anshulagarwal12 wrote:What is the median number of employees assigned per projects at Company Z?

1. 25% of the projects at Company Z have 4 or more employees assigned to each project
2. 35% of the projects at Company Z have 2 or fewer employees assigned to each project

I'd go with C here. Here's why:

Statement 1: If 25% have 4 or more, that means that the entire 75% remaining could be 2 or 3, so the median would be 2 or 3.

Statement 2: Can be eliminated for the same reasons as statement 1.

Combined: Lets say 2.5 are 4's or more (25%), and 3.5 are 2's or less (35%) that means there must be 4 3's (30%). Now write these is order (you can even round up the 2.5 and 3.5!):
22223333444 <-- If you eliminate one at a time from each side you will find that the middle number is 3. Thus, your median must be 3. So combined they are sufficient. Choose C.
Does this help? What the OA?

Hi Alex,

Thanks for your reply.

But how can you assume this : If 25% is 4 or more, than 75% will be 2 or 3 ONLY . It can be 5 or 6 or any number

Yes...your answer is correct. But still I am not comfortable with the assumption you made.

anshulagarwal12 wrote:Hi Alex,

Thanks for your reply.

But how can you assume this : If 25% is 4 or more, than 75% will be 2 or 3 ONLY . It can be 5 or 6 or any number

Yes...your answer is correct. But still I am not comfortable with the assumption you made.

It cannot be 5,6,7... because that is greater than 4!...and the statement says only 25% is greater than 4. So the remaining 75% must be less than 4.
But you're right. It could be any number (less than 4), I was just using 3 and 2 as an example to show that we do not know what the median is. If you have 2 (or more) possible solutions, the answer is insufficient. But yes... I could be 1, 0, -1, ect...