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# Tough Overlapping Set Problem- Veritas

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Tough Overlapping Set Problem- Veritas [#permalink]  11 Jul 2010, 10:51
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Question Stats:

31% (02:56) correct 68% (01:27) wrong based on 1 sessions
Of a group of people, 10 play piano, 11 play guitar, 14 play violin, 3 play all the instruments, 20 play only one instrument. How many play 2 instruments?

A. 3
B. 6
C. 9
D. 12
E. 15

Dont know how the OA is derived?
[Reveal] Spoiler: OA

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Re: Tough Overlapping Set Problem- Veritas [#permalink]  11 Jul 2010, 11:18
I computed 3

Still want to know how I did it?
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Re: Tough Overlapping Set Problem- Veritas [#permalink]  11 Jul 2010, 11:20
Explanation plssssssssssss!!!!!!!!!
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Re: Tough Overlapping Set Problem- Veritas [#permalink]  11 Jul 2010, 11:34
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Sure!

Sorry, I can't do diagrams. I'll illustrate:

So you have a three circle venn diagram. You have the intersection of all three and that equals to three. You have three criteria; let's set up equations for them.

For Piano students:

A = students taking only piano
X = students taking piano and guitar
Z = students taking piano and violin

A + X + Z = 10 - 3 <---- we subtract the three already, from the intersection

For Guitar students:

B = students taking guitar only
X = students taking guitar and piano
Y = students taking guitar and violin

B + X + Y = 11 - 3

For Violin students:

C = students taking violin only
Y = students taking violin and guitar
Z = students taking violin and piano

We know that there are 20 students taking only one instrument, as such:

A + B + C = 20

When we combine the first three derived formulas, we would have:

A + B + C + 2X + 2Y + 2Z = 26

Subtracting A + B + C, we'd have

2X + 2Y + 2Z = 6
2 ( X + Y + Z ) = 6
X + Y + Z = 3

X + Y + Z is the number of students taking two instruments. As such, I choose A, but that's not the OA.
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Re: Tough Overlapping Set Problem- Veritas [#permalink]  11 Jul 2010, 11:37
I checked the results, it seems consistent. For example:

Three students with three instruments.

We have three students taking two instruments. Let's say, X = 1, Y = 1, and Z = 1 for illustration purposes.

A + 1 + 1 + 3 = 10 ===> A = 5
B + 1 + 1 + 3 = 11 ===> B = 6
C + 1 + 1 + 3 = 14 ===> C = 9

A + B + C = 20
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Re: Tough Overlapping Set Problem- Veritas [#permalink]  11 Jul 2010, 12:02
3
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This can be explained easily with a Venn Diagram.
Attachment:

Venn Diagram.jpg [ 23.96 KiB | Viewed 1232 times ]

Piano: x
Guitar: y
Violin: z
Piano and Guitar: a
Piano and Violin: b
Guitar and Violin: c

To find: a+b+c + 3

Given Conditions:

x+a+b+3 = 10
a+b = 7-x (1)

a+y+3+c = 11
a+c = 8-y (2)

b+c+z+3 = 14
b+c = 11-z (3)

x+y+z = 20

Adding 1, 2 and 3, we get :

a+b+b+c+a+c = 7-x+8-y+11-z = 26-(x+y+z) = 26-20 = 6

2(a+b+c) = 6
(a+b+c) = 3

Final answer: Number of people who play 2 instruments = Number of people who play ONLY two + # who play 3 = 3+3 = 6

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Re: Tough Overlapping Set Problem- Veritas [#permalink]  11 Jul 2010, 12:06
Whiplash - diagram meister! Yeah, I'd think the OA is wrong - we got the same answers.
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Re: Tough Overlapping Set Problem- Veritas [#permalink]  11 Jul 2010, 12:07
Oh, crap, forgot out the three students. Hats off to you whiplash!
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Re: Tough Overlapping Set Problem- Veritas [#permalink]  12 Jul 2010, 06:13
OA is "B". Thanks for the input.
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Re: Tough Overlapping Set Problem- Veritas [#permalink]  12 Jul 2010, 06:14
RGM wrote:
Oh, crap, forgot out the three students. Hats off to you whiplash!

I disagree about that. We cant add 3 students because these 3 students play all of enstrument. We need the students who only play 2 enstruments.

The right answer should be 3 not 6. Any opinion why are we adding 3 to a+b+C
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Re: Tough Overlapping Set Problem- Veritas [#permalink]  12 Jul 2010, 06:23
Sorry, forgot to award kudoos!!
What RGM & I did wrong was we didn't realize the difference between the following two sentences.
How many play 2 instruments?
How many play ONLY 2 instruments? We both assumed that the word "only" is used in the question, when infact its not.
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Re: Tough Overlapping Set Problem- Veritas [#permalink]  12 Jul 2010, 06:25
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fatihaysu wrote:
I disagree about that. We cant add 3 students because these 3 students play all of enstrument. We need the students who only play 2 enstruments.

The right answer should be 3 not 6. Any opinion why are we adding 3 to a+b+C

Please read the question. It asks for the number of people who play 2 instruments, not the number of people who play ONLY two instruments. The people who play 3 instruments obviously play two instruments as well, and hence can be counted in the group that plays two instruments. Had the question asked for people who play only two instruments, then, yes, the answer would be 3.
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Re: Tough Overlapping Set Problem- Veritas [#permalink]  25 Jul 2010, 04:31
+1 Kudos to whiplash2411 for excellent explanation
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Re: Tough Overlapping Set Problem- Veritas [#permalink]  25 Jul 2010, 05:25
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One instruments = A+B+C - 2(AB+BC+AC) +3ABC

20 = 11+14+10 -2X +9

=> X= 17

Two instruments = AB+BC+AC - 3ABC = 17 -9 =6

Hence B
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Re: Tough Overlapping Set Problem- Veritas   [#permalink] 25 Jul 2010, 05:25
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