SETS

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pre-gmat: Master | Next Rank: 500 Posts; Posts: 105; Joined: Tue Jul 15, 2008 7:03 pm; Thanked: 5 times

by pre-gmat » Wed Aug 27, 2008 10:55 am

Found this problem but not sure bout the solution

54 gardeners grow roses, dhalias or chrysantemums. At the local flower show, 22 display roses, 26 display dhalias and 29 display chrysantemums; 5 show all three flowers. How many display roses ans dhalias only but not chrysanthemums?

I have some solution to it...not sure if that's the correct one.

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Source: Beat The GMAT — Problem Solving | Wed Aug 27, 2008 10:55 am

Stuart@KaplanGMAT: GMAT Instructor; Posts: 3225; Joined: Tue Jan 08, 2008 2:40 pmFollowed by: 614 members; Location: Toronto; Thanked: 1710 times; GMAT Score:800

Re: SETS

by Stuart@KaplanGMAT » Wed Aug 27, 2008 11:06 am

pre-gmat wrote:Found this problem but not sure bout the solution

54 gardeners grow roses, dhalias or chrysantemums. At the local flower show, 22 display roses, 26 display dhalias and 29 display chrysantemums; 5 show all three flowers. How many display roses ans dhalias only but not chrysanthemums?

I have some solution to it...not sure if that's the correct one.

We don't have enough info to solve.

We can figure out how many display exactly 2 flowers, but not how many display those two flowers.

The general equation for 3 overlapping sets is:

True # of objects = (total # in group 1) + (total # in group 2) + (total # in group 3) + (total # in none of the groups) - (# in only groups 1&2) - (# in only groups 1&3) - (# in only groups 2&3) - 2(# in all 3 groups)

Appling the formula in this question, we can determine that:

54 = 22 + 26 + 29 + 0 - (# in only groups 1&2) - (# in only groups 1&3) - (# in only groups 2&3) - 2(5)

54 = 77 - 10 - (# in only groups 1&2) - (# in only groups 1&3) - (# in only groups 2&3)

and rearranging a bit get to:

(# in only groups 1&2) + (# in only groups 1&3) + (# in only groups 2&3) = 13

In other words, we know that 13 gardeners grow exactly 2 types of flowers, but we can't break it down any further without more information.

Was this a problem solving or data sufficiency question? Where did you get it?

Stuart Kovinsky | Kaplan GMAT Faculty | Toronto

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pre-gmat: Master | Next Rank: 500 Posts; Posts: 105; Joined: Tue Jul 15, 2008 7:03 pm; Thanked: 5 times

by pre-gmat » Wed Aug 27, 2008 11:09 am

This is aproblem solving and found this on Testmagic...

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kuroneko1313: Senior | Next Rank: 100 Posts; Posts: 35; Joined: Thu Aug 21, 2008 2:55 pm; Thanked: 1 times

by kuroneko1313 » Wed Aug 27, 2008 11:27 am

Thanks for the explanation Stuart. However, I'm still a little confused about the true # of object formula:

True # of objects = (total # in group 1) + (total # in group 2) + (total # in group 3) + (total # in none of the groups) - (# in only groups 1&2) - (# in only groups 1&3) - (# in only groups 2&3) - 2(# in all 3 groups)

My question is, why do you need to multiple the # in all 3 groups by two here? Would subtracting it just be enough?

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Stuart@KaplanGMAT: GMAT Instructor; Posts: 3225; Joined: Tue Jan 08, 2008 2:40 pmFollowed by: 614 members; Location: Toronto; Thanked: 1710 times; GMAT Score:800

by Stuart@KaplanGMAT » Wed Aug 27, 2008 11:37 am

kuroneko1313 wrote:Thanks for the explanation Stuart. However, I'm still a little confused about the true # of object formula:

True # of objects = (total # in group 1) + (total # in group 2) + (total # in group 3) + (total # in none of the groups) - (# in only groups 1&2) - (# in only groups 1&3) - (# in only groups 2&3) - 2(# in all 3 groups)

My question is, why do you need to multiple the # in all 3 groups by two here? Would subtracting it just be enough?

3 set questions are actually a lot easier to understand with Venn diagrams, but those are tough to reproduce on this site. Someone needs to create a Venn diagram template!

If an object appears in all 3 groups, then if we merely add up the 3 separate totals we've counted it 3 times. To get the true number of objects we only want to count it once, so we need to subtract it twice.

If an object appears in 2 of the 3 groups then we've counted it twice, so we only need to subtract it once to get the true number of objects.

