It is currently 22 Jun 2017, 19:06

### GMAT Club Daily Prep

#### 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

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

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# In a survey about the popularity of magazines A,B and C

Author Message
Intern
Joined: 19 Jul 2008
Posts: 23
In a survey about the popularity of magazines A,B and C [#permalink]

### Show Tags

24 Aug 2008, 21:45
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.

SVP
Joined: 29 Aug 2007
Posts: 2473

### Show Tags

24 Aug 2008, 22:08
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.

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
_________________

Gmat: http://gmatclub.com/forum/everything-you-need-to-prepare-for-the-gmat-revised-77983.html

GT

Intern
Joined: 19 Jul 2008
Posts: 23

### Show Tags

24 Aug 2008, 22:23
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.

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 ....
SVP
Joined: 29 Aug 2007
Posts: 2473

### Show Tags

24 Aug 2008, 23:21
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.

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.
_________________

Gmat: http://gmatclub.com/forum/everything-you-need-to-prepare-for-the-gmat-revised-77983.html

GT

SVP
Joined: 07 Nov 2007
Posts: 1800
Location: New York

### Show Tags

24 Aug 2008, 23:50
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.

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?
_________________

Smiling wins more friends than frowning

Manager
Joined: 15 Jul 2008
Posts: 206

### Show Tags

25 Aug 2008, 06:15
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.

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.
Manager
Joined: 15 Jul 2008
Posts: 206

### Show Tags

25 Aug 2008, 06:16
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.

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
SVP
Joined: 07 Nov 2007
Posts: 1800
Location: New York

### Show Tags

25 Aug 2008, 06:28
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.

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.
_________________

Smiling wins more friends than frowning

Senior Manager
Joined: 16 Jul 2008
Posts: 289

### Show Tags

25 Aug 2008, 06:39
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
_________________

http://applicant.wordpress.com/

Last edited by Nerdboy on 25 Aug 2008, 06:43, edited 1 time in total.
Manager
Joined: 15 Jul 2008
Posts: 206

### Show Tags

25 Aug 2008, 06:42
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"
SVP
Joined: 07 Nov 2007
Posts: 1800
Location: New York

### Show Tags

25 Aug 2008, 07:06
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
_________________

Smiling wins more friends than frowning

Manager
Joined: 15 Jul 2008
Posts: 206

### Show Tags

25 Aug 2008, 07:58
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.
SVP
Joined: 07 Nov 2007
Posts: 1800
Location: New York

### Show Tags

25 Aug 2008, 08:02
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.
_________________

Smiling wins more friends than frowning

Intern
Joined: 24 Jul 2008
Posts: 3

### Show Tags

25 Aug 2008, 10:28
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
Senior Manager
Joined: 16 Jul 2008
Posts: 289

### Show Tags

25 Aug 2008, 11:17
Only B can have 0 unique readers, because it has the smallest number of readers.
_________________

http://applicant.wordpress.com/

Intern
Joined: 24 Jul 2008
Posts: 3

### Show Tags

25 Aug 2008, 12:21
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
Senior Manager
Joined: 31 Jul 2008
Posts: 296

### Show Tags

25 Aug 2008, 12:24
boy !! ; this was real good
Manager
Joined: 15 Jul 2008
Posts: 206

### Show Tags

25 Aug 2008, 12:59
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
Manager
Joined: 22 Jul 2008
Posts: 151

### Show Tags

26 Aug 2008, 13:35
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
Intern
Joined: 03 Mar 2008
Posts: 47

### Show Tags

26 Aug 2008, 19:00
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
Re: another set question   [#permalink] 26 Aug 2008, 19:00
Display posts from previous: Sort by