AMC 8 Daily Practice Round 8

Complete problem set with solutions and individual problem pages

Problem 28 Hard

Addy and Brandon are working together on a project. The original plan was for them to work alternately, one person per day, in the order Addy first, then Brandon, and so on, and the work would be finished in an exact whole number of days. If they instead worked in the order Brandon first, then Addy, alternating one day each, it would take half a day longer than the original plan. Now, if Addy and Brandon work together every day, they can finish the project in 8\frac{2}{3} days. If they follow the original plan, how many days will it take to complete the entire task?

  • A.

    15

  • B.

    16

  • C.

    17

  • D.

    18

  • E.

    19

Answer:C

In the original plan, the last day must have been worked by Addy.

If the order is changed to Brandon–Addy–Brandon–Addy \cdots, then on what was originally the last day (Addy’s workday), Brandon would work for a full day, and then Addy would work for half a day to finish the job.

Thus, \text{Addy} = \text{Brandon} + \frac{1}{2} \text{Addy} which means \text{Addy} = 2 \times \text{Brandon}.

Since \text{Addy} + \text{Brandon} = \frac{1}{8\frac{2}{3}} = \frac{3}{26},

Addy’s and Brandon’s work rates are \frac{1}{13} and \frac{1}{26} respectively.

In the original plan, since the last day is worked by Addy, the days worked by Addy and Brandon before the last day are equal.

Thus, each of them works: \frac{1 - \frac{1}{13}}{\frac{1}{13} + \frac{1}{26}} = 8 \ \text{days}.

Therefore, under the original plan, the total time to complete the task is: 8 \times 2 + 1 = 17 \ \text{days}.