Author 
Message 
TAGS:

Hide Tags

Manager
Joined: 22 May 2007
Posts: 211

Among 200 people, 56% like strawberry jam, 44% like apple jam, and 40% [#permalink]
Show Tags
05 Jul 2008, 07:05
Question Stats:
62% (01:54) correct 38% (01:35) wrong based on 617 sessions
HideShow timer Statistics
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 M087
Official Answer and Stats are available only to registered users. Register/ Login.



VP
Joined: 03 Apr 2007
Posts: 1242

Re: Among 200 people, 56% like strawberry jam, 44% like apple jam, and 40% [#permalink]
Show Tags
Updated on: 05 Jul 2008, 10:29
Attachment:
sheetal_venn.GIF [ 12.71 KiB  Viewed 15075 times ]
Originally posted by goalsnr on 05 Jul 2008, 08:38.
Last edited by goalsnr on 05 Jul 2008, 10:29, edited 2 times in total.



Senior Manager
Joined: 29 Mar 2008
Posts: 341

Re: Among 200 people, 56% like strawberry jam, 44% like apple jam, and 40% [#permalink]
Show Tags
06 Jul 2008, 09:26
Like strawberry = 112 Like Apple= 88 Like Rasberry = 80 Like straberry and apple = 60 As shown in the diagram: (52y)+(60x)+(28z)+(80xyz)+x+y+z=200 (1) We need (80xyz) = Like Rasberry but do not like either Straberry or Apple. From (1), (80xyz) = 200(52+60+28) =60 Hence (B)
Attachments
venn.jpg [ 13.93 KiB  Viewed 15022 times ]
_________________
To find what you seek in the road of life, the best proverb of all is that which says: "Leave no stone unturned." Edward Bulwer Lytton



CEO
Joined: 17 Nov 2007
Posts: 3486
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth)  Class of 2011

Re: Among 200 people, 56% like strawberry jam, 44% like apple jam, and 40% [#permalink]
Show Tags
06 Jul 2008, 11:53
Fast way. In order to transform sum 112+88+80=280 to 200 we have to exclude double counting: 28060x=200 > x=20 (raspberry jam + any other jam) > 8020=60 only raspberry jam
_________________
HOT! GMAT TOOLKIT 2 (iOS) / GMAT TOOLKIT (Android)  The OFFICIAL GMAT CLUB PREP APP, a musthave app especially if you aim at 700+  PrepGame



Math Expert
Joined: 02 Sep 2009
Posts: 46302

Among 200 people, 56% like strawberry jam, 44% like apple jam, and 40% [#permalink]
Show Tags
22 Jul 2014, 01:40
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
M087 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+8860=140 (on the diagram it the area covered by Strawberry and Apple). So there are TOTAL of 200140=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. Attachment:
Jams.png [ 12.55 KiB  Viewed 14894 times ]
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Manager
Joined: 28 Apr 2014
Posts: 248

Re: Among 200 people, 56% like strawberry jam, 44% like apple jam, and 40% [#permalink]
Show Tags
22 Jul 2014, 06:30
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
M087 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+8860=140 (on the diagram it the area covered by Strawberry and Apple). So there are TOTAL of 200140=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. Bunuel how did you assume in this question that there would be atleast some people liking all the three jams ?



Math Expert
Joined: 02 Sep 2009
Posts: 46302

Re: Among 200 people, 56% like strawberry jam, 44% like apple jam, and 40% [#permalink]
Show Tags
22 Jul 2014, 06:50
himanshujovi wrote: 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
M087 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+8860=140 (on the diagram it the area covered by Strawberry and Apple). So there are TOTAL of 200140=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. Bunuel how did you assume in this question that there would be atleast some people liking all the three jams ? Saying that there are no such overlap would be an assumption... Which part of my solution are you referring exactly?
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Manager
Joined: 28 Apr 2014
Posts: 248

Re: Among 200 people, 56% like strawberry jam, 44% like apple jam, and 40% [#permalink]
Show Tags
22 Jul 2014, 19:14
Quote: 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+8860=140 (on the diagram it the area covered by Strawberry and Apple). So there are TOTAL of 200140=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. I am referring to the venn diagram. Basically the main premise of the solution that these people comprise of three intersecting sets. In other words , for all overlapping sets question do we assume that the distribution as per the above Venn diagram , unless ofcourse mentioned otherwise



