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m04q19- Six students study Russian, Ukrainian & Hebrew [#permalink]
 13 Apr 2008, 15:52
Question Stats:
78% (01:55) correct
21% (01:12) wrong based on 1 sessions
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 Source: GMAT Club Tests - hardest GMAT questions
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Re: students in a group : fastest way to solve [#permalink]
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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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 The picture is attached.
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VENN DIAGRAM.JPG [ 31.09 KiB | Viewed 2700 times ]
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Re: students in a group : fastest way to solve [#permalink]
14 Apr 2008, 07:01
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Re: overlapping sets [#permalink]
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]
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]
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"  @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]
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]
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]
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]
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]
14 Apr 2008, 23:53
bkk145 wrote: That was neat thread bkk145. Kudos.
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Re: students in a group : fastest way to solve [#permalink]
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]
17 Apr 2008, 17:41
bkk145 wrote: Thanks for the link.
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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]
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]
05 Sep 2008, 01:05
Thanks, alpha_plus_gamma. I forgot to include ( x + y + z + a) and was getting incorrect answer....Duh  .....Should stop making these kind of mistakes.
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Re: Sets- Language [#permalink]
05 Sep 2008, 01:11
I got A as well. Why is it incorrect?
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Re: Sets- Language [#permalink]
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]
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
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05 Sep 2008, 03:40
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