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m04q19- Six students study Russian, Ukrainian & Hebrew

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m04q19- Six students study Russian, Ukrainian & Hebrew [#permalink]

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New post 13 Apr 2008, 15:52
6 students in a group study different languages as specified:

* Russian: 4
* Ukrainian: 3
* Hebrew: 2

Each student studies at least 1 language. If it is also known that exactly 3 students study exactly 2 languages, how many students are studying all three languages?

(A) 0
(B) 1
(C) 2
(D) 3
(E) 4

[Reveal] Spoiler: OA
A

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Re: students in a group : fastest way to solve [#permalink]

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New post 13 Apr 2008, 17:04
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Use Set Theory.

Total = A + B + C - (AB+BC+CA) + ABC
Given A = 4, B = 3, C = 2, AB+BC+CA = 3, Total = 6
So ABC = Total - A - B - C + (AB+BC+CA) = 6 - 4 - 3 - 2 + 3 = 0

Answer A.

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Re: m04 q19 [#permalink]

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New post 26 Jan 2009, 13:52
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ConkergMat wrote:
6 students in a group study different languages as specified:
Russian: 4
Ukrainian: 3
Hebrew: 2
Each student studies at least 1 language. It is also known that exactly 3 students learn exactly 2 languages. How many students are studying all languages?

Can somebody explain with Venn diagram if possible?


r + u + h - (ru+uh+hr) - 2(rhu) + none = 6
4 + 3 + 2 - (3) - 6 + 0 = 2 (rhu)
2 (rhu) = 0


no. of students studying all languages (rhu) = 0

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Re: students in a group : fastest way to solve [#permalink]

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New post 14 Apr 2008, 07:01
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watch out for "exactly two" question...

http://www.gmatclub.com/forum/7-p374736 ... n+#p374736

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Re: overlapping sets [#permalink]

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New post 04 Jan 2011, 01:19
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A for me!

The question says that 3 students study exactly 2 languages each. THerefore from the total of languages being studied by the 6 individual, 9

9-6=3

There are 3 people left for 3 languages left. Moreover, every person has to study at least one language so it is no feasible to have one person studying 3 languages and the other 2 studying none.

Dunno if it is clear ;-)

Ans A. 0

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Re: students in a group : fastest way to solve [#permalink]

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New post 13 Apr 2008, 19:54
abhijit_sen wrote:
Use Set Theory.

Total = A + B + C - (AB+BC+CA) + ABC
Given A = 4, B = 3, C = 2, AB+BC+CA = 3, Total = 6
So ABC = Total - A - B - C + (AB+BC+CA) = 6 - 4 - 3 - 2 + 3 = 0

Answer A.


I tried this and messed up somewhere.

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Re: students in a group : fastest way to solve [#permalink]

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New post 13 Apr 2008, 23:51
kyatin wrote:
abhijit_sen wrote:
Use Set Theory.

Total = A + B + C - (AB+BC+CA) + ABC
Given A = 4, B = 3, C = 2, AB+BC+CA = 3, Total = 6
So ABC = Total - A - B - C + (AB+BC+CA) = 6 - 4 - 3 - 2 + 3 = 0

Answer A.


I tried this and messed up somewhere.


@Abhijit
The colored is not "plus" :lol:

@kyatin,
Total = A+B+C -2*(ABC) - (AB+BC+CA)
Let try :)
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Re: students in a group : fastest way to solve [#permalink]

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New post 14 Apr 2008, 06:35
Sondenso + is correct.

It should be written more like this


Total = A + B + C - { (AB+BC+CA) - 3(ABC) }

I believe Abhijit did miss 3 in the equation but has it in the final calculation.

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Re: students in a group : fastest way to solve [#permalink]

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New post 14 Apr 2008, 08:40
kyatin wrote:
Sondenso + is correct.

It should be written more like this


Total = A + B + C - { (AB+BC+CA) - 3(ABC) }

I believe Abhijit did miss 3 in the equation but has it in the final calculation.


I have used the correct formula. 3 is not coming from multiplication but from the value mentioned in the question (as given "It is also known that exactly 3 students learn exactly 2 languages").

Also see my explanation earlier, in which I have mentioned what all of these values mentioned in formula should be and we are getting value of ABC and not Total.

