Asif123 wrote:
A car traveled 462 miles per tankful of gasoline on the highway and 336 miles per tankful of gasoline in the city. If the car traveled 6 fewer miles per gallon in the city than on the highway, how many miles per gallon did the car travel in the city?
(A) 14
(B) 16
(C) 21
(D) 22
(E) 27
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