If 38 percent of the departments in a certain research organization

Question

If 38 percent of the departments in a certain research organization have 14 or fewer members each, what is the median number of members per department in this organization?

(1) 58 percent of the departments have 15 or fewer members each.
(2) 42 percent of the departments have 16 or more members each.

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gmatophobia · Accepted Answer

A very straightforward question. If 38 percent of the departments in a certain research organization have 14 or fewer members each Inference : Out of every 100 departments, 38 departments in the research organization have 14 or fewer members each. Question : What is the median number of members per department in this organization? Pre-Analysis : Let's assume that the number of departments in the organization is 100. The departments are ranked D_1 , D_2 ... D_{100} based on the number of members in that department. D_1 ⇒ Represents the department with the least number of members D_2 ⇒ Represents the department with the highest number of members The median number of members per department is the average of the number of members of departments represented by D_{50} & D_{51} Screenshot 2024-01-28 193104.png Statement 1 (1) 58 percent of the departments have 15 or fewer members each. From the question stem we know that departments D_1 to D_{14} have 14 or fewer members. Statement 1 provides additional information that 58 percent of the departments have 15 or fewer members each. Hence, departments D_{1} to D_{58} , have 15 or fewer members. We already know that the number of members until D_{14} is at max 14. Hence, the number of members from D_{15} to D_{58} must be 15 . Therefore we can conclude that departments D_{50} and D_{51} both have exactly 15 members. The median number of members per department in this organization is \frac{15+15}{2} = 15 Screenshot 2024-01-28 193512.png The statement alone is sufficient to arrive at a unique answer to the question asked. We can eliminate B, C, and E. Statement 2 (2) 42 percent of the departments have 16 or more members each. If 42 percent of the departments have 16 or more members each, 58 percent (i.e. 100-42) percent of the departments have 15 or fewer members each. This is obvious as the number of members cannot be in fractions. Hence, Statement 2 conveys the same information as that of Statement 1. We have already determined that Statement 1 is sufficient. Hence, Statement 2 is also sufficient. Option D

Bunuel · Answer

I see the confusion.

The statement "38% of the departments have 14 or fewer members each" implies that there is at least one department with exactly 14 members. Therefore, it differs from saying "38% of the departments have fewer than 14 members each." Similarly, when it is stated that "58% of the departments have 15 or fewer members each," it suggests that there is at least one department with exactly 15 members.

Bunuel · Answer

The question asks for the median number of members per department in the organization, not the median of the departments themselves, which would not make any sense. For example, assume there are 100 departments and consider the departments and the number of employees in each as follows:

D1 = 1, D2 = 1, D3 = 7, ..., D49 = 15, D50 = 15, ..., D100 = 91.

Since the median of an even count of values is the average of the two middle values when the values are in order, then the median number of members per department for our example would be the average of the two middle values. Thus, it would be the average of D49 and D50, making the median equal to 15.

Hope it's clear.

stne · Answer

Let the # of departments be  100.

To find the median we need the average of the total members in the  50  and  51  departments, when the departments are arranged is ascending order according to the # of members present.

We know  38  departments have  14  or fewer members.

(1) 58 percent of the departments have 15 or fewer members each.

58 departments have  15  or fewer members.
38 departments have  14  or fewer members.

Hence departments  39  to  58  must have  15  members.

Thus  50   and  51  departments have  15  members each. Thus the median is the avg. of these two  =15.

SUFF.

(2) 42 percent of the departments have 16 or more members each.

Thus  58  percent of the departments have  15  or fewer members.

This basically tells the same thing as statement (1)

We can follow the same reasoning as shown in statement (1)

SUFF.

Ans D

Hope it helped.

anish777 · Answer

if 42 percent has more than or equal to 16 members, how come you came to this conclusion that 58 percent must have 15 or less members ?
Can you please elaborate on this ?

stne · Answer

Hi anish777,

Suppose there are  100   departments. It is given that  42  departments have  16  or more members. Then  100-42= 58  departments must have less than  16  members. Hence these  58  departments will have  15   or fewer members.

Hope it clearer now.

Sixcarbs · Answer

Hi, I am new here, and far from an expert, but I find the answer troubling.

I know out of 100 we want to find the average of #50 and #51.

It says 38% have 14 or fewer, and then we get 58% have 15 or fewer.

How do we know that the 38% don't have 1 each, or 2 each?

And how do we know that the 15 or fewer doesn't mean 3 or 4 people each?

Thank you

Chiragdoshi · Answer

Thank you for posting this. I also have the same question. 15 or less can mean any number less than 15. Can someone please explain

unraveled · Answer

You are going in the right direction to point out that but missing an essential part of the median concept. Median simply means an specific number. 
While the question seems to gives us numbers that vary from 0 to 14 or 1 to 15, the middle departments can't say that they have 14 or fewer or 15 or fewer and certainly not 0 to 14 < 15.

