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Of a group of people, 10 play piano, 11 play guitar, 14 play

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Of a group of people, 10 play piano, 11 play guitar, 14 play [#permalink] New post 30 Oct 2009, 20:32
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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
[Reveal] Spoiler: OA

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Re: Overlapping Sets Problem [#permalink] New post 30 Oct 2009, 20:48
slingfox wrote:
Can someone tell explain how they go about solving this one?

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


Let single piano players = p
Let single guitar players = g
Let single violin players = v

Given, p + g + v + 3 + 20 = 10 + 11 + 14.
Hence, p + g + v = 35 - 23 = 12

D.
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Re: Overlapping Sets Problem [#permalink] New post 30 Oct 2009, 20:52
35 with overlaps. 35-20=15 two or more. 15-3=12 two instruments. (D)
Venn diagrams helps in these cases. 3 circles, each for Piano, Guitar and Violin. Shade the intersecting areas (the place where all three circles meet is 3, rest of the overlap shaded area would be 10+14+11-3-20... edit here as I forgot subtracting 20)
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Re: Overlapping Sets Problem [#permalink] New post 30 Oct 2009, 22:24
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gmattokyo wrote:
slingfox wrote:
The OA from Veritas is B.

I keep getting A whereas two people have gotten D :shock:


Did Veritas give any explanation? Unless there is a hidden trap (q looks straightforward enough) which makes the venn diagram wrong, I still don't see how it could be 6.



Here is one approach to get B... shoot!!
3 people play all three instruments. So they got counted in 10 of Piano, 11 of Guitar and 14 of Violin.
So people who play 1 or 2 instruments is 35-3x3=26
Given 20 people play 1 instrument only... people who play 2 instruments= 26-20=6

:cry:
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Re: Overlapping Sets Problem [#permalink] New post 30 Oct 2009, 22:40
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This problem was included in an in-class handout, so there is no explanation, only an answer. Here is how I did the problem:

P = 10
G = 11
V = 14
PGV = 3

Let PG + PV + GV = X

Number of People in Exactly One Set = 20 = P + G + V -2(X) + 3*PGV
Solving for this you get: 2X = (10 + 11 +14) + 3*3 - 20 = 24
So X = 12

Formula for people in exactly two sets = X - 3*PGV = 12 - 3*3 = 3

My Answer: 3

FYI: formulae-for-3-overlapping-sets-69014.html
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Re: Overlapping Sets Problem [#permalink] New post 30 Oct 2009, 22:56
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slingfox, I'm trying to understand how you derived, meanwhile here is the venns I used
Attachments

vennSaviour.GIF
vennSaviour.GIF [ 11.33 KiB | Viewed 8904 times ]

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Re: Overlapping Sets Problem [#permalink] New post 31 Oct 2009, 00:12
slingfox wrote:
GmatTokyo, take a look at iCandy's post in this thread (it is really helped me understand the formulas for 3 set venn diagrams): formulae-for-3-overlapping-sets-69014.html


slingfox, yes I had a look at that thread... great formulas. But yet to figure out what is wrong where... something's gotta give :roll: will go for a run and break, and return with fresh perspective!
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Re: Overlapping Sets Problem [#permalink] New post 31 Oct 2009, 10:31
gmattokyo wrote:
gmattokyo wrote:
slingfox wrote:
The OA from Veritas is B.

I keep getting A whereas two people have gotten D :shock:


Did Veritas give any explanation? Unless there is a hidden trap (q looks straightforward enough) which makes the venn diagram wrong, I still don't see how it could be 6.



Here is one approach to get B... shoot!!
3 people play all three instruments. So they got counted in 10 of Piano, 11 of Guitar and 14 of Violin.
So people who play 1 or 2 instruments is 35-3x3=26
Given 20 people play 1 instrument only... people who play 2 instruments= 26-20=6

:cry:


That bold st. is the key. The 3 people who play all three instruments is included in each of these, "10 play piano, 11 play guitar, 14 play violin" so it has to be subtracted from each.

It should have been:

Let single piano players = p
Let single guitar players = g
Let single violin players = v

Given, p + 3 + g + 3 + v + 3 + 20 = 10 + 11 + 14.
Hence, p + g + v = 35 - 29 = 6
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Re: Overlapping Sets Problem [#permalink] New post 15 Mar 2010, 23:48
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Jivana wrote:
gmattokyo wrote:
gmattokyo wrote:
The OA from Veritas is B.

I keep getting A whereas two people have gotten D :shock:

Did Veritas give any explanation? Unless there is a hidden trap (q looks straightforward enough) which makes the venn diagram wrong, I still don't see how it could be 6.



