of rate miles/gallons in the city (C) and miles/gallons in the highway (H).
We know that in typical weeks Jon drives 20miles in the city (Dc) and 60 miles on highway (Dh) (Going + Coming Back from his parent's house).
The rates of the number of gallons per mile are calculated as follows:
[tex]\begin{gathered} C=\frac{20}{G_c}=\frac{\text{Distance}}{\text{Gallons used}}\to G_C=\frac{20}{C} \\ \\ H=\frac{60}{G_H}=\frac{\text{Distance}}{\text{Gallons used}}\to G_H=\frac{60}{H} \end{gathered}[/tex]Anthe total number of gallons is equal to the sum of each case:
[tex]\begin{gathered} G=G_H+G_C \\ \\ G=\frac{20}{C}+\frac{60}{H} \end{gathered}[/tex]Now, to solve the problem, we just need to substitute the values as follows:
A - I)
The present car has the given rates:
[tex]\begin{gathered} H=22\text{mpg} \\ C=16\text{mpg} \end{gathered}[/tex]Substituting:
[tex]\begin{gathered} G_1=\frac{20}{16}+\frac{60}{22} \\ G_1=1.25+2.73 \\ G_1=3.98\text{ Gallons} \end{gathered}[/tex]A-II)
The new values are:
[tex]\begin{gathered} H=35\text{mpg} \\ C=28mpg \end{gathered}[/tex]Substituting, we have:
[tex]\begin{gathered} G_2=\frac{20}{28}+\frac{60}{35} \\ G_2=0.72+1.71 \\ \\ G_2=2,43\text{ Gallons} \end{gathered}[/tex]A-III
The difference for 52-week year is equal to the difference for a single week times 52, as it is calculated as follows:
[tex]\begin{gathered} \Delta G=(\frac{20}{16}+\frac{60}{22})-(\frac{20}{28}+\frac{60}{35})_{} \\ \Delta G=1.55\text{ Gallon} \end{gathered}[/tex][tex]\begin{gathered} \Delta G_{\text{Total}}=52\times\Delta G \\ \\ \Delta G_{\text{Total}}=80.53\text{ Gallons} \end{gathered}[/tex]