2011 AMC 10A Problem 15

Let the trip be $x$ miles. The gasoline portion is $x - 40$ miles, consuming $0.02(x - 40)$ gallons. The battery portion uses no gasoline.

The trip's average is total miles divided by total gallons:
$$\frac{x}{0.02(x - 40)} = 55.$$
Multiply out: $x = 55 \cdot 0.02\,(x - 40) = 1.1(x - 40) = 1.1x - 44$. So $0.1x = 44$ and $x = 440$.

Check: $400$ gasoline miles use $8$ gallons, and $440/8 = 55$.

The answer is $\boxed{\textbf{(C)}\ 440}$.

"Miles per gallon" over a whole trip is an overall ratio, total distance over total fuel, not an average of rates. The battery miles inflate the distance without adding fuel, which is why the average ($55$) exceeds the gasoline-only rate of $1/0.02 = 50$ mpg. Set up the ratio with the correct totals and the equation is linear.

Test the choices. For $440$ miles, the gasoline miles are $400$, using $400 \cdot 0.02 = 8$ gallons, and $440/8 = 55$ exactly. For $240$: $200$ miles use $4$ gallons, $240/4 = 60$, too high; for $640$: $600$ miles use $12$ gallons, $640/12 \approx 53.3$, too low.

• Computing gasoline used from the whole trip length rather than only the miles driven after the first 40 battery miles.
• Solving $\frac{x - 40}{0.02(x-40)} = 55$, which is impossible since the gasoline-only rate is fixed at $50$ mpg, and then guessing.