In a class of 345 students, equal number of students enrolled in Engli

We can create an equation using the following formula:

Total = N(E) + N(M) + N(S) - N(E and M) - N(M and S) - N(E and S) + N(all three) + N(none)

345 = n + n + n - 30 - 26 - 28 + 14 + 43

345 = 3n - 84 + 57

345 = 3n - 27

372 = 3n

124 = n

Now, we can use another formula:

N(Exactly one set) = N(E) + N(M) + N(S) - 2[N(E and M) + N(M and S) + N(E and S)] + 3*N(all three)

Thus, the number of students enrolled in exactly one club is:

124 + 124 + 124 - 2(30 + 26 + 28) + 3(14)

372 - 168 + 42 = 246

Answer: D

Bunuel, what is the trigger here to know that the Number of students taking all three classes needs to be subtracted from the number of students listed for taking two classes each? Is it because it doesn't include the word exactly? The way it reads in my mind is that each 2 set overlap is as stated and then you have the 3 set overlap. So, instead of 16-12-14 for the two set overlaps, I used what was stated in the prompt.

hi GMATWhizTeam

can you please explain why are you adding up sums of three clubs a + b + c where as we are looking for number of student in one club. i am kinda confused with question stem...it asks "how many students enrolled only in one club?" so if we are looking for a number of students only in one of three clubs then why are we summing up three clubs ?

i thought correct answer was 124 by simply applying this formula: Total = A+B+C - (three groups of 2 overlaps) + (section of three group overlap) + neither

thanks ­

Hey Dave,

In the diagram, "a" is the number of students enrolled in "only" English club, "b" is the number of students enrolled "only" in Math club, and "c" is the number of students enrolled in only Science club.

Since, the question is asking us, how many students enrolled in only one club, we need to add a + b + c, because these people are enrolled "only" in either English "or" Math "or" Science club.


Note: You can use your formula to get the same answer also.

Enrolled in two clubs = (30 + 26 + 28)/2 = 42 [note I am dividing this by 2, because we are counting each cases twice and we need it only once]

Enrolled in three clubs = 14

Total = Enrolled only in one club + enrolled in two clubs + enrolled in three clubs + not enrolled in any club

345 = Enrolled only in one club + 42 + 14 + 43

Enrolled only in one club = 246­

• Number of students who enrolled in at least one of the clubs = 345 – 43 = 302
• Number of students enrolled in only English and math club = 30 – 14 =16
• Number of students enrolled in only Math and Science club = 26 – 14 =12
• Number of students enrolled in only Science and English club = 28 – 14 =14
Then the No of students in English club=E+16+14+14
No of students in Science club= S+12+14+14
in math club= M+16+12+14
Total students in any club will be
E+M+S+16+14+12+14=302
E+M+S=302-56
=246

Answer D

345= only E+ only M + only S + 16+ 12+14+14+43.
Only E+ only M + only S = 345-99= 246. Answer option D.

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