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# The members of a club were asked whether they speak

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The members of a club were asked whether they speak [#permalink]

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05 Aug 2007, 04:32
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The members of a club were asked whether they speak Cantonese, Mandarin and Japanese. 100 said that they spoke Cantonese, 150 said that they spoke Mandarin and 200 said that they spoke Japanese. 120 said that they spoke exactly two of the three languages. How many members does the club have?

(1) There are twice as many members who speak none of the languages as there are who speak all three languages.
(2) Half of the members who speak Japanese and Cantonese also speak Mandarin.
[Reveal] Spoiler: OA

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Re: The members of a club were asked whether they speak [#permalink]

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05 Aug 2007, 08:12
kevincan wrote:
The members of a club were asked whether they speak Cantonese, Mandarin and Japanese. 100 said that they spoke Cantonese, 150 said that they spoke Mandarin and 200 said that they spoke Japanese. 120 said that they spoke exactly two of the three languages. How many members does the club have?

(1) There are twice as many members who speak none of the languages as there are who speak all three languages.
(2) Half of the members who speak Japanese and Cantonese also speak Mandarin.

A for me.
Set x = number of people who speak all three languages.
(1) (100+150+200) - 120 - 2x + Neither = Total
Neither = 2x
We can cancel out the 2x in the equation; thus, find the total.
SUFFICIENT.

(2) Don't know anything about "Neither" information. INSUFFICIENT.

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Re: The members of a club were asked whether they speak [#permalink]

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05 Aug 2007, 08:34
kevincan wrote:
The members of a club were asked whether they speak Cantonese, Mandarin and Japanese. 100 said that they spoke Cantonese, 150 said that they spoke Mandarin and 200 said that they spoke Japanese. 120 said that they spoke exactly two of the three languages. How many members does the club have?

(1) There are twice as many members who speak none of the languages as there are who speak all three languages.
(2) Half of the members who speak Japanese and Cantonese also speak Mandarin.

I take A here

Given that

Cantonese = 100
Mandarin = 150
Japanese = 200

Let x,y and z represent the ONLY 2 lanuages sections in the Venn diagram and "t " represent all three

So

Only Cantonsese = 100 - (x+y+t)
Only Mandarin = 150 - (x+z+t)
Only Japanese = 200 - (z+y+t)

Formula now goes as

Total = Only A + Only B + Only C + Only AB+ Only BC+ Only CA + ABC + None

From 1 we have
None = 2*t

Substitute the rest in above eq.

100 - (x+y+t) + 150 - (x+z+t) + 200 - (z+y+t) + 120 + t + 2*t

= 330

From(2) I dont think we can decipher anything..

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Re: The members of a club were asked whether they speak [#permalink]

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06 Aug 2007, 13:46
dahcrap wrote:
bkk145 wrote:
kevincan wrote:
The members of a club were asked whether they speak Cantonese, Mandarin and Japanese. 100 said that they spoke Cantonese, 150 said that they spoke Mandarin and 200 said that they spoke Japanese. 120 said that they spoke exactly two of the three languages. How many members does the club have?

(1) There are twice as many members who speak none of the languages as there are who speak all three languages.
(2) Half of the members who speak Japanese and Cantonese also speak Mandarin.

A for me.
Set x = number of people who speak all three languages.
(1) (100+150+200) - 120 - 2x + Neither = Total
Neither = 2x
[b]
I think the highlighted sign shud be + and not -[/b]
We can cancel out the 2x in the equation; thus, find the total.
SUFFICIENT.

(2) Don't know anything about "Neither" information. INSUFFICIENT.

The answer has to be E

I am quite sure it is negative there...if you have sum of all the numbers, you have to subtract all the "repeats", then add neither to equal to total.

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Re: The members of a club were asked whether they speak [#permalink]

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06 Aug 2007, 17:32
Here is why i say we add the triple..

see when we calculate the doubles..it already takes into account the triple...i.e we have already minused the triple to get the doubles... so we cant deduct triple again cause we have already taken em into account..

bkk145 wrote:
fresinha12 wrote:
dont we add the singles, minus the doubles and add the triples????

kevincan wrote:
OA=A

If you can explain why you should add the triple, I'll see if I can say something...

