m04q19- Six students study Russian, Ukrainian & Hebrew : Retired Discussions [Locked]
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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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13 Apr 2008, 14:52
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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13 Apr 2008, 16:04
4
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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

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26 Jan 2009, 12:52
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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

The picture is attached.
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VENN DIAGRAM.JPG [ 31.09 KiB | Viewed 3228 times ]

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

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14 Apr 2008, 06:01
1
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watch out for "exactly two" question...

http://www.gmatclub.com/forum/7-p374736 ... n+#p374736
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04 Jan 2011, 00:19
1
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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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13 Apr 2008, 18:54
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.
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Re: students in a group : fastest way to solve [#permalink]

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13 Apr 2008, 22:51
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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Re: students in a group : fastest way to solve [#permalink]

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14 Apr 2008, 05:35
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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14 Apr 2008, 07:40
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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14 Apr 2008, 10:57
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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14 Apr 2008, 19:36
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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14 Apr 2008, 22:53
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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15 Apr 2008, 18:03
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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17 Apr 2008, 16:41
bkk145 wrote:
watch out for "exactly two" question...

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

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04 Sep 2008, 23:42
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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04 Sep 2008, 23:57
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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05 Sep 2008, 00:05
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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05 Sep 2008, 00:11
I got A as well. Why is it incorrect?
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05 Sep 2008, 01:08
I got 0 as well....but using formula....AUBUC = A + B + C - AB - BC - CA + ABC.
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05 Sep 2008, 02:40
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

Re: Sets- Language   [#permalink] 05 Sep 2008, 02:40

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# m04q19- Six students study Russian, Ukrainian & Hebrew

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