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Re: students in a group : fastest way to solve [#permalink]

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New post 14 Apr 2008, 11:57
Abhijit,

You are correct.

\(Total = A + B + C - [ (AB+BC+CA) - (ABC) ]\)

is the right formula.

Thanks
Kyatin

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Re: students in a group : fastest way to solve [#permalink]

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New post 14 Apr 2008, 20:36
the way I approached this..

4+3+2-3-2(abc)=6

6-2(abc)=6

2(abc)=0..therefore no one takes all 3 courses..

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Re: students in a group : fastest way to solve [#permalink]

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New post 14 Apr 2008, 23:53
bkk145 wrote:
watch out for "exactly two" question...

http://www.gmatclub.com/forum/7-p374736 ... n+#p374736


That was neat thread bkk145. Kudos.

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Re: students in a group : fastest way to solve [#permalink]

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New post 15 Apr 2008, 19:03
abhijit_sen wrote:
kyatin wrote:
Sondenso + is correct.

It should be written more like this


Total = A + B + C - { (AB+BC+CA) - 3(ABC) }

I believe Abhijit did miss 3 in the equation but has it in the final calculation.


I have used the correct formula. 3 is not coming from multiplication but from the value mentioned in the question (as given "It is also known that exactly 3 students learn exactly 2 languages").

Also see my explanation earlier, in which I have mentioned what all of these values mentioned in formula should be and we are getting value of ABC and not Total.


At the moment, I think, I am being confused. The comment I made is from GmatCAt. And now I am advived by the link provided by bkk145. I think I need a thorough material about this concept.

Friend, do you have a good material about the Set theory and may you share with me?
Many thank and appreciation!
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Re: students in a group : fastest way to solve [#permalink]

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New post 17 Apr 2008, 17:41
bkk145 wrote:
watch out for "exactly two" question...

http://www.gmatclub.com/forum/7-p374736 ... n+#p374736


Thanks for the link.

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Sets- Language [#permalink]

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New post 05 Sep 2008, 00:42
6 students in a group study different languages as specified:
Russian: 4
Ukrainian: 3
Hebrew: 2
Each student studies at least 1 language. It is also known that exactly 3 students learn exactly 2 languages. How many students are studying all languages?
a) 0
b) 1
c) 2
d) 3
e) 4
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Re: Sets- Language [#permalink]

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New post 05 Sep 2008, 00:57
leonidas wrote:
6 students in a group study different languages as specified:
Russian: 4
Ukrainian: 3
Hebrew: 2
Each student studies at least 1 language. It is also known that exactly 3 students learn exactly 2 languages. How many students are studying all languages?
a) 0
b) 1
c) 2
d) 3
e) 4


A.

I solved using a venn diagram, consider x, y and z to be students studying 2 languages and let a be num. of students who study all 3.

then

2 - x -y -a + 4 -x -z -a + 3 -y -z -a + x + y + z + a = 6

9 - (x +y+z ) -2a = 6

We know that x + y + z = 3

giving us a =0

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Re: Sets- Language [#permalink]

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New post 05 Sep 2008, 01:05
Thanks, alpha_plus_gamma.

I forgot to include ( x + y + z + a) and was getting incorrect answer....Duh :oops: .....Should stop making these kind of mistakes.
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Re: Sets- Language [#permalink]

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New post 05 Sep 2008, 01:11
I got A as well. Why is it incorrect?
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Re: Sets- Language [#permalink]

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New post 05 Sep 2008, 02:08
I got 0 as well....but using formula....AUBUC = A + B + C - AB - BC - CA + ABC.

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Re: Sets- Language [#permalink]

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New post 05 Sep 2008, 03:40
I got A as well:

Total = U + R + H - 3(because 3 people study exactly 2 languages, 3 has been counted twice so you need to subtract it once) - 2x (because the number of people studying all the 3 languages have been counted 3 times, you want to subtract by twice the amount so that you can count it only once instead of 3 times).

so:

6= H+R+U - 3 - 2x

6 = 2+4+3 - 3 -2x

6= 9 - 3 - 2x

6 = 6 - 2x

0 = -2x

x = 0

Answer A

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Re: Sets- Language   [#permalink] 05 Sep 2008, 03:40

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