SVP
Status: The Best Or Nothing
Joined: 27 Dec 2012
Posts: 1837
Location: India
Concentration: General Management, Technology
WE: Information Technology (Computer Software)

Re: Among 200 people, 56% like strawberry jam, 44% like apple jam, and 40% [#permalink]
Show Tags
22 Jul 2014, 20:29
Refer diagram below: We require to find the maximum value of the pink shaded region Setting up the equation: 112 + 28b + b + 80  (a+b+c) = 200 80  (a+b+c) = 60 Answer = B
Attachments
rasp.png [ 9.78 KiB  Viewed 12799 times ]
_________________
Kindly press "+1 Kudos" to appreciate



Math Expert
Joined: 02 Sep 2009
Posts: 46302

Re: Among 200 people, 56% like strawberry jam, 44% like apple jam, and 40% [#permalink]
Show Tags
23 Jul 2014, 02:33
himanshujovi wrote: Quote: 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+8860=140 (on the diagram it the area covered by Strawberry and Apple). So there are TOTAL of 200140=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. I am referring to the venn diagram. Basically the main premise of the solution that these people comprise of three intersecting sets. In other words , for all overlapping sets question do we assume that the distribution as per the above Venn diagram , unless ofcourse mentioned otherwise____________ Yes. How else?
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Manager
Joined: 28 Apr 2014
Posts: 248

Re: Among 200 people, 56% like strawberry jam, 44% like apple jam, and 40% [#permalink]
Show Tags
23 Jul 2014, 12:08
Somethin else as in say Set A and set B having some common intersection with set C being totally a sub set of A/B or set C being a altogether different set with no intersection
Sent from my iPhone using Tapatalk



Math Expert
Joined: 02 Sep 2009
Posts: 46302

Re: Among 200 people, 56% like strawberry jam, 44% like apple jam, and 40% [#permalink]
Show Tags
23 Jul 2014, 12:15



Manager
Joined: 10 Mar 2014
Posts: 212

Re: Among 200 people, 56% like strawberry jam, 44% like apple jam, and 40% [#permalink]
Show Tags
08 Aug 2014, 02:13
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
M087 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+8860=140 (on the diagram it the area covered by Strawberry and Apple). So there are TOTAL of 200140=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. HI Bunuel, Just one query. Cant I say People who like only strawberry = 11260 = 52 people who like only applejam = 8860 = 28 so here 80 people who like either strawberry or applejam. i did it in this way and got answer 80. Could you please clarify why cant we use this logic? Thanks



SVP
Status: The Best Or Nothing
Joined: 27 Dec 2012
Posts: 1837
Location: India
Concentration: General Management, Technology
WE: Information Technology (Computer Software)

Re: Among 200 people, 56% like strawberry jam, 44% like apple jam, and 40% [#permalink]
Show Tags
08 Aug 2014, 03:16
PathFinder007 wrote: 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
HI Bunuel,
Just one query.
Cant I say
People who like only strawberry = 11260 = 52 people who like only applejam = 8860 = 28
so here 80 people who like either strawberry or applejam.
i did it in this way and got answer 80.
Could you please clarify why cant we use this logic?
Thanks Not Bunuel, but lets try If 30% of the people like both strawberry and apple jamThe highlighted portion is ignored in this calculation
_________________
Kindly press "+1 Kudos" to appreciate



Manager
Joined: 10 Mar 2014
Posts: 212

Re: Among 200 people, 56% like strawberry jam, 44% like apple jam, and 40% [#permalink]
Show Tags
08 Aug 2014, 03:38
HI Paresh,
here i am subtracting 30 % OF 200 = 60 FROM 112 and 88 so i am considering this highlighted statement. So What you mean that The highlighted portion is ignored in this calculation. please clarify this
Thanks



SVP
Status: The Best Or Nothing
Joined: 27 Dec 2012
Posts: 1837
Location: India
Concentration: General Management, Technology
WE: Information Technology (Computer Software)

