Check GMAT Club Decision Tracker for the Latest School Decision Releases https://gmatclub.com/AppTrack

 It is currently 28 May 2017, 10:15

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# m04q19- Six students study Russian, Ukrainian & Hebrew

Author Message
Intern
Joined: 16 Jun 2011
Posts: 33
Followers: 0

Kudos [?]: 1 [0], given: 6

Re: 6 students in a group study different languages [#permalink]

### Show Tags

25 Jun 2011, 15:23
Substitution is fast but inelegant. Sub 0 for all three - 0 students used. Then subtract the 3 paired students from 4 Russian and 3 Ukrainian leaving 1 Russian and 2 Hebrew student. 3 pairs + 1 Russian + 1 Hebrew = 6 students. 0 yields 6=6 students
Senior Manager
Joined: 17 May 2010
Posts: 291
GMAT 1: 710 Q47 V40
Followers: 4

Kudos [?]: 53 [0], given: 7

Re: 6 students in a group study different languages [#permalink]

### Show Tags

13 Jul 2011, 19:43
Use formula

Total = A + B + C + neither - 2*all3 - atleast2

Neither = 0.

Plug it in and we get answer as 0.
_________________

If you like my post, consider giving me KUDOS!

Senior Manager
Status: mba here i come!
Joined: 07 Aug 2011
Posts: 264
Followers: 45

Kudos [?]: 1124 [0], given: 48

Re: 6 students in a group study different languages [#permalink]

### Show Tags

27 Aug 2011, 12:48
Total = A+B+C -(exactly 2) -2(exactly 3) + Neither

you'll get A
_________________

press +1 Kudos to appreciate posts

Intern
Joined: 13 Jul 2010
Posts: 2
Followers: 0

Kudos [?]: 10 [0], given: 4

### Show Tags

12 Sep 2011, 19:03
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?

0
1
2
3
4
Senior Manager
Joined: 24 Mar 2011
Posts: 450
Location: Texas
Followers: 5

Kudos [?]: 184 [0], given: 20

### Show Tags

12 Sep 2011, 19:49
gokoli wrote:
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?

0
1
2
3
4

i dont think we need a venn diagram here..

from given -
1) each student studies atleast 1 language, so 6 students studies 6 languages atleast...
2) 3 students studies exactly 2 languages.. = 6
so alltogether from the info, 6 students studies 6+3 = 9 languages (we are adding 3 because 1 language isalready considered before)..

now its also given that total 4+3+2 =9 languages are studied..
so none student studies all 3 languages. answer A.
Senior Manager
Joined: 08 Nov 2010
Posts: 409
Followers: 8

Kudos [?]: 116 [0], given: 161

### Show Tags

12 Sep 2011, 20:56
I try always to use the formula:

Total = A+B+C-Both-2*all

so:

6=4+3+2-3-2x -----> x=0
_________________
Intern
Joined: 24 Jul 2009
Posts: 29
Followers: 0

Kudos [?]: 5 [0], given: 1

### Show Tags

12 Sep 2011, 21:10
agdimple333 wrote:
gokoli wrote:
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?

0
1
2
3
4

I believe this is correct, but please help me find the flaw in my reasoning here...if it exists

1. From the info we know all 6 students have at least 1 language and 3 have exactly two. Therefore we have (3+6) =9 languages at minimum. Because we know 3 have exactly 2, the number studying all three must be 3 or less, so rule out d.

2. Because the numbers next to the languages added together represent the total of all possible combinations, when we add them we get the same number as with the requirements above. So, we know that the requirements must be 1 language for 3 and 2 languages for 3. Therefore, we have 0 students studying all 3.

Manager
Status: Prepping for the last time....
Joined: 28 May 2010
Posts: 187
Location: Australia
Concentration: Technology, Strategy
GMAT 1: 630 Q47 V29
GPA: 3.2
Followers: 0

Kudos [?]: 29 [0], given: 21

### Show Tags

12 Sep 2011, 21:21
(4+3+2) -3 -2x = 6
=> x=0
_________________

Two great challenges: 1. Guts to Fail and 2. Fear to Succeed

Intern
Joined: 13 Oct 2011
Posts: 4
Followers: 0

Kudos [?]: 0 [0], given: 0

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

### Show Tags

14 Oct 2011, 06:22
I Have a special formula for this case.....
S-X=TWO+2*THREE
Here s is the sum of sets which equals to 4+3+2=9
x is the no.of people who study at least one language which is 6 bcoz six people r there nd each reads at least one language so 6*1=6
nd exactly 3 students study 2 language..
so S=9,X=6,No.of student exactly studying two language=3
so 9-6=3+2*3 which equals to 0...
Manager
Joined: 16 Sep 2010
Posts: 220
Location: United States
Concentration: Finance, Real Estate
GMAT 1: 740 Q48 V42
Followers: 9

Kudos [?]: 117 [0], given: 2

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

### Show Tags

16 Nov 2011, 23:16
A. Just work with the three who study two and you see its impossible for one to study three.
Director
Joined: 28 Jun 2011
Posts: 890
Followers: 92

Kudos [?]: 232 [0], given: 57

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

### Show Tags

24 Nov 2011, 14:57
kyatin wrote:
Abhijit,

You are correct.

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

is the right formula.

Thanks
Kyatin

This is the right way to solve.........
_________________
Senior Manager
Joined: 15 Sep 2009
Posts: 268
Followers: 11

Kudos [?]: 74 [0], given: 6

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

### Show Tags

15 Oct 2012, 06:26
I almost never advocate memorizing formulas but the 3-set Venn calls for an exception.

The formula is fast, clean, and highly recommended.

A+B+C - (partcipants in exactly two activities) - 2(participants in all three activities)= Total number of participants.

Applying this formula to the given problem:

4+3+2-3-2x=6
Where x=participants in all three activities

-2x=6-6=0
x=0

Cheers,
Der alte Fritz.

Posted from my mobile device
_________________

+1 Kudos me - I'm half Irish, half Prussian.

Re: m04q19- Six students study Russian, Ukrainian & Hebrew   [#permalink] 15 Oct 2012, 06:26

Go to page   Previous    1   2   3   [ 52 posts ]

Similar topics Replies Last post
Similar
Topics:
At the end of the term, a group of three students (m25#25) 0 15 Feb 2012, 07:00
2 If there are 85 students in a statistics class and we assume 1 05 Jun 2010, 07:59
5 At Springfield High, three-fourths of the male students and 7 27 Jun 2014, 01:55
Display posts from previous: Sort by

# m04q19- Six students study Russian, Ukrainian & Hebrew

Moderator: Bunuel

 Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.