Cannot get a proper solution for this one

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Cannot get a proper solution for this one

by Architj » Wed Apr 22, 2015 7:35 pm
Club Number of Students
Chess 40
Drama 30
Math 25

The table above shows the number of students in three clubs at McAuliffe School. Although no student is in all three clubs, 10 students are in both chess and drama, 5 students are in both chess and math, and 6 students are in both drama and math. How many different students are in the three clubs?
(A) 68
(B) 69
(C) 74
(D) 79
(E) 84

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by sidceg » Wed Apr 22, 2015 9:11 pm
Answer is C

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by GMATGuruNY » Wed Apr 22, 2015 9:16 pm
Architj wrote:Club Number of Students
Chess 40
Drama 30
Math 25

The table above shows the number of students in three clubs at McAuliffe School. Although no student is in all three clubs, 10 students are in both chess and drama, 5 students are in both chess and math, and 6 students are in both drama and math. How many different students are in the three clubs?
(A) 68
(B) 69
(C) 74
(D) 79
(E) 84
One formula for 3 overlapping groups:

T = A + B + C - (AB + AC + BC) - 2(ABC)

The big idea with overlapping group problems is to SUBTRACT THE OVERLAPS.
When we add together everyone in A, everyone in B, and everyone in C:
Those in exactly 2 of the groups (AB+AC+BC) are counted twice, so they need to be subtracted from the total ONCE.
Those in all 3 groups (ABC) are counted 3 times, so they need to be subtracted from the total TWICE.
By subtracting the overlaps, we ensure that no one is overcounted.

In the problem above:
Let total = T.
Chess = 40.
Drama = 30.
Math = 25.
Exactly 2 of the groups = 10+5+6.
All 3 groups = 0.

Plugging these values into the formula, we get:
T = 40 + 30 + 25 - (10+5+6) - 2*0
T = 74.

The correct answer is C.

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