Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized for You
we will pick new questions that match your level based on your Timer History
Track Your Progress
every week, we’ll send you an estimated GMAT score based on your performance
Practice Pays
we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
It appears that you are browsing the GMAT Club forum unregistered!
Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club
Registration gives you:
Tests
Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.
Applicant Stats
View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more
Books/Downloads
Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!
Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:
In a survey about the popularity of magazines A,B and C [#permalink]
24 Aug 2008, 20:45
1
This post was BOOKMARKED
This topic is locked. If you want to discuss this question please re-post it in the respective forum.
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
Re: another set question [#permalink]
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 -ab-bc-ca - 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 _________________
Re: another set question [#permalink]
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 -ab-bc-ca - 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 ....
Re: another set question [#permalink]
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 -ab-bc-ca - 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. _________________
Re: another set question [#permalink]
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 -30-10= 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= 100-80=20
What is OA? _________________
Your attitude determines your altitude Smiling wins more friends than frowning
Re: another set question [#permalink]
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.
Re: another set question [#permalink]
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 -30-10= 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= 100-80=20
What is OA?
I think ABC should be 11... to maximize the # of non-liking pups
Re: another set question [#permalink]
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 -30-10= 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= 100-80=20
What is OA?
I think ABC should be 11... to maximize the # of non-liking 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. _________________
Your attitude determines your altitude Smiling wins more friends than frowning
Re: another set question [#permalink]
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:
Re: another set question [#permalink]
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"
Re: another set question [#permalink]
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 _________________
Your attitude determines your altitude Smiling wins more friends than frowning
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. _________________
Your attitude determines your altitude Smiling wins more friends than frowning
Re: another set question [#permalink]
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?
Re: another set question [#permalink]
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?
Re: another set question [#permalink]
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 A-only by increasing ABC-only, then you will run into -ve readers for mag B-only
Re: another set question [#permalink]
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