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

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Bunuel
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There are enough available spaces on a school team to select at most [#permalink] New post   11 Mar 2020, 03:21
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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
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D
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BrentGMATPrepNow
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Re: There are enough available spaces on a school team to select at most [#permalink] New post   11 Mar 2020, 06:27
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Bunuel wrote:
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


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
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Re: There are enough available spaces on a school team to select at most [#permalink] New post   11 Mar 2020, 08:31
Bunuel wrote:
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


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
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Re: There are enough available spaces on a school team to select at most [#permalink] New post   11 Mar 2020, 09:59
Bunuel wrote:
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

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)
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Re: There are enough available spaces on a school team to select at most [#permalink] New post   14 Mar 2020, 10:03
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Bunuel wrote:
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


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

Answer: D
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Re: There are enough available spaces on a school team to select at most [#permalink] New post   20 Sep 2020, 17:12
BrentGMATPrepNow wrote:
Bunuel wrote:
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


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


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? :think:
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BrentGMATPrepNow
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Re: There are enough available spaces on a school team to select at most [#permalink] New post   20 Sep 2020, 17:39
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rafa90 wrote:

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? :think:


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?"
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Re: There are enough available spaces on a school team to select at most [#permalink] New post   20 Sep 2020, 22:23
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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
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Re: There are enough available spaces on a school team to select at most [#permalink] New post   04 Apr 2022, 03:07
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