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rschunti
 
 

Set A, B, C have some elements in common. if 16 elements are

by rschunti Fri Dec 14, 2007 4:34 am

Set A, B, C have some elements in common. if 16 elements are in both A and B, 17 elements are in both A and C, and 18 elements are in both B and C, how many elements do all three of the sets A, B, and C have in common?

1). of the 16 elements that are in both A and B, 9 elements are also in C

2). A has 25 elements, B has 30 elements, and C has 35 elements

This is GMATPREP question. What is the best way to solve it?
RonPurewal
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by RonPurewal Fri Dec 14, 2007 5:34 am

i apologize for the screams of anguish that will result from doing this, but:

i can't draw a venn diagram on the page, and don't currently have the equipment to hand-draw one and scan it in. so: click here for a reference venn diagram.

* the upper left circle (pdr1-3) will stand for set a.
* the upper right circle (pdr1*gad) will stand for set b.
* the bottom circle (pdre) will stand for set c.

--

(1) unless i'm reading this wrong, this statement is actually telling you the number you want. the 9 elements mentioned are in all three sets, and they're the only ones that are (think about this for a sec and it should sink in - if anything is in all three, then it's got to be one of those 16 in the first place, so you can be sure that you're not missing anything here). so that's definitely sufficient.

(2) because of statement (1) we know that 9 common elements will definitely work. if we put a 9 in the middle circle of the venn diagram above, then we have the following:
1, 7, 5 (imagine these numbers where you see '2, 1, 3' in the link)
8, 9, 9 (imagine these numbers where you see '0, 23, 4' in the link)
9 (where you see '191' in the link)

but, if you put an 8 in the middle circle, you get the following, which also works:
0, 8, 4
9, 8, 10
8

so, insufficient.
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by Guest Fri Dec 14, 2007 2:52 pm

Anyone know any good references for Venn problems? I've gone over the strategy guide about solving from inside out and I still don't get it. Very frustrating.
RonPurewal
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by RonPurewal Sat Dec 15, 2007 5:34 am

hth:
http://www.regentsprep.org/Regents/math ... acVenn.htm
http://www.math.tamu.edu/~kahlig/venn/venn.html

that ought to keep you busy for a little while.

we could generalize the idea of 'working from the inside out' to something like 'start by filling in the portions of the diagram that you actually can fill in with the given information'. it just so happens that the middle of the diagram happens to be one of those portions; hence our advice.

the reason you can't normally fill in the outer portions of a venn diagram until the end of a problem is that those portions, when expressed in words, are nitpicky and awful. for instance, if the three circles in a venn diagram are sets a, b, and c, then the outer part of circle a is 'those elements that are in set a, but not in set b or set c'. that's a bit too artificial to be given in most problems. on the other hand, the middle circle - 'elements that are in all three sets' - is a natural thing to cite.
urvigupta
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Re: Set A, B, C have some elements in common. if 16 elements are

by urvigupta Wed Sep 23, 2009 11:18 am

hi ron,
part 1: 9 elements are in all the 3 sets, I don't understand how the 9 can be part of the 16.
A intersection B intersection C = 9
A intersection B = 9 + 16 = 25
right? wrong?
RonPurewal
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Re: Set A, B, C have some elements in common. if 16 elements are

by RonPurewal Thu Oct 01, 2009 7:24 am

urvigupta Wrote:hi ron,
part 1: 9 elements are in all the 3 sets, I don't understand how the 9 can be part of the 16.
A intersection B intersection C = 9
A intersection B = 9 + 16 = 25
right? wrong?


first of all, A int. B int. C is ALWAYS, ALWAYS a part of A int. B. if you don't see why this is so, construct a venn diagram.

or think about a real-life situation: "students who play football, rugby, and baseball" are clearly included in the group "students who play football and rugby".

... not only that, but statement 1 actually TELLS you that the 9 are part of the 16.
it says: (emphasis mine)
of the 16 elements that are in both A and B, 9 elements are also in C
-- that "of" absolutely guarantees you that the 9 are part of the 16.
so does the "also".

