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Sets - Could someone please outline a time efficient method

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Sets - Could someone please outline a time efficient method

by beater » Sat Jan 17, 2009 2:41 pm
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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by sonu_thekool » Sat Jan 17, 2009 3:10 pm
I am getting 0 (A)

This is one damn lengthy calculation though...I know you were asking for an efficient method but I am just happy that I could solve it after 4 mts..There must be a faster way.

Is 0 right ?
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Re: Sets - Could someone please outline a time efficient met

by piyush_nitt » Sat Jan 17, 2009 10:25 pm
beater 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
Not sure if this is efficient approach but what I did is

Total number of languages as subjects = 9 (4+3+2)
Total number of students = 6
3 students know exactly 2 languages , therefore total lang as subjects consumed = 6

remaining languages= 9 - 6 = 3.

and condition given is that every student is learning atleast 1 language , so remaining 3 students must be learning 1 language each .

hence IMO A

does that make any sense?
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by beater » Sun Jan 18, 2009 7:25 am
I think for problems such as these it may be easier to follow a conceptual approach versus an algebraic.

I'm confused as to what this means: 3 students know exactly 2 languages. Could some please explain. Thanks!
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by arzanr » Sun Jan 18, 2009 9:10 am
beater wrote:I think for problems such as these it may be easier to follow a conceptual approach versus an algebraic.
This question should be solved using algebra.

The formula for a three set problem is as follows:

n(AuBuC) = n(A) + n(B) + n(C) - n(AnB) - n(AnC) - n(BnC) + n(AnBnC)

now..

n(A) = number of students studying Russian = 4
n(B) = number of students studying Ukranian = 3
n(A) = number of students studying Hebrew = 2
[n(AnB) + n(AnC) + n(BnC)] = number of students that learn exactly 2 languages = 3
and let n(AnBnC), which is what you're solving for, be equal to X
therefore,

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

X = 0
beater wrote: I'm confused as to what this means: 3 students know exactly 2 languages. Could some please explain. Thanks!
This means that exactly 3 students know just 2 languages. So maybe 3 students know both Russian and Hebrew ONLY, while 0 know Rusian AND Ukranian ONLY or Hebrew AND Ukranian ONLY. Another example, 1 knows Russian AND Hebrew ONLY, 1 knows HEBREW AND Ukranian ONLY and 1 knows Russian AND Ukranian ONLY. If you see the formula above, it doesn't matter what the combination is as long as you know the total number of people knowing only 2 languages.
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by beater » Sun Jan 18, 2009 11:00 am
great explanation. Thanks!
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by aroon7 » Sun Jan 18, 2009 8:22 pm
if we remember the formula below we can do it quickly

n(AuBuC) = n(A) + n(B) + n(C) - n(AnB) - n(AnC) - n(BnC) + n(AnBnC)
--------------------------
i am back!
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by resepulv » Mon Jan 19, 2009 10:41 am
if we remember the formula below we can do it quickly

n(AuBuC) = n(A) + n(B) + n(C) - n(AnB) - n(AnC) - n(BnC) + n(AnBnC)
This formula is badly used in this problem, the answer is right just because the intersection n(AnBnC) is 0.

in this formula, n(AnB) includes n(AnBnC) so:

[n(AnB) + n(AnC) + n(BnC)] = number of students that learn exactly 2 languages + 3 times the students that are studying all languages= 3+3*n(AnBnC)
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by sjd00d » Mon Jan 19, 2009 10:30 pm
resepulv wrote:
if we remember the formula below we can do it quickly

n(AuBuC) = n(A) + n(B) + n(C) - n(AnB) - n(AnC) - n(BnC) + n(AnBnC)
This formula is badly used in this problem, the answer is right just because the intersection n(AnBnC) is 0.

in this formula, n(AnB) includes n(AnBnC) so:

[n(AnB) + n(AnC) + n(BnC)] = number of students that learn exactly 2 languages + 3 times the students that are studying all languages= 3+3*n(AnBnC)
This has been a major source of confusion for me for a long time and an expert advise here would be much appreciated.

Basically i've known the aforementioned formula for a long time but never questioned the validity of it. I don't really know where it comes from and where it really applies but when i draw the venn diagram, here's the formula i came up with.

p(AUBUC) = P(A) + P(B) + P(C) - p(AnB) - p(BnC) - p(CnA) - 2p(AnBnC)

The bold portion is critical and source of confusion but i believe it is correct. Basically the p(AnB) etc. takes care of removing the two overlapping portions, what you still have is 3 3-overlapping portion and you must subtract two of those (include just one) to know the total occurence of AUBUC
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by sjd00d » Mon Jan 19, 2009 10:48 pm
sjd00d wrote:
resepulv wrote:
if we remember the formula below we can do it quickly

n(AuBuC) = n(A) + n(B) + n(C) - n(AnB) - n(AnC) - n(BnC) + n(AnBnC)
This formula is badly used in this problem, the answer is right just because the intersection n(AnBnC) is 0.

in this formula, n(AnB) includes n(AnBnC) so:

[n(AnB) + n(AnC) + n(BnC)] = number of students that learn exactly 2 languages + 3 times the students that are studying all languages= 3+3*n(AnBnC)
This has been a major source of confusion for me for a long time and an expert advise here would be much appreciated.

Basically i've known the aforementioned formula for a long time but never questioned the validity of it. I don't really know where it comes from and where it really applies but when i draw the venn diagram, here's the formula i came up with.

p(AUBUC) = P(A) + P(B) + P(C) - p(AnB) - p(BnC) - p(CnA) - 2p(AnBnC)

The bold portion is critical and source of confusion but i believe it is correct. Basically the p(AnB) etc. takes care of removing the two overlapping portions, what you still have is 3 3-overlapping portion and you must subtract two of those (include just one) to know the total occurence of AUBUC

Actually ignore my post as i just realized that it creates more confusion, basically resepulv's assertion is correct. To tackle this problem (and its cousin where the p(AnBnC) is given but we are suppose to find the p(AnB) + p(AnC) ....) I'll use the orignal formula (not the one i revised but where we add p(AnBnC) ), arrive at p(AnB) + p(BnC) + p(CnA) and then simply deduct 3p(AnBnC) from that if we are specifically told so in the question (i.e. find ONLY number of folks that do exactly 2 things OR like here)
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