AMC 10 Weekly Practice Round 2

Complete problem set with solutions and individual problem pages

Problem 30 Medium

A group of workers is assigned to load and unload a batch of goods, with each worker working at the same rate.

If all workers work together from the start, the job can be completed in 10 hours.

Now, the method is changed: one worker starts alone, and then every t hours (where t is an integer) one more worker joins.

Each worker who joins continues working until the job is finished.

The last worker added works for a time equal to one quarter of the first worker’s total working time.

What is the maximum number of workers that can be used to complete the job?

  • A.

    7

  • B.

    11

  • C.

    13

  • D.

    15

  • E.

    19

Answer:C

Let the loading and unloading work take x hours to complete.

The first worker works for x hours, and the last worker works for \frac{x}{4} hours.

Together, they work for \left( x + \frac{x}{4} \right) hours, so the average time worked per person is \frac{1}{2} \left( x + \frac{x}{4} \right) hours.

From the problem statement, we know:

The second worker and the second-to-last worker, the third worker and the third-to-third-last worker, and so on, each also work an average of \frac{1}{2} \left( x + \frac{x}{4} \right) hours.

Therefore:\frac{1}{2} \left( x + \frac{x}{4} \right) = 10

Solving gives: x = 16

With the new loading and unloading method, the total time is 16 hours.

Let y be the total number of workers.

Since one worker is added every t hours, the last worker works (y - 1)t hours less than the first worker.

From the problem, we have:

16 - (y - 1)t = 16 \times \frac{1}{4}

(y - 1)t = 12

Since y and t are both positive integers:

\begin{cases} y = 2, & t = 12 \\y = 3, & t = 6 \\y = 4, & t = 4 \\y = 5, & t = 3 \\y = 7, & t = 2 \\y = 13, & t = 1\end{cases}

Thus, the number of workers could be 2, 3, 4, 5, 7, or 13.

Therefore, the maximum number of workers is 13.