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01 Apr 2013, 11:43
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
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57% (02:18) correct
43%
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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 .
T = A + B + C - (AB + AC + BC) - 2(ABC) ... I am getting a wrong answer
thanks !
A) 3
B) 6
C) 9
D) 18
E)20
could any one show me the solution using 3 overlapping groups formula .
T = A + B + C - (AB + AC + BC) - 2(ABC) ... I am getting a wrong answer
thanks !
01 Apr 2013, 12:52
Kudos
Bookmarks
This could be solved easier:
(1) There are 14+10+11 = 35 seats to fill (some students occupy only 1 seat, others 2, others 3).
(2) 20 students occupy only 1 seat: 35-20 = 15... now we have 15 seats to fill
(3) 3 students occupy 3 seats: 15-9 = 6 seats left
(4) We have 6 seats for students that occupy 2 seats ---> solution is 3 students!!
Answer A
(1) There are 14+10+11 = 35 seats to fill (some students occupy only 1 seat, others 2, others 3).
(2) 20 students occupy only 1 seat: 35-20 = 15... now we have 15 seats to fill
(3) 3 students occupy 3 seats: 15-9 = 6 seats left
(4) We have 6 seats for students that occupy 2 seats ---> solution is 3 students!!
Answer A
02 Apr 2013, 10:11
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If we put up a Venn diagram of the situation, we get something like this :
venn.jpg [ 55.63 KiB | Viewed 11023 times ]
With :
A : the number of students only attending math class ;
B : the number of students only attending english class ;
C : the number of students only attending PE class ;
x : the number of students attending both math and english ;
y : the number of students attending both math and PE ;
z : the number of students attending both english and PE.
If we use the data we're given, we get the following system :
14 students attend math classes => \(x + y + 3 + A = 14\) (1)
10 students attend english classes => \(x + z + 3 + B = 10\) (2)
11 students attend PE classes =>\(y + z + 3 + C = 11\) (3)
We also know that there are only 20 students attending only one class, which means that : \(A + B + C = 20\) (4)
If we add equations (1), (2) and (3) and take into account equation (4) we get : \(2*(x + y + z) + 9 + 20 =35\)
The quantity \((x + y + z)\) represents the total number of students attending 2 classes, which is what we're looking for. So completing the computation, we get :
\(2*(x + y + z) + 9 + 20 =35\) => \(2*(x + y + z) = 6\) => \((x + y + z) = 3\) which is answer choice A.
Hope that helped
Attachment:
venn.jpg [ 55.63 KiB | Viewed 11023 times ]
A : the number of students only attending math class ;
B : the number of students only attending english class ;
C : the number of students only attending PE class ;
x : the number of students attending both math and english ;
y : the number of students attending both math and PE ;
z : the number of students attending both english and PE.
If we use the data we're given, we get the following system :
14 students attend math classes => \(x + y + 3 + A = 14\) (1)
10 students attend english classes => \(x + z + 3 + B = 10\) (2)
11 students attend PE classes =>\(y + z + 3 + C = 11\) (3)
We also know that there are only 20 students attending only one class, which means that : \(A + B + C = 20\) (4)
If we add equations (1), (2) and (3) and take into account equation (4) we get : \(2*(x + y + z) + 9 + 20 =35\)
The quantity \((x + y + z)\) represents the total number of students attending 2 classes, which is what we're looking for. So completing the computation, we get :
\(2*(x + y + z) + 9 + 20 =35\) => \(2*(x + y + z) = 6\) => \((x + y + z) = 3\) which is answer choice A.
Hope that helped
