lucyqin wrote:
rpmodi wrote:
jimmyjamesdonkey wrote:
Chess 40
Drama 30
Math 25
The table above shows the number of students in 3 clubs at McAuliffe School. Although no student is in all 3 clubs, 0 students are in both Chess and Drama, 5 are in both chess and math, and 6 are in both drama and math. How many different students are in the 3 clubs.
A 68
B 69
C 74
D 79
E 84
I am getting 73 ,which is not an option here !
I got 73 as well.
35 students in Chess are taking only Chess, or Chess and Drama.
20 students in Math are taking only Math, or Math and Drama.
24 students in Drama are taking only Drama, or Drama and Chess.
14 students in Math are taking only Math.
Since 0 students are taking both Chess and Drama, add up 35 + 24 + 14 = 73.
If this is not the answer, then can someone explain what I did wrong? Thx
Hello,
I'm finding E here.
IMO you just count here the number of students in just
one club.
Trying to correct what you did:
35 students in Chess are taking only Chess, or Chess and Drama => here only 35 Chess students only OK
14 students in Math are taking only Math. => OK with this
24 students in Drama are taking only Drama, or Drama and Chess => OK with this only 24 Drama students since C and D =0
Now we are left to find the students involved in several clubs.
Since C and D and M = 0, where and is the intersection operator
C and D =0 => (C and M), (D and M) give different students!!
So we add 73 to the number of these different students (5+
6) => 84.
close
