Ans: : Median is the middle value. Statements 1 and 2 alone give us insufficient data but when we combine both of them we see that 25 % have 4 or more and 35% have 2 or less therefore 40 %have 3 employees. Therefore the median would be 3 and the answer is (C).

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.

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