What is the median number of employees assigned per project

TheAlchemist36 you're proposing a scenario that is possible, even with both statements combined (that the top 25% of projects all have 5 employees assigned to them). So, that's good.
Now, you're suggesting that since is scenario is possible, the correct answer is E.
The problem with your post is that you haven't provided any justification for choosing answer choice E.
This makes it very difficult to help you find your mistake.
So: why is it that you believe answer choice E is appropriate, given that it's possible that the top 25% of projects all have 5 employees assigned to them?

What is the median number of employees assigned per project for the projects at Company Z?

(1) 25 percent of the projects at Company Z have 4 or more employees assigned to each project. Not sufficient on its own.

(2) 35 percent of the projects at Company Z have 2 or fewer employees assigned to each project. Not sufficient on its own.

(1)+(2) Since 35% of of the projects have 2 or fewer (\(\leq{2}\))employees and 25% of the projects have 4 or more (\(\geq{4}\)) employees, then 100%-(25%+35%)=40% of the projects have exactly 3 employees assigned to each of them. So, the median number of employees assigned per project is 3. Sufficient.

Answer: C.

To elaborate more: consider there are 100 projects: \(\{p_1, \ p_2, \ ... , \ p_{100}\}\). The values of \(p_1\) to \(p_{35}\) will be 0, 1, or 2; the values of \(p_{36}\) to \(p_{75}\) will be exactly 3; the values of \(p_{76}\) to \(p_{100}\) will be 4 or more. \(Median=\frac{p_{50}+p_{51}}{2}=\frac{3+3}{2}=3\).

For example list can be: \(\{2, \ 2, \ 2, \ ..., \ (p_{35}=2), \ (p_{36}=3), \ 3, \ ..., \ (p_{75}=3), \ (p_{76}=4), \ 4, \ ..., \ (p_{100}=4)\}\);
OR:
\(\{0, \ 0, \ 1, \ 1, \ 1, \ 2, \ 2, \ ..., \ (p_{35}=2), \ (p_{36}=3), \ 3, \ ..., \ (p_{75}=3), \ (p_{76}=4), \ 5, \ 7, \ 27, \ ..., \ (p_{100}=10000)\}\) (of course there are a lot of other breakdowns).

In any case median=3.

Hope it's clear.

P.S. Please post PS questions in the PS subforum: https://gmatclub.com/forum/gmat-problem-solving-ps-140/ and DS questions in the DS subforum: https://gmatclub.com/forum/gmat-data-sufficiency-ds-141/ No posting of PS/DS questions is allowed in the main Math forum.

Bunuel your answer is dependent on the assumption that the company has 3 employees assigned to some projects. However no such information has been provided in the question stem. It is also possible that the company has 40% projects with 5 employees. Therefore the answer to this question should be E and not C.

Solution:

Question Stem Analysis:


We need to determine the median number of employees assigned per project for the projects at Company Z.

Statement One Alone:

We know the upper 25 percent of the projects at Company Z have 4 or more employees assigned to each project. However, we can’t determine the median number of employees assigned per project because it could be 3, 2, or even 1 person.

Statement Two Alone:

We know the lower 35 percent of the projects at Company Z have 2 or fewer employees assigned to each project. However, we can’t determine the median number of employees assigned per project because it could be 3, 4, or even more people.

Statements One and Two Together:

We know the upper 25 percent of the projects at Company Z have 4 or more employees assigned to each project and the lower 35 percent of the projects at Company Z have 2 or fewer employees assigned to each project, which means the middle 40 percent of the projects at Company Z must have exactly 3 employees assigned to each project. Since the middle 40 percent of the projects contains the median number, the median number of employees assigned per project must be 3.

Answer: C

ScottTargetTestPrep

Thank you for your helpful explanation.

When you mention "since the middle 40% of the projects contains the median number, the median number of employees assigned per project must be 3."

I was basically thinking that basically 100%-35%-25%=40%. The median of 100% is 50%, and within that 40% of values, the 50% must lie. Does that thinking make sense?