--

if it had said 16 elements are in ONLY A and B, on the other hand, then you would be absolutely correct.
michail.palagaschwili
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Re: Set A, B, C have some elements in common. if 16 elements are

by michail.palagaschwili Thu Jan 21, 2016 8:11 am

Hi dear math experts, I'm just trying to refresh my skills for 3-Way-Venn-Diagram, would appreciate some comments on my solution. Thanks.

I won't draw a venn diagram, but for (1) it's pretty straightforward solution, as a 2-Group overlap already includes an overlap for all three groups.

(1) This gives us straight the solution. A,b,c have 9 elements in common. Sufficient
(2) Clearly not sufficient, as we have no info about the TOTAL and the elements in group NEITHER (see formula: Total=a+b+c-Sum of 2-Group overlaps+All 3+Neither)

Answer A
RonPurewal
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Re: Set A, B, C have some elements in common. if 16 elements are

by RonPurewal Sun Jan 24, 2016 2:58 pm

i'm a bit confused, because you wrote this ...
michail.palagaschwili Wrote:I'm just trying to refresh my skills for 3-Way-Venn-Diagram


...but then this...

I won't draw a venn diagram


interesting.
...?

in any case, your analysis is mostly on point, except for your consideration of the "none" group in statement (2).
the "none" group is irrelevant to statement 2, since that group doesn't figure into anything in either statement 2 or the question prompt.
AbhishekM943
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Re: Set A, B, C have some elements in common. if 16 elements are

by AbhishekM943 Tue Nov 22, 2016 5:14 am

Ron, I am unable to access the image from the link (in your first post). Could you please check if its still working?
RonPurewal
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Re: Set A, B, C have some elements in common. if 16 elements are

by RonPurewal Sun Nov 27, 2016 8:42 am

it's just supposed to be a blank venn diagram.

Like this
https://cartesianproduct.files.wordpres ... e_venn.png
ajaym8
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Re:

by ajaym8 Sun Jan 22, 2017 11:42 am

RonPurewal Wrote:hth:
http://www.regentsprep.org/Regents/math ... acVenn.htm
http://www.math.tamu.edu/~kahlig/venn/venn.html

that ought to keep you busy for a little while.

we could generalize the idea of 'working from the inside out' to something like 'start by filling in the portions of the diagram that you actually can fill in with the given information'. it just so happens that the middle of the diagram happens to be one of those portions; hence our advice.

the reason you can't normally fill in the outer portions of a venn diagram until the end of a problem is that those portions, when expressed in words, are nitpicky and awful. for instance, if the three circles in a venn diagram are sets a, b, and c, then the outer part of circle a is 'those elements that are in set a, but not in set b or set c'. that's a bit too artificial to be given in most problems. on the other hand, the middle circle - 'elements that are in all three sets' - is a natural thing to cite.


Hi Ron,
I was working with the links you pasted above & got stuck on the first problem I chose to do.
Here's the link - http://www.math.tamu.edu/~kahlig/venn/p ... tants.html

I could not figure out the following (Please refer to the notes below, too) :
1. regions 2 & 5 individually, although I know the sum
2. regions 7 & 5 individually, although I know the sum

Note 1:
I am referring to regions numbered as per the image here - https://www.google.co.in/search?q=venn+ ... YQCLouM%3A

Note 2:
I considered the left circle for crude, the right one for phosphorus & the bottom circle for sulphur.

Can you please share how to go about this ?

Thanks,
ajaym8
RonPurewal
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Re: Re:

by RonPurewal Sun Jan 29, 2017 9:37 pm

• the 177 is outside all three circles, so just subtract that from 1000... so, the three circles contain a total of 823 rivers.

• 101 is region 1
• 37 is region 4
• 137 is region 3
• 289 is regions 3 and 6 together, so, 289 – 137 = 152 is region 6

• let region 5 = x
• 439 is regions 2 + 3 + 5 + 6, so, (region 2) + 137 + x + 152 = 439. ...thus (region 2) = 150 – x
• 463 is regions 4 + 5 + 6 + 7, so, 37 + x + 152 + (region 7) = 463. ...thus (region 7) = 274 – x

• finally, you haven't used the 823 yet (...which comes from the total of 1000 -- you can also look at it as "I haven't used that 1000 yet")
add up all 7 regions to this quantity:
101 + (150 – x) + 137 + 37 + x + 152 + (274 – x) = 823
...this will let you solve for x, and then you have everything.