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23 Jun 2025, 01:48
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Maximum: 60
Minimum: 20
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
35%
(medium)
Question Stats:
71% (01:46) correct
29%
(02:16)
wrong
based on 849
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At a university media club with 120 members, 2/3 of the students are enrolled in a video editing course, and 50% are enrolled in a graphic design course.
In the table below, select Maximum for the maximum number of students who enrol in both Video editing and Graphic design courses and select Minimum for the minimum number of students who enrol in both Video editing and Graphic design courses. Make only two selections, one in each column.
Source: http://www.GMATinsight.com
In the table below, select Maximum for the maximum number of students who enrol in both Video editing and Graphic design courses and select Minimum for the minimum number of students who enrol in both Video editing and Graphic design courses. Make only two selections, one in each column.
Source: http://www.GMATinsight.com
| Maximum | Minimum | |
| 0 | ||
| 20 | ||
| 30 | ||
| 40 | ||
| 60 | ||
| 80 |
ShowHide Answer
Official Answer
Maximum: 60
Minimum: 20
23 Jun 2025, 23:06
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Given:
To find: Maximum and Minimum possible overlap of students in both courses.
Solution:
Correct Answer:
Maximum – 60
Minimum – 20
Shweta Koshija
GMAT, GRE, SAT Coach for 10+ years
- Total members = 120
- Enrolled in Video Editing (V) = 2/3 of 120 = 80
- Enrolled in Graphic Design (G) = 50% of 120 = 60
To find: Maximum and Minimum possible overlap of students in both courses.
Solution:
- Maximum overlap
- Visualize two circles in a Venn diagram.
- The maximum overlap between the two circles will be when the smaller circle is completely contained within the bigger circle.
- That is, when G (60) is completely within V (80). (Interpretation: each student enrolled in Graphic design is also enrolled in Video editing).
- So, maximum number enrolled in both = Minimum of 80 and 60 = 60
- Minimum overlap:
- Ideally, minimum overlap between two circles would be when they are completely disjoint.
- But then we would have at least 80 + 60 = 140 members (not considering "neither" at all).
- And we know this is not possible since we only have 120 members.
- So, minimum overlap is the EXTRA = 140 - 120 = 20
Correct Answer:
Maximum – 60
Minimum – 20
Shweta Koshija
GMAT, GRE, SAT Coach for 10+ years
17 Apr 2026, 08:11
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Deconstructing the Question
There are \(120\) students total.
Of these, \(\frac{2}{3}\) are enrolled in Video Editing, and \(50\%\) are enrolled in Graphic Design.
So:
\(\frac{2}{3}\cdot 120=80\)
and
\(50\%\cdot 120=60\)
We need the maximum and minimum number of students enrolled in both courses.
This is an overlap question.
The maximum overlap happens when the smaller group is completely contained inside the larger group.
The minimum overlap happens when the two groups overlap as little as possible, while still fitting inside the total of \(120\) students.
Step-by-step
Let the Video Editing group be \(80\) and the Graphic Design group be \(60\).
For the maximum number in both courses, make the smaller group entirely part of the larger group.
So the maximum overlap is:
\(\min(80,60)=60\)
So the maximum is \(60\).
For the minimum number in both courses, use:
\(80+60-120=20\)
This is the smallest forced overlap, because the two groups together exceed the total by \(20\).
So the minimum is \(20\).
Answer: Maximum = 60, Minimum = 20
There are \(120\) students total.
Of these, \(\frac{2}{3}\) are enrolled in Video Editing, and \(50\%\) are enrolled in Graphic Design.
So:
\(\frac{2}{3}\cdot 120=80\)
and
\(50\%\cdot 120=60\)
We need the maximum and minimum number of students enrolled in both courses.
This is an overlap question.
The maximum overlap happens when the smaller group is completely contained inside the larger group.
The minimum overlap happens when the two groups overlap as little as possible, while still fitting inside the total of \(120\) students.
Step-by-step
Let the Video Editing group be \(80\) and the Graphic Design group be \(60\).
For the maximum number in both courses, make the smaller group entirely part of the larger group.
So the maximum overlap is:
\(\min(80,60)=60\)
So the maximum is \(60\).
For the minimum number in both courses, use:
\(80+60-120=20\)
This is the smallest forced overlap, because the two groups together exceed the total by \(20\).
So the minimum is \(20\).
Answer: Maximum = 60, Minimum = 20
