AMC 8 Daily Practice Round 8

Complete problem set with solutions and individual problem pages

Problem 26 Medium

In a factory production team working on a batch of parts, if each worker stays at their original position, the task can be completed in 9 hours. If workers A and B exchange positions while the efficiency of the other workers remains unchanged, the task can be completed 1 hour earlier. If workers C and D exchange positions while the efficiency of the other workers remains unchanged, the task can also be completed 1 hour earlier. If A and B exchange positions and C and D also exchange positions at the same time, with the efficiency of the other workers unchanged, how many minutes earlier can the task be completed?

  • A.

    90

  • B.

    100

  • C.

    105

  • D.

    108

  • E.

    115

Answer:D

Divide the total task into 72 parts.

Originally, the team completed \frac{72}{9} = 8 parts per hour, so each part took \frac{60}{8} = 7.5 minutes.

After exchanging A and B, the team completed \frac{72}{8} = 9 parts per hour, which is 1 more part per hour than before.

Since the efficiency of the other workers remains unchanged, this extra part is entirely due to the improved combined efficiency of A and B.

Similarly, after exchanging C and D, they also completed 1 more part per hour than before, due to the improved combined efficiency of C and D.

If both exchanges are made at the same time, A and B and C and D each contribute an extra part per hour.

Therefore, the team completes 8 + 1 + 1 = 10 parts per hour, meaning each part takes \frac{60}{10} = 6 minutes to complete.

Thus, for each part, the time saved is 7.5 - 6 = 1.5 minutes, and for all 72 parts, the total time saved is 72 \times 1.5 = 108 minutes.

Answer: The task can be completed 108 minutes earlier.