There are enough available spaces on a school team to select at most

Question

There are enough available spaces on a school team to select at most 1/3 of the 50 students trying out for the team. What is the greatest number of students that could be rejected while still filling all available spaces for the team?

(A) 16
(B) 17
(C) 33
(D) 34
(E) 35

(A) 16
(B) 17
(C) 33
(D) 34
(E) 35

BrentGMATPrepNow · Answer

1/3 of 50 = (1/3)(50) = 50/3 = 16 2/3
Since the number of available spaces cannot be greater than 16 2/3, there must be 16 available spaces.
So, 16 students (among the original 50 students) can be selected to join the team, which means the other 34 students must be rejected.

Answer: D

Cheers,
Brent

Archit3110 · Answer

number selected or rejected has to be an integer so factor <50 is 48 ; 1/3*48 ; 16 so rejected would be
50-16 ; 34
IMO B

Abhishek009 · Answer

Maximum members that can be selected is 1/3*50 = 16

So, Greatest number of students that can be rejected is 50 - 16 = 34 , Answer must be (D)

ScottTargetTestPrep · Answer

To fill the space on the team, a maximum of 16 people could be selected, so 50 - 16 = 34 would be rejected.

Answer: D

rafa90 · Answer

Hi BrentGMATPrepNow Bunuel ScottTargetTestPrep,

This was my approach.

The maximum available spots in the team are 16. Also, the problem doesn't state that I need to fill up the team. Hence I can select 16 or even 15 or no students at all from the group of 50 students.

So if we select 16 students (max possible), we reject the remaining 34. We can still select 15 students and reject the rest of the 35.

This makes me think at the problem should be worded as "what is the least number of students that could be rejected". Am I missing anything here?

BrentGMATPrepNow · Answer

The problem is highlighted above. 
The question asks: "There are enough available spaces on a school team to select at most 1/3 of the 50 students trying out for the team. What is the greatest number of students that could be rejected   while still filling all available spaces   for the team?"

MathRevolution · Answer

Atmost ( \frac{1}{3} ) of 50 = 16.xx

For the number of students, it has to be an integer value. Hence,16

Rejected: 50 - 16 = 34

Answer D

Anonymous · Answer

Hello BrentGMATPrepNow and GMAT Experts  Bunuel  ,  ScottTargetTestPrep  ,  avigutman  ,  GMATCoachBen

I have the same issue as  rafa90 . We know that  when we want to maximize one variable, we minimize another  (and vice versa). If there is enough space for  at most  16 students, then there would be enough for, let's say, 10 students, which lead to 50 - 10 = 40 rejected students, which is more than 34. Could you clarify this, please?

Krunaal · Answer

Imo, the catch is the last sentence of the question,  "...   while still filling  all available spaces  for the team",  this statement acts as a constraint. We know the all available spaces are 16, and filling  all  available spaces - which means no space in the team is empty - the maximum students that can be rejected are 34.

There are enough available spaces on a school team to select at most

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