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# Among 200 people, 56% like strawberry jam, 44% like apple jam, and 40%

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Manager
Joined: 14 Jul 2014
Posts: 94
Re: Among 200 people, 56% like strawberry jam, 44% like apple jam, and 40%  [#permalink]

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27 Mar 2015, 23:26
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.

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.

Hi Bunuel

Can I use Gclub formula & is this approach correct

Total = S + A + R - (Sum of Exactly 2 group overlaps) - 2 (All 3)

s+a+r-sa-sr-ar-2sar = 100 %

thus r-sr-ar-2sar = 100-[56+44-30] = 30

thus 30% of 200 = 60.

Thanks
Current Student
Joined: 17 Oct 2015
Posts: 28
Re: Among 200 people, 56% like strawberry jam, 44% like apple jam, and 40%  [#permalink]

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20 Jan 2016, 08:57
EMPOWERgmatRichC

Could you please help here? I've tried to use the tic tac toe board, but it didn't work...

I've consider S and A as only one at the first row and let the R alone at the first line.

Regards!
Manager
Joined: 01 Mar 2014
Posts: 129
Schools: Tepper '18
Re: Among 200 people, 56% like strawberry jam, 44% like apple jam, and 40%  [#permalink]

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17 Mar 2016, 01:43
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.

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???
Intern
Joined: 03 Feb 2017
Posts: 25
Re: Among 200 people, 56% like strawberry jam, 44% like apple jam, and 40%  [#permalink]

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06 Feb 2017, 21:24
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

If we can find the total number of people who like either strawberry jam or apple jam or both and subtract it from total number of people, it gives us the number of people who like only raspberry jam. No equation is required but a Venn diagram could be a great help to solve this.

Total number of people who like either strawberry jam or apple jam or both = 112 + 88 - 60 = 140 (60 being common in both strawberry and apple so we subtract it )
No of people left ie. No of people who don't like strawberry jam or apple jam = 200 -140 = 60 B
Intern
Joined: 09 May 2017
Posts: 1
Re: Among 200 people, 56% like strawberry jam, 44% like apple jam, and 40%  [#permalink]

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14 Jul 2017, 04:24
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

The questions asks "Largest possible number of people". Hence, according to my logic, in order to maximise the number of people who like raspberry jam, we should make all the 60 people like strawberry and Apple. Hence, the answer comes as 80 rather than 60. Can someone tell if there is anything wrong with my logic??
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Math Expert
Joined: 02 Sep 2009
Posts: 47157
Re: Among 200 people, 56% like strawberry jam, 44% like apple jam, and 40%  [#permalink]

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14 Jul 2017, 04:35
LatentProdigy 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

The questions asks "Largest possible number of people". Hence, according to my logic, in order to maximise the number of people who like raspberry jam, we should make all the 60 people like strawberry and Apple. Hence, the answer comes as 80 rather than 60. Can someone tell if there is anything wrong with my logic??

The case you are considering is not possible because the total number of people in this case will be more than 200: 112 + 88 - 60 + 80 = 220.

There are several correct solution on previous two pages. For example, here: https://gmatclub.com/forum/among-200-pe ... l#p1385279
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Joined: 28 Jun 2015
Posts: 296
Concentration: Finance
GPA: 3.5
Re: Among 200 people, 56% like strawberry jam, 44% like apple jam, and 40%  [#permalink]

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14 Jul 2017, 05:35
Let Strawberry = S, Apple = A, Raspberry = R, Total = N.

S = 112, A = 88, R = 80
Overlap = (112+88+80) - 200 = 280 - 200 = 80.

S&A = 60.

R only: 280-60-(R&A and R&S) = 200
R&A and R&S = 20

R only = R - (R&A and R&S) = 80 - 20 = 60. Ans - B.
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Intern
Joined: 29 Aug 2016
Posts: 3
Re: Among 200 people, 56% like strawberry jam, 44% like apple jam, and 40%  [#permalink]

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26 Jul 2017, 08:47
Hi Bunuel

I got the answer correct. I just need a little clarification. The raspberry circle consists of 80 ppl. If 60 can be the highest in the gray portion, do 20 ppl fall in that portion of the raspberry circle that overlapping with strawberry and/or apple circle?
Math Expert
Joined: 02 Sep 2009
Posts: 47157
Re: Among 200 people, 56% like strawberry jam, 44% like apple jam, and 40%  [#permalink]

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26 Jul 2017, 08:58
priyankachhabra01 wrote:
Hi Bunuel

I got the answer correct. I just need a little clarification. The raspberry circle consists of 80 ppl. If 60 can be the highest in the gray portion, do 20 ppl fall in that portion of the raspberry circle that overlapping with strawberry and/or apple circle?

Yes, the remaining 20 could be in strawberry and/or apple circle.
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Manager
Joined: 12 Feb 2015
Posts: 56
Location: India
GPA: 3.84
Among 200 people, 56% like strawberry jam, 44% like apple jam, and 40%  [#permalink]

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10 Aug 2017, 11:47
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:

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.

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.

Attachment:
Jams.png

Senior Manager
Joined: 02 Apr 2014
Posts: 486
GMAT 1: 700 Q50 V34
Among 200 people, 56% like strawberry jam, 44% like apple jam, and 40%  [#permalink]

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01 Feb 2018, 12:59
First let us find number of people either strawberry or apple or both = 56 + 44 - 30 = 70%

Line diagram should help us find the answer.

S + A (from left)
|<------------70---------------><---30--->|
|<------------60-----><-10><--30--->| 40 (raspberry from right), as we can see minimum overlap can be 10%, 30% of people can like rasp but neither S nor A.
30% of 200 = 60 => (B)
Among 200 people, 56% like strawberry jam, 44% like apple jam, and 40% &nbs [#permalink] 01 Feb 2018, 12:59

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