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In a survey about the popularity of magazines A,B and C [#permalink]
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24 Aug 2008, 20:45
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In a survey about the popularity of magazines A,B and C among 100 pupils, 39 pupils like A, 36 like B, 45 like C, 14 like A and B only, 11 like B and C only, 5 like A and C only. Find the maximum number of pupils who do not like any of the magazines.
Please help to explain the steps to solve this. Thanks



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Re: another set question [#permalink]
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24 Aug 2008, 21:08
gmatcraze wrote: In a survey about the popularity of magazines A, B and C among 100 pupils, 39 pupils like A, 36 like B, 45 like C, 14 like A and B only, 11 like B and C only, 5 like A and C only. Find the maximum number of pupils who do not like any of the magazines.
Please help to explain the steps to solve this. Thanks this is good one to have maximum people disliking nither a nor b nor c, people liking all should equal to 0. all = a+b+c abbcca  2(all) + neither a nor b nor c 100 = 39 + 36+45  14  11 15  2(0) + neither a nor b nor c neither a nor b nor c = 10
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Re: another set question [#permalink]
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24 Aug 2008, 21:23
GMAT TIGER wrote: gmatcraze wrote: In a survey about the popularity of magazines A, B and C among 100 pupils, 39 pupils like A, 36 like B, 45 like C, 14 like A and B only, 11 like B and C only, 5 like A and C only. Find the maximum number of pupils who do not like any of the magazines.
Please help to explain the steps to solve this. Thanks this is good one to have maximum people disliking nither a nor b nor c, people liking all should equal to 0. all = a+b+c abbcca  2(all) + neither a nor b nor c 100 = 39 + 36+45  14  11 15  2(0) + neither a nor b nor c neither a nor b nor c = 10 this was exactly my approach to solve this question .... but unfortunately, the OA for this question seems to be something different .... so am wondering if this approach is right? Got this question from another forum ....



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Re: another set question [#permalink]
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24 Aug 2008, 22:21
gmatcraze wrote: GMAT TIGER wrote: gmatcraze wrote: In a survey about the popularity of magazines A, B and C among 100 pupils, 39 pupils like A, 36 like B, 45 like C, 14 like A and B only, 11 like B and C only, 5 like A and C only. Find the maximum number of pupils who do not like any of the magazines.
Please help to explain the steps to solve this. Thanks this is good one to have maximum people disliking nither a nor b nor c, people liking all should equal to 0. all = a+b+c abbcca  2(all) + neither a nor b nor c 100 = 39 + 36+45  14  11 15  2(0) + neither a nor b nor c neither a nor b nor c = 10 this was exactly my approach to solve this question .... but unfortunately, the OA for this question seems to be something different .... so am wondering if this approach is right? Got this question from another forum .... i guess i missed someting. work tomorrow.
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Re: another set question [#permalink]
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24 Aug 2008, 22:50
gmatcraze wrote: In a survey about the popularity of magazines A,B and C among 100 pupils, 39 pupils like A, 36 like B, 45 like C, 14 like A and B only, 11 like B and C only, 5 like A and C only. Find the maximum number of pupils who do not like any of the magazines.
Please help to explain the steps to solve this. Thanks Nof people like magzine= a+b+c (ab+bc+ca)  2(abc) = 39+36+45 (14+11+5)2(5) = 120 3010= 80 [ to get max number of pupils not like magzine > min number of pupils like the magzine abc should be maximum.. possible maximum is "5" 5 pupils like A and C ] Max people doesn't like magzine= 10080=20 What is OA?
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Re: another set question [#permalink]
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25 Aug 2008, 05:15
gmatcraze wrote: In a survey about the popularity of magazines A,B and C among 100 pupils, 39 pupils like A, 36 like B, 45 like C, 14 like A and B only, 11 like B and C only, 5 like A and C only. Find the maximum number of pupils who do not like any of the magazines.
Please help to explain the steps to solve this. Thanks Is the OA 32 ? If that is correct, will explain BTW... it is a damn good trap question. Only because you mentioned in one of the posts that the OA is something else, i even thought of it the way i did.



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Re: another set question [#permalink]
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25 Aug 2008, 05:16
x2suresh wrote: gmatcraze wrote: In a survey about the popularity of magazines A,B and C among 100 pupils, 39 pupils like A, 36 like B, 45 like C, 14 like A and B only, 11 like B and C only, 5 like A and C only. Find the maximum number of pupils who do not like any of the magazines.
Please help to explain the steps to solve this. Thanks Nof people like magzine= a+b+c (ab+bc+ca)  2(abc) = 39+36+45 (14+11+5)2(5) = 120 3010= 80 [ to get max number of pupils not like magzine > min number of pupils like the magzine abc should be maximum.. possible maximum is "5" 5 pupils like A and C ] Max people doesn't like magzine= 10080=20 What is OA? I think ABC should be 11... to maximize the # of nonliking pups



