A stationary engine has enough fuel to run 12 hours when its tank is $\frac{4}{5}$ full. How long will it run when the tank is $\frac{1}{3}$ full?

### Concept & Logic This problem is based on direct proportionality. The runtime of the engine is directly proportional to the amount of fuel in the tank. If we establish the total runtime for a completely full tank (1 whole unit), we can easily calculate the runtime for any other fractional capacity. ### Step-by-Step Solution * **Given:** Runtime for $\frac{4}{5}$ of the tank = 12 hours. * **Calculation:** First, find the runtime for a completely full tank. Let total runtime be $T$. $$ \frac{4}{5} \times T = 12 $$ $$ T = 12 \times \frac{5}{4} $$ $$ T = 3 \times 5 = 15 \text{ hours} $$ * Now, calculate the runtime when the tank is $\frac{1}{3}$ full. $$ \text{New Runtime} = \frac{1}{3} \times T $$ $$ \text{New Runtime} = \frac{1}{3} \times 15 $$ $$ \text{New Runtime} = 5 \text{ hours} $$ ### Exam Strategy & Shortcut Use the **Unitary Method Shortcut**. If 4 parts out of 5 give 12 hours, then 1 part gives $12 \div 4 = 3$ hours. The whole tank (5 parts) gives $5 \times 3 = 15$ hours. A tank that is $\frac{1}{3}$ full will run for $\frac{1}{3}$ of the total 15 hours, which is 5 hours. This mental path completely bypasses formal algebraic setup. ### Common Pitfall A common mistake is inverting the fraction improperly when solving for the total time (e.g., multiplying 12 by $\frac{4}{5}$ instead of $\frac{5}{4}$), which incorrectly reduces the total full-tank runtime to 9.6 hours. Always logically verify: a full tank MUST run longer than a partially full one. ### Final Answer Therefore, the correct answer is **5 hours**.