Bunuel wrote:
aaron22197 wrote:
Among 200 people, 56% like strawberry jam, 44% like apple jam, and 40% like raspberry jam. If 30% of the people like both strawberry and apple jam, what is the largest possible number of people who like raspberry jam but do not like either strawberry or apple jam?
A. 20
B. 60
C. 80
D. 86
E. 92
M08-7
Look at the diagram below:
Attachment:
Jams.png
Notice that "30% of the people like both strawberry and apple jam" doesn't mean that among these 30% (60) can not be some people who like raspberry as well. Both strawberry and apple jam is the intersection of these two groups, if we refer to the diagram it's the yellow segment in it.
Next, no formula is needed to solve this question: 112 like strawberry jam, 88 like apple jam, 60 people like both strawberry and apple jam. So the # of people who like either strawberry or apple (or both) is 112+88-60=140 (on the diagram it the area covered by Strawberry and Apple). So there are TOTAL of 200-140=60 people left who "do not like either strawberry or apple jam". Can ALL these 60 people like raspberry? As \(Raspberry=80\geq{60}\), then why not! So, maximum # of people who like raspberry and don't like either strawberry or apple jam is 60 (grey segment on the diagram). Notice here that in this case the # of people who like none of the 3 jams (area outside three circles) will be zero.
Answer: B.
Side note: minimum # of people who like raspberry and don't like either strawberry or apple jam would be zero (consider Raspberry circle inside Strawberry and/or Apples circles). In this case those 60 people (who "do not like either strawberry or apple jam") will be those who like none of the 3 jams.
This is great. i also got the correct answer but used a different approach :
A = no. of people who like only Strawberry
B = no. of people who like only apple
C = no. of people who like only raspberry
D = no. of people who like both strawberry and apple but not raspberry
E = no. of people who like both strawberry and raspberry but not apple
F = no. of people who like both apple and raspberry but not strawberry
G = no. of people who like all 3
A+D+E+G =112
B+D+G+F = 88
C+E+G+F = 80
D+G = 60
A+B+C+2(D+E+F) + 3G = 280
A+B+C+D+E+F+G =200
Subtracting the 2 ->
(D+E+F) +2G = 80, which can be written as -> (D+G) +(E+F) + G =80 -> 60 +E+F+G = 80
=> E+F+G = 20
Also from above -> C+E+F+G = 80 => C+ 20 = 8
=> C= 60...
Is this method correct???