guerrero wrote:A school has 3 classes, math class has 14 students. English class has 10 students, PE class has 11 students. There are 20 students taking only one class, 3 students are taking all three classes. How many students are taking two classes?
A) 3
B) 6
C) 9
D) 18
E)20
could any one show me the solution using 3 overlapping groups formula .Mitch's way ..
T = A + B + C - (AB + AC + BC) - 2(ABC) ... I am getting a wrong answer
OA
A
Here is the 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 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:
A = Math = 14.
B = English = 10.
C = PE = 11.
Since the number of students taking exactly 2 classes is unknown, AB + AC + BC = x.
Since 3 students take all three classes, ABC = 3.
Plugging these values into the formula, we get:
T = 14 + 10 + 11 - x - 2(3)
T = 29 - x.
The following must also be true:
T = (students taking exactly 1 class) + (students taking exactly 2 classes) + (students taking all 3 classes).
Since 20 students are taking exactly 1 class, x students are taking exactly 2 classes, and 3 students are taking all 3 classes, we get:
T = 20 + x + 3.
T = 23 + x
Since T = 29 - x and T = 23 + x, we get:
29 - x = 23 + x
6 = 2x
x = 3.
The correct answer is
A.
Check here for another problem about triple-overlapping groups:
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