2008 AMC 10A Problem 1

1. First find how long the machine has been running. From 8:30 AM to 11:10 AM is $2$ hours $40$ minutes.
2. In that time it did one third of the job. The machine runs at a steady speed, so the whole job takes three times as long:
$$3 \times (2\text{ h } 40\text{ min}) = 6\text{ h } 120\text{ min}.$$
3. Since $120$ minutes is $2$ hours, the whole job takes $8$ hours.
4. The whole job started at 8:30 AM, so count the $8$ hours from there: 8:30 AM plus $8$ hours is 4:30 PM.

The answer is $\boxed{\textbf{(D)}\ \text{4:30 PM}}$.

When a machine works at a steady speed, time and work grow together: one third of the work takes one third of the time. The only care needed is to add the total time to the moment the job started, not to the moment you were told about.

Count only what is left. Two thirds of the job remain, which takes twice as long as the first third: $2 \times (2\text{ h } 40\text{ min}) = 5\text{ h } 20\text{ min}$. Starting from 11:10 AM, add $5$ hours to get 4:10 PM, then $20$ more minutes to get 4:30 PM.