Re: Among 200 people, 56% like strawberry jam, 44% like apple jam, and 40% [#permalink]
Show Tags
09 Aug 2014, 05:20
PathFinder007 wrote: HI Paresh,
here i am subtracting 30 % OF 200 = 60 FROM 112 and 88 so i am considering this highlighted statement. So What you mean that The highlighted portion is ignored in this calculation. please clarify this
Thanks Hello PathFinder007, Kindly refer to Bunuel's diagram. (Yellow shaded) I've attached herewith 30% (60) is overlap of BOTH Strawberry & Apple; so it cannot be subtracted as stated in your answer So, only strawberry would be less than 52, only apple would be less than 28. How much less?? Its unknown, that's why variables "a" & "b" are taken respectively (Refer my earlier post diagram) Also common to all "c" has to be considered. That's why only Raspberry would be less than 80 In this problem, as three products are give, 3! + 1 = 7 possibilities have to be considered 1. Only Strawberry 2. Only Apple 3. Only Raspberry 4. Strawberry+Apple 5. Strawberry+Raspberry 6. Apple+Raspberry 7. Apple+Raspberry+Strawberry All possibilities have to be considered with caution to avoid duplicationFor me, Venn diagram helps a lot
Attachments
rasp.png [ 9.78 KiB  Viewed 12331 times ]
Jams.png [ 12.55 KiB  Viewed 12351 times ]
_________________
Kindly press "+1 Kudos" to appreciate



Intern
Joined: 15 Apr 2014
Posts: 10
Concentration: Strategy, Marketing
WE: Marketing (Advertising and PR)

Re: Among 200 people, 56% like strawberry jam, 44% like apple jam, and 40% [#permalink]
Show Tags
08 Dec 2014, 23:54
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
M087 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+8860=140 (on the diagram it the area covered by Strawberry and Apple). So there are TOTAL of 200140=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. Please confirm this...either  or implies that atleast one of the condition must hold. In other words, both condition can also be true. either not A or not B implies only one condition needs to be false and not both. In other words, any one one condition can only be false. Is this the logic taken in above question??
_________________
May everyone succeed in their endeavor. God Bless!!!



Math Expert
Joined: 02 Sep 2009
Posts: 46302

Re: Among 200 people, 56% like strawberry jam, 44% like apple jam, and 40% [#permalink]
Show Tags
09 Dec 2014, 08:45
enders wrote: 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
M087 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+8860=140 (on the diagram it the area covered by Strawberry and Apple). So there are TOTAL of 200140=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. Please confirm this...either  or implies that atleast one of the condition must hold. In other words, both condition can also be true. either not A or not B implies only one condition needs to be false and not both. In other words, any one one condition can only be false. Is this the logic taken in above question?? _____________ Yes, that's true.
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Intern
Joined: 25 Jun 2014
Posts: 40

Re: Among 200 people, 56% like strawberry jam, 44% like apple jam, and 40% [#permalink]
Show Tags
16 Dec 2014, 19:36
Hi Bunuel, Is my approach correct? a is number of people who like both Strawberry and Rasberry b is number of people who like both Apple and Rasberry c is number of people who like all three We have 100% = 56% + 44% + 40%  30% ab +c => 10+c = a+ b In order to have maximum of people who only like Rasberry we have to minimize a,b,c => a+b = 10 and c =0 => Answer is (40%  10% )*200 = 60 Thank you



Director
Joined: 07 Aug 2011
Posts: 565
Concentration: International Business, Technology

Re: Among 200 people, 56% like strawberry jam, 44% like apple jam, and 40% [#permalink]
Show Tags
03 Jan 2015, 19:59
walker wrote: Fast way.
In order to transform sum 112+88+80=280 to 200 we have to exclude double counting: 28060x=200 > x=20 (raspberry jam + any other jam) > 8020=60 only raspberry jam walker Shouldnt it be 280 (60+ x ) + ( All three) why did not you count ALL THREE ?
_________________
Thanks, Lucky
_______________________________________________________ Kindly press the to appreciate my post !!




Re: Among 200 people, 56% like strawberry jam, 44% like apple jam, and 40%
[#permalink]
03 Jan 2015, 19:59



Go to page
1 2
Next
[ 32 posts ]



