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AD2GMAT
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Of the students in a certain class, 10 had taken a course in 22 May 2017, 04:27
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There are two type of formula we need to understand - I applied the first one, skipping "exactly 2" from the consideration. Second, I assumed that A+B+C = 20, hence got the wrong answer. The correct answer should be A.
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Navpatel909
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Of the students in a certain class, 10 had taken a course in 28 Feb 2020, 22:42
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Tot is 35. Minus 3 from each ABC, leaves 26. 20 is single so 6/2 = 3

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sayan640
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Of the students in a certain class, 10 had taken a course in 31 May 2024, 20:05
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Lets say , total no of students = N
Students opting for at least 2 courses = N-20
Studetns opting for exactly 2 courses = N-20 - 3 = N -23

Total students = No. of students opting for course A + No. of students opting for course B + No. of students opting for course C -
( No. of students opting for Exactly 2 courses ) - 2* ( No. of students opting for All 3 courses )

N = 35 - (N-23) - 6

2n= 52
N= 26­

Hence , No. of studetns opting for exactly 2 courses = N-20 - 3 = N -23 = 26-23 =3 Ans.
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Of the students in a certain class, 10 had taken a course in 15 Jul 2026, 22:01
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I = Exactly 1
II = Exactly 2
III = Exactly 3

When we add the number of people who have taken course A (visualize circle 1), the number of people who have taken course B (visualize circle 2), and the number of people who have taken course C (visualize circle C), we can see that the II portions get counted twice (i.e., those who have taken exactly 2 courses gets counted twice. Similarly, the III portion gets counted thrice (once for each of the 3 circles).

Thus -> I + 2II + 3III = 10+11+14 = 35

Now, we are given that
I = 20
III = 3

So,

20 + 2II + 3(3) = 35

=> 2II = 6

=> II = 3. This is our answer.

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Harsha
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