Work and Rate Problems

What we see: A work-rate question. Two machines share one job, and we want how long one of them takes alone.

Step 1. Turn each time into a speed.
What we see: we are given times for one full job (1,000 nails): 12 hours for X, 5 hours for X and Y together.
What it means: when machines work together, their speeds add up. Their times do not.
What we do and why: we write each one as "jobs per hour" so we can add and subtract them. X does \(\frac{1}{12}\) of the job each hour. Together they do \(\frac{1}{5}\) of the job each hour.

Step 2. Find Y's speed.
What we see: the together speed is X's speed plus Y's speed.
What it means: Y's speed is the together speed minus X's speed.
What we do and why: \(\frac{1}{5} - \frac{1}{12} = \frac{12}{60} - \frac{5}{60} = \frac{7}{60}\). Y does \(\frac{7}{60}\) of the job each hour.

Step 3. Turn Y's speed back into a time.
What we see: Y does \(\frac{7}{60}\) of the job per hour.
What it means: the time for one full job is 1 divided by the speed.
What we do and why: \(1 \div \frac{7}{60} = \frac{60}{7} = 8\frac{4}{7}\) hours.

The trap: subtracting the times, \(12 - 5 = 7\) hours (choice C). Times never add or subtract; only speeds do.
Faster way: when you know one machine's time \(A\) and the together time \(T\), the other machine takes \(\frac{A \times T}{A - T} = \frac{12 \times 5}{12 - 5} = \frac{60}{7}\) hours.
Takeaway: in work problems, add or subtract speeds (jobs per hour), never times.

Answer: E