2011 AMC 10A Problem 15

"Averaged 55 miles per gallon" means: total miles divided by total gallons of gas equals $55$. So we need both totals.

1. Let the trip be $x$ miles long.
2. The first $40$ miles use no gas. The rest of the trip is $x - 40$ miles, and each of those miles uses $0.02$ gallons. So the total gas is $0.02(x - 40)$ gallons.
3. Set total miles divided by total gallons equal to $55$:
$$\frac{x}{0.02(x - 40)} = 55.$$
4. Multiply both sides by the bottom: $x = 55 \times 0.02 \times (x - 40)$. Since $55 \times 0.02 = 1.1$, this is
$$x = 1.1(x - 40) = 1.1x - 44.$$
5. Subtract $x$ from both sides: $0 = 0.1x - 44$, so $0.1x = 44$ and $x = 440$.

Check: the gas part is $440 - 40 = 400$ miles, using $400 \times 0.02 = 8$ gallons. And $440 \div 8 = 55$. ✓

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

Miles per gallon for a whole trip is one big ratio: all the miles over all the gallons. On gas alone the car gets $1 \div 0.02 = 50$ miles per gallon. The free battery miles add distance without using gas, which is why the trip average ($55$) is higher than $50$. Once you write both totals correctly, the equation is a simple one.

Test the answer choices, since each one takes only a moment.

• $240$ miles: $200$ gas miles use $4$ gallons, and $240 \div 4 = 60$. Too high.
• $440$ miles: $400$ gas miles use $8$ gallons, and $440 \div 8 = 55$. ✓
• $640$ miles: $600$ gas miles use $12$ gallons, and $640 \div 12 \approx 53.3$. Too low.

Only $440$ gives exactly $55$.

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 $\boxed{\textbf{(C)}\ 440}$.