Here is one approach to get B... shoot!!
3 people play all three instruments. So they got counted in 10 of Piano, 11 of Guitar and 14 of Violin.
So people who play 1 or 2 instruments is 35-3x3=26
Given 20 people play 1 instrument only... people who play 2 instruments= 26-20=6

:cry:


That bold st. is the key. The 3 people who play all three instruments is included in each of these, "10 play piano, 11 play guitar, 14 play violin" so it has to be subtracted from each.

It should have been:

Let single piano players = p
Let single guitar players = g
Let single violin players = v

Given, p + 3 + g + 3 + v + 3 + 20 = 10 + 11 + 14.
Hence, p + g + v = 35 - 29 = 6


3 people play all three instruments. So they got counted in 10 of Piano, 11 of Guitar and 14 of Violin.
ppl playing Piano & Guitar but not Violin counted twice; in 10 and 11
ppl playing Guitar & Violin but not Piano counted twice; in 11 and 14
ppl playing Violin & Piano but not Guitar counted twice; in 14 and 10

ie ppl playing only two instruments are counted twice if we add 10 , 11 , 14

ie 10 + 11 + 14 = 20 + 2B + 3 x 3
2B = 6
B = 3 ,

The Qn is how many ppl play two instruments

= ppl playing only two instruments + ppl playing 3 instruments
3 + 3 = 6

-V
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Re: Overlapping Sets Problem [#permalink] New post 16 Mar 2010, 07:19
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slingfox wrote:
Can someone tell explain how they go about solving this one?

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


Lets' say PG+GV+VP = S where PG is intersection of piano and guitar, GV is intersection of guitar and violin and VP is intersection of violin and piano

No. of players who play at least one instrument = 10+11+14 - S +3 or (P+G+V-(PG+GV+VP)+PGV)
=38 - S

No. of players who play at least one instrument also = 20+s-6 or (No.of people who play only one instrument + PG+GV+VP - 2 * PGV)
14+s = 38 - s
=>2S = 24 => s = 12

No. of people who play only 2 instruments = PG+GV+VP - 2 * PGV = 12 - 2*3 = 6

So, answer is B
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Re: Overlapping Sets Problem [#permalink] New post 13 May 2010, 06:59
IMO A.

I am assuming that question is asking how many playing only 2 instruments.

Let's say
a = playing P only
b = playing G only
c = playing V only, and (a+b+c) = 20

d = playing P and G
e = playing G and V
f = playing V and P
g = playing P, G and V = 3

Therefore,
10 = a+d+e+g
11 = b+d+f+g
14 = c+e+f+g
------------------------------
35 = (a+b+c) + 2(d+e+f) + 3g
------------------------------

35 = 20 + 2(d+e+f) + 3x3 = 20 + 2(d+e+f) + 9 = 29 + 2(d+e+f)

Therefore, (d+e+f) = 3


Please tell me where I am wrong.
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Re: Tough Overlapping Set Problem- Veritas [#permalink] New post 11 Jul 2010, 10:34
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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] New post 11 Jul 2010, 10:37
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] New post 11 Jul 2010, 11:02
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This can be explained easily with a Venn Diagram.
Attachment:
Venn Diagram.jpg
Venn Diagram.jpg [ 23.96 KiB | Viewed 3116 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

Answer: B
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Re: Tough Overlapping Set Problem- Veritas [#permalink] New post 12 Jul 2010, 05:14
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] New post 12 Jul 2010, 05:23
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] New post 12 Jul 2010, 05:25
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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] New post 25 Jul 2010, 04:25
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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: Overlapping Sets Problem [#permalink] New post 10 Aug 2010, 00:10
If the question asks (which in fact it doesn't) how many people play only 2 instruments, then the answer is 3 (I used the same approach as ykaiim). If we want to find the number of people who play two instruments (i.e., 2 or more), then we have to add 3 (the number of people who play all three instruments) to the previous number. So we get 3+3 = 6.
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Re: Overlapping Sets Problem [#permalink] New post 26 Aug 2011, 02:07
vin2010 wrote:
ie 10 + 11 + 14 = 20 + 2B + 3 x 3
2B = 6
B = 3 ,

The Qn is how many ppl play two instruments

= ppl playing only two instruments + ppl playing 3 instruments
3 + 3 = 6

-V


you got the exact point the question is not asking how many play ONLY TWO instrument
but TWO instrument (which includes the no of people playing 3 instruments too)
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Re: Overlapping Sets Problem   [#permalink] 26 Aug 2011, 02:07
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