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# m04q19- Six students study Russian, Ukrainian & Hebrew

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Senior Manager
Joined: 29 Jan 2007
Posts: 441
Location: Earth
m04q19- Six students study Russian, Ukrainian & Hebrew [#permalink]

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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

Source: GMAT Club Tests - hardest GMAT questions
Director
Joined: 10 Sep 2007
Posts: 938
Re: students in a group : fastest way to solve [#permalink]

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13 Apr 2008, 17:04
4
KUDOS
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

Senior Manager
Joined: 29 Jan 2007
Posts: 441
Location: Earth
Re: students in a group : fastest way to solve [#permalink]

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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

I tried this and messed up somewhere.
SVP
Joined: 04 May 2006
Posts: 1892
Schools: CBS, Kellogg
Re: students in a group : fastest way to solve [#permalink]

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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

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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Senior Manager
Joined: 29 Jan 2007
Posts: 441
Location: Earth
Re: students in a group : fastest way to solve [#permalink]

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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.
VP
Joined: 10 Jun 2007
Posts: 1439
Re: students in a group : fastest way to solve [#permalink]

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14 Apr 2008, 07:01
1
KUDOS
watch out for "exactly two" question...

http://www.gmatclub.com/forum/7-p374736 ... n+#p374736
Director
Joined: 10 Sep 2007
Posts: 938
Re: students in a group : fastest way to solve [#permalink]

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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.
Senior Manager
Joined: 29 Jan 2007
Posts: 441
Location: Earth
Re: students in a group : fastest way to solve [#permalink]

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14 Apr 2008, 11:57
Abhijit,

You are correct.

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

is the right formula.

Thanks
Kyatin
Current Student
Joined: 28 Dec 2004
Posts: 3357
Location: New York City
Schools: Wharton'11 HBS'12
Re: students in a group : fastest way to solve [#permalink]

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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..
Senior Manager
Joined: 29 Jan 2007
Posts: 441
Location: Earth
Re: students in a group : fastest way to solve [#permalink]

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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.
SVP
Joined: 04 May 2006
Posts: 1892
Schools: CBS, Kellogg
Re: students in a group : fastest way to solve [#permalink]

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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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Director
Joined: 06 Jan 2008
Posts: 546
Re: students in a group : fastest way to solve [#permalink]

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

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

Senior Manager
Joined: 29 Mar 2008
Posts: 348

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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
_________________

To find what you seek in the road of life, the best proverb of all is that which says:
"Leave no stone unturned."
-Edward Bulwer Lytton

Director
Joined: 14 Aug 2007
Posts: 727

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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
Senior Manager
Joined: 29 Mar 2008
Posts: 348

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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.
_________________

To find what you seek in the road of life, the best proverb of all is that which says:
"Leave no stone unturned."
-Edward Bulwer Lytton

Senior Manager
Joined: 16 Jul 2008
Posts: 289

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05 Sep 2008, 01:11
I got A as well. Why is it incorrect?
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SVP
Joined: 17 Jun 2008
Posts: 1547

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05 Sep 2008, 02:08
I got 0 as well....but using formula....AUBUC = A + B + C - AB - BC - CA + ABC.
SVP
Joined: 21 Jul 2006
Posts: 1510

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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

Senior Manager
Joined: 29 Mar 2008
Posts: 348

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05 Sep 2008, 07:58
Nerdboy wrote:
I got A as well. Why is it incorrect?

I mentioned earlier that, I made a silly mistake and lost my ability to think further at 1 am and posted the message
_________________

To find what you seek in the road of life, the best proverb of all is that which says:
"Leave no stone unturned."
-Edward Bulwer Lytton

VP
Joined: 17 Jun 2008
Posts: 1381

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05 Sep 2008, 21:36
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

Even i got A
Good question !!
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cheers
Its Now Or Never

Re: Sets- Language   [#permalink] 05 Sep 2008, 21:36

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# m04q19- Six students study Russian, Ukrainian & Hebrew

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