So we're looking for:

\(\frac{miles}{gallon}\)

and we're given:

\(\frac{miles}{tank}\), where the tank consists of a fixed maximum number of gallons.


To find out the number of gallons in the tank, we can think of it as a ratio:

\(\frac{gallons}{tank}\)


Writing this out, we get:

\(\frac{462 miles}{tank}\) on the highway.

\(\frac{336 miles}{tank}\) in the city.

We also know that the car in the city drove 6 miles less than the car did on the highway. Let \(x\) be the number of miles driven on the highway. It follows that \(x-6\) miles we're driven in the city. This can be written as:

\(\frac{(x)miles}{gallons}\) on the highway.

\(\frac{(x-6)miles}{gallons}\) in the city.


To find the gallons per tank we can simply divide miles/tank by miles/gallons to eliminate the miles:

\(\frac{462 miles}{tank} / \frac{(x) miles}{gallons}\)

\(\frac{336 miles}{tank} / \frac{(x-6) miles}{gallons}\)

This simplifies to:

\((\frac{462 miles}{tank}) * (\frac{gallons}{(x) miles})\)

\(\frac{462 miles}{(x) miles}\) \(\frac{gallons}{tank}\)


And:

\(\frac{336 miles}{tank} * \frac{gallons}{(x-6) miles}\)

\(\frac{336 miles}{(x-6) miles}\) \(\frac{gallons}{tank}\)

The miles cancel and we're left with:

\(\frac{462}{x}\) \(\frac{gallons}{tank}\)

\(\frac{336}{(x-6)}\) \(\frac{gallons}{tank}\)

Now we can simply equal them to eachother to solve for \(x\):

\(\frac{462}{x}\) \(=\) \(\frac{336}{(x-6)}\)

\(462x - 2772 = 336x\)

\(-2772 = -126x\)

\(x = 22\)

Since we're looking for the number of miles per gallon incurred in the city, we would plug this into the miles per gallon equation written above for the city:

\(\frac{(x-6)miles}{gallon}\) in the city.

\(\frac{(22-6)miles}{gallon}\) in the city.

\(\frac{16miles}{gallon}\) in the city.

So the answer is B