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pre-gmat: Master | Next Rank: 500 Posts; Posts: 105; Joined: Tue Jul 15, 2008 7:03 pm; Thanked: 5 times

by pre-gmat » Wed Aug 27, 2008 11:43 am

I was checking formulas for SETS somewhere....

for AUBUC, shud that formula be followed in this question?

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kuroneko1313: Senior | Next Rank: 100 Posts; Posts: 35; Joined: Thu Aug 21, 2008 2:55 pm; Thanked: 1 times

by kuroneko1313 » Wed Aug 27, 2008 11:47 am

Stuart Kovinsky wrote: If an object appears in 2 of the 3 groups then we've counted it twice, so we only need to subtract it once to get the true number of objects.

That helps alot

thanks alot Stuart!

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Stuart@KaplanGMAT: GMAT Instructor; Posts: 3225; Joined: Tue Jan 08, 2008 2:40 pmFollowed by: 614 members; Location: Toronto; Thanked: 1710 times; GMAT Score:800

by Stuart@KaplanGMAT » Wed Aug 27, 2008 11:48 am

pre-gmat wrote:I was checking formulas for SETS somewhere....

for AUBUC, shud that formula be followed in this question?

That's the formula that I applied, I just used words instead of symbols.

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pre-gmat: Master | Next Rank: 500 Posts; Posts: 105; Joined: Tue Jul 15, 2008 7:03 pm; Thanked: 5 times

by pre-gmat » Wed Aug 27, 2008 12:33 pm

Thanks. This is kinda complicated question for me...but ur explanantion above is very good.

Thanks.

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pre-gmat: Master | Next Rank: 500 Posts; Posts: 105; Joined: Tue Jul 15, 2008 7:03 pm; Thanked: 5 times

by pre-gmat » Tue Oct 21, 2008 1:13 pm

Stuart,

I was not able to get to the answer to the following question applying the formula above:

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.

Could you guide me through this.

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dmateer25: Community Manager; Posts: 1049; Joined: Sun Apr 06, 2008 5:15 pmFollowed by: 27 members; Location: Pittsburgh, PA; Thanked: 113 times; GMAT Score:710

by dmateer25 » Tue Oct 21, 2008 3:15 pm

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.

I will try to explain. You have 100 pupils so that is the total #.

39 like A
36 like B
45 like C
14 like AB
11 like BC
5 like AC
0 like ABC
# of pupils that don't like any = ???

Let X = # of pupils that like none of the magazines

Formula will look like this:

100 = 39 + 36 + 45 +X - 14 -11 - 5 - 2(0)

100 = 120 + X - 30

10 = X

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pre-gmat: Master | Next Rank: 500 Posts; Posts: 105; Joined: Tue Jul 15, 2008 7:03 pm; Thanked: 5 times

by pre-gmat » Tue Oct 21, 2008 3:16 pm

Nopes, the answer is 11.

You gotta find the maximum possible number

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dmateer25: Community Manager; Posts: 1049; Joined: Sun Apr 06, 2008 5:15 pmFollowed by: 27 members; Location: Pittsburgh, PA; Thanked: 113 times; GMAT Score:710

by dmateer25 » Tue Oct 21, 2008 3:37 pm

I see my mistake. I made an assumption that none liked all three and I shouldn't have.

But 11 would be the maximum that like all three magazines.

32 would be the maximum that like none.

Number that like A alone = 39 - 14 - 5 = 20
Number that like B alone = 36 - 14 - 11 = 11
Number that like C alone = 45 - 11 - 5 = 29

So since only 11 like B alone that means the maximum that like all three is 11.

Now apply this to the formula.

100 = 39 + 36 + 45 + n - 14 - 11 - 5 - 2(11)

100 = 120 + n - 52

32 = n

32 is the maximum that like none of the magazines.

Where did you get the OA 11 from? What is the source of this question?

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pre-gmat: Master | Next Rank: 500 Posts; Posts: 105; Joined: Tue Jul 15, 2008 7:03 pm; Thanked: 5 times

by pre-gmat » Tue Oct 21, 2008 5:16 pm

Yes sorry, 32 is the correct answer.

I got this question from Testmagic, but the explanation was not that clear and I wanted to see if the formula works.

Here's another one if you wanna try.

In a class of 217 students, the number of students who passed in English only is 43, in Math only is 47, in commerce only is 56. The number who passed in English and Maths is 14, Maths and Commerce is 26 and English and Commerce is 21 and the number who passed in all the subjects is 6. Find the number of students who failed in all the subjects.

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cramya: Legendary Member; Posts: 2467; Joined: Thu Aug 28, 2008 6:14 pmFollowed by: 11 members; Thanked: 331 times

by cramya » Tue Oct 21, 2008 6:41 pm

4 is what I get

OA?

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