The ratio of the numbers of the workers in teams of X and Y is 3:4, and the ratio of the working efficiencies for the workers in teams X and Y is 5:4. Both teams do the projects in which the workload and the conditions are identical. However, Y finishes the project 9 days earlier than X. Later, two new teams are composed. M consists of *2/3*of the workers in X and 1/3 of the workers in Y, and N consist of the rest of workers in both X and Y. Both teams (M and N) do the projects in which the work load and the conditions are identical. However, N finishes the project six days earlier than M. What is the ratio of the workloads in the two projects
I am having trouble answering this question, and I don't have any work to show. Hints or solution is welcome. Thank you very much.
Let the total number of workers be W Number of workers in X = 3/7 W Number of workers in Y = 4/7 W
Let the efficiency of a worker in X be E units of work per day Efficiency of a worker in Y = 4/5 E
Let t days be taken by Y to complete the project and t+9 be taken by X
Since workloads is the same in both projects undertaken by X and Y, {Number of workers x Efficiency of a worker x Time in days} is the same for both X and Y
Equation 1: 3/7 W E (t+9) = 4/7 W 4/5 E t
After the new teams M and N are formed, we can assume that the efficiency of each worker remains the same in the newly formed teams also.
For the newly assigned projects to M and N, let us assume that N completes the work in t' days. M completes it in (t' + 6) days
Since the workloads for the newly assigned projects to M and N are the same, we have:
Equation 2: {2/3 (3/7 W E) + 1/3 (4/7 W 4/5 E)} (t'+6) = {1/3 (3/7 W E) + 2/3 (4/7 W 4/5 E)} t'
You can solve Equation 1 to find t. You can solve equation 2 to find t'
Ratio of the workload in the two projects = [3/7 W E (t+9)] / [{2/3 (3/7 W E) + 1/3 (4/7 W 4/5 E)} (t'+6)]
Since t and t' are known and W and E can be cancelled, you can then calculate this ratio.
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