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Re: another set question [#permalink]
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25 Aug 2008, 05:28
bhushangiri wrote: x2suresh wrote: gmatcraze wrote: In a survey about the popularity of magazines A,B and C among 100 pupils, 39 pupils like A, 36 like B, 45 like C, 14 like A and B only, 11 like B and C only, 5 like A and C only. Find the maximum number of pupils who do not like any of the magazines.
Please help to explain the steps to solve this. Thanks Nof people like magzine= a+b+c (ab+bc+ca)  2(abc) = 39+36+45 (14+11+5)2(5) = 120 3010= 80 [ to get max number of pupils not like magzine > min number of pupils like the magzine abc should be maximum.. possible maximum is "5" 5 pupils like A and C ] Max people doesn't like magzine= 10080=20 What is OA? I think ABC should be 11... to maximize the # of nonliking pups thats impossible.. 5 like A and C only .. how come 11 people like A and B and C only that leads to >11 like A and C only which is not true.
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Re: another set question [#permalink]
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25 Aug 2008, 05:39
The answer should be 32, although no way I could answer this on the real test in 2 mins. So, magazine B has the smallest number of readers. We have to assume that B has no 'unique' readers, i.e., all 36 read at least one more magazine. 11 read B and C, 14 read B and A, therefore 11 read all three. From this: abc=11 ab=14 bc=11 ac=5 a=9 c=18 b=0
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Re: another set question [#permalink]
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25 Aug 2008, 05:42
x2suresh wrote: thats impossible.. 5 like A and C only .. how come 11 people like A and B and C only that leads to >11 like A and C only which is not true. 5 like A and C only. That does not say anything about how many like A,b, and C. There is a difference between " 5 like A and C only" and " Only 5 like A and C"



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Re: another set question [#permalink]
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25 Aug 2008, 06:06
bhushangiri wrote: x2suresh wrote: thats impossible.. 5 like A and C only .. how come 11 people like A and B and C only that leads to >11 like A and C only which is not true. 5 like A and C only. That does not say anything about how many like A,b, and C. There is a difference between " 5 like A and C only" and " Only 5 like A and C" Agreed... I am too smart today.. How did you come up with 11 why not >11
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Re: another set question [#permalink]
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25 Aug 2008, 06:58
x2suresh wrote: How did you come up with 11 why not >11
#B is the smallest and = 36. And we need to attain #B with max shared likes. The min Bonly is 0. So ABConly + ABonly + BConly + Bonly = 36 ABConly + 14 + 11 + Bonly = 36 Minimum of Bonly and Max of ABConly will miminize the total number who like atleast 1. So ABC can't be more than 11 unless Bonly can be negative. For some reason i am unable to attach the venn diagram.



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Re: another set question [#permalink]
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25 Aug 2008, 07:02
bhushangiri wrote: x2suresh wrote: How did you come up with 11 why not >11
#B is the smallest and = 36. And we need to attain #B with max shared likes. The min Bonly is 0. So ABConly + ABonly + BConly + Bonly = 36 ABConly + 14 + 11 + Bonly = 36 Minimum of Bonly and Max of ABConly will miminize the total number who like atleast 1. So ABC can't be more than 11 unless Bonly can be negative. For some reason i am unable to attach the venn diagram. Got it. Yeap!! looks like there is some problem with site.. I don't see avatars are populated and kudos.. not populated... Hope administrators will fix the issue ASAP.
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Re: another set question [#permalink]
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25 Aug 2008, 09:28
Can you explain again real quick how you initally know that it is mag B that must have 0 unique readers? Is it just because it has the smallest number of total readers and so, of the three, it is easiest to make the 0 unique readers assumption for the one with the smallest total, hence B?
Thanks



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Re: another set question [#permalink]
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25 Aug 2008, 10:17
Only B can have 0 unique readers, because it has the smallest number of readers.
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Re: another set question [#permalink]
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25 Aug 2008, 11:21
So when I was looking at this problem earlier, I tried setting the other 2 mags to 0 unique readers and making the same list of different groups of readers according to the same method and those didn't work because you run into negative readers at some point which obviously isn't possible. Am I seeing the same thing?
Thanks



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Re: another set question [#permalink]
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25 Aug 2008, 11:24
boy !! ; this was real good



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Re: another set question [#permalink]
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25 Aug 2008, 11:59
Saywhat75 wrote: So when I was looking at this problem earlier, I tried setting the other 2 mags to 0 unique readers and making the same list of different groups of readers according to the same method and those didn't work because you run into negative readers at some point which obviously isn't possible. Am I seeing the same thing?
Thanks Yes. If you try to make 0 unique for mag Aonly by increasing ABConly, then you will run into ve readers for mag Bonly



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Re: another set question [#permalink]
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26 Aug 2008, 12:35
39 like A. Lets see how we can maximize overlaps so that we can maximize the no. of pupils who do not like any magazines. 14 like A/B. 5 like A/C. This leaves us with 39(14 + 5) = 20 that like A only. How can we further reduce this number 20. By assuming that some pupils like A/B/C. However, we have to be careful in choosing that number because we do not want to exceed the allotted numbers to each magazine. 36 like B. Out of this 14 like A/B and 11 like A/C. Therefore, only 11 would like B alone. 45 like C. Out of this 5 like A/C and 11 like B/C. Therefore only 29 could possibly like C alone. Now we can manipulate a bit. Lets say, 11 like A/B/C. This implies, 0 pupils like B alone; 18 pupils like C alone; 9 pupils like A alone. Therefore, the max. no. of pupils who do not like any mag. is 100  (9+0+18+14+11+11+5) = 32



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Re: another set question [#permalink]
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26 Aug 2008, 18:00
bhushangiri wrote: x2suresh wrote: How did you come up with 11 why not >11
#B is the smallest and = 36. And we need to attain #B with max shared likes. The min Bonly is 0. So ABConly + ABonly + BConly + Bonly = 36 ABConly + 14 + 11 + Bonly = 36 Minimum of Bonly and Max of ABConly will miminize the total number who like atleast 1. So ABC can't be more than 11 unless Bonly can be negative. For some reason i am unable to attach the venn diagram. wow!!! good one kudos to bhushangiri  good explanation kudos to gmatcraze for the question too bad the kudos button is invisible




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