This concludes that that specific number must be 15. So, all the departments frm 39th to 58th have 15 members each.

HTHs.

rebinh1982 · Answer

What is the median number of members per department in this organization?

Does this question ask the median of each department in the organization? Or the median of the departments in the organization?

The phrase per department in the question stem is very confusing.

anikpait17 · Answer

I see someone asked the question that why cant we have 1 or 2 members each but I did not get clarity from the answers given. Lets say we have 100 departments. 38 departments have 14 or fewer members, lets say each of the 38 departments have 2 members. 58 departments have 15 or fewer members, lets say each of the (58-38=20) departments have 3 members each. If this scenario is the case, median shall be avg of  50th and 51st value which is 3. Then how are we concluding the fact that median is exactly 15 (it could be one of the cases but not the only case)?

ramiav · Answer

I agree that logically the solution is D but there could be a flaw on the question

<= 15 could be 14 and 15 ...

so we know 38% <= 14,
and the 1st statement says: 58% have 15 or less,
so imagine that we have 100
the lowest 38, we don't care about the value
but from 39 to 58 included,
we can ONLY affirm that at least one of them is 15 and the rest have 14. so in that case the median is 14, but if all of them but 1 are 15 ( the exception is 14) then the answer is 15.

So with this we can say that statement 1 is  insuff

Then with B we have something similar

the lowest 38 are 1...14 who cares

the greatest 42 they are 16 or more

but in the middle we have from 39 to 58 but this could be 14, 15 or 16 so we cannot ensure the answer again.

putting all of them tgtr we can say that the value is 14 or 15 but without security.

so the answer should be E.

If we add on the statements that the limit (with the equal) includes all the cases, so this number is excluded on the rest, then the answer is D

42 percent of the departments have 16 or more members each.

SergejK · Answer

I agree with you, even if it is an official question from mock 4. I also think that 14, 15 and even 14.5 (here, we are not halving a person, just providing a data point, which is an average between 50 and 51 % of the departments

) are possible. Same with B. EDIT: I think I understand it a bit better now. The 38% percent is a hard cut-off point for 14 members. So the 39th percentile can't be 14 any longer as then, we would have said that 39% of the departments have <=14 people in it. Looking now at S1, it tells us that 58% are <=15, which in turn means, considering that we know that 38% are at most 14, 39-58% of the departments must all be 15. The information that 58% are <= 15 just combines the 2 ranges of 1-38% and 39-58%, and hence we are saying that <=58% must be <=15. Again, the range between 39-58 is exactly 15. The other interpretation could only be valid if we are told that less or equal to 58 percent have 15 or less people per department. Then, we wouldn't know if 50th to 51st percentile actually has 15 or if it is already 16 or more. Took me also a while to understand.

SudsMeister · Answer

How I thought about it in my head:

Question states that 38% have 14 or lesser, and then adds to that that 58% have 15 or lesser.

It is evident that the latter 58% group contains the former 38% group.

However, the extra 20% departments cannot have 14 or lesser because it will make the original statement false.

Because then the question won't state that 38% of the departments have 14 or lesser, it will state that at least 38% and at most 58% of the departments have 14 or lesser or some other form of such a statement.

So, 38% can have 14 or lesser, we don't need exact distribution.

The next 20% must have exactly 15.

KarishmaB · Answer

Number of members must be integers. We are given that 38% depts have members from 0 to 14. Median will be obtained at the 50% mark. (1) 58 percent of the departments have 15 or fewer members each. Screenshot 2025-09-16 at 11.46.36 AM.png If 38% depts have 14 or fewer members and 58% depts have 15 or fewer members, it means 20% depts (from 38% to 58%) will have 15 members. So median will be at 15. Sufficient alone. (2) 42 percent of the departments have 16 or more members each. Screenshot 2025-09-16 at 11.46.41 AM.png If 38% depts have 14 or fewer members and 42% depts have 16 or more members then 20% depts (from 38% to 58%) will have 15 members. So median will be at 15. Sufficient alone. Answer (D)

nikita.nanda1810 · Answer

KarishmaB  I have question about statement 1 -

(1) 58 percent of the departments have 15 or fewer members each.

Is it not possible that dept 57, 58 have 15 each and dept 50, 51 have 14. Does 15 or fewer have to imply that all after 38% till 58% will have 15 each?

I am having difficulty in interpreting the 'or' kind of statement -- should we consider that each of the terms can be 15 and they can also be all less than 15 or few have to be 15 and few have to be less than 15.

Appreciate your help. Thanks!

If 38 percent of the departments in a certain research organization

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If 38 percent of the departments in a certain research organization

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615 to 715 in 15 Days (Ria's Strategies)

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