The number covered by A and B will be 17+28 - 12 = 33. (Minus 12 since it's included twice in the 17 and the 28.)
The number covered by A, B and T will be 33 + 3 = 36, leaving 4 outside A, B and T.
Easy from there.
Problem:
In a class of 40 students, 17 have ridden an airplane, 28 have ridden a boat. 10 have ridden a train, 12 have ridden both an airplane and a boat. 3 have ridden a train only and 4 have ridden an airplane only. Some students in the class have not ridden any of the three modes of transportation and an equal number have taken all three.
Let Sets: A = airplane, B= Boat, T = train
I understand some of the number of students to be on the parts of the circles intersecting... there is a little bit confusing on the example given and the solution also... i tried to fix it on (A intersection B )
in the (A intersection B) = the number of students has ridden A and B...
my problem is how it was being separated in to 8 and 4... i could not get the rationale over that part...
thanks for your help mathematicians..
The number covered by A and B will be 17+28 - 12 = 33. (Minus 12 since it's included twice in the 17 and the 28.)
The number covered by A, B and T will be 33 + 3 = 36, leaving 4 outside A, B and T.
Easy from there.
Looks logical to me, and let me know if you happen to find something easier.
Once you know that there are 4 outside A, B and T you know that there are 4 in the common intersection of A, B and T, (which is what you asking).