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Re: The members of a club were asked whether they speak [#permalink]

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06 Aug 2007, 18:12
fresinha12 wrote:
Here is why i say we add the triple..

see when we calculate the doubles..it already takes into account the triple...i.e we have already minused the triple to get the doubles... so we cant deduct triple again cause we have already taken em into account..

bkk145 wrote:
fresinha12 wrote:
dont we add the singles, minus the doubles and add the triples????

kevincan wrote:
OA=A

If you can explain why you should add the triple, I'll see if I can say something...

No, double doesn't take triple into account. They are different. If the question say something like "two or more languages", then it would take triple into account. But this problem said specifically that exactly two languages.

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Re: The members of a club were asked whether they speak [#permalink]

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07 Aug 2007, 09:26
OK let me understand...

in a 3 set venn diagram...say C for chinese, M for Mandarian and J for Japense

120 speak exactly 2 languages...is equal to the sum of intersect of C n J, C n M and J n M...?? right if thats the case then 120 has already minused the number of people who speak C+J+M, correct?

bkk145 wrote:
fresinha12 wrote:
Here is why i say we add the triple..

see when we calculate the doubles..it already takes into account the triple...i.e we have already minused the triple to get the doubles... so we cant deduct triple again cause we have already taken em into account..

bkk145 wrote:
fresinha12 wrote:
dont we add the singles, minus the doubles and add the triples????

kevincan wrote:
OA=A

If you can explain why you should add the triple, I'll see if I can say something...

No, double doesn't take triple into account. They are different. If the question say something like "two or more languages", then it would take triple into account. But this problem said specifically that exactly two languages.

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Re: The members of a club were asked whether they speak [#permalink]

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07 Aug 2007, 13:10
fresinha12 wrote:
OK let me understand...

in a 3 set venn diagram...say C for chinese, M for Mandarian and J for Japense

120 speak exactly 2 languages...is equal to the sum of intersect of C n J, C n M and J n M...?? right if thats the case then 120 has already minused the number of people who speak C+J+M, correct?

bkk145 wrote:
fresinha12 wrote:
Here is why i say we add the triple..

see when we calculate the doubles..it already takes into account the triple...i.e we have already minused the triple to get the doubles... so we cant deduct triple again cause we have already taken em into account..

bkk145 wrote:
fresinha12 wrote:
dont we add the singles, minus the doubles and add the triples????

kevincan wrote:
OA=A

If you can explain why you should add the triple, I'll see if I can say something...

No, double doesn't take triple into account. They are different. If the question say something like "two or more languages", then it would take triple into account. But this problem said specifically that exactly two languages.

That's right. Let's look at something more simple and neglect "neither" at this point.
1) Say 2 people speak Chinese, 2 people speaker Mandarin, and 2 people speak Japanese. Say 3 speak "exactly two languages". Let x = people who speaker all three languages, we have:
2+2+2-3-2x = people who speak one language

2) If 3 people speak "two languages or more", then 2+2+2-3-x = people who speak only one language.

Notice the "2" factor is missing in the second one. The idea here is that if you speak two languages, then it overlap one time. If you speak all, then it overlap two times.

I hope I didn't confuse you even more. I think my reasoning is right, but feel free to correct if anyone disagree.

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Re: The members of a club were asked whether they speak [#permalink]

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27 Nov 2007, 10:52
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The members of a club were asked whether they speak Cantonese, Mandarin, and Japanese. 100 said that they spoke Cantonese, 150 said that they spoke Mandarin and 200 said that they spoke Japanese. 120 said that they spoke exactly 2 of the 3 languages. How many members does the club have?

1. there are twice as many members who speak none of the languages as there are who speak all 3 languages.

2. half of the members who speak Japanese and Cantonese also speak Mandarin.

I don't have the answer...in my opinion that's E....does smbd thinks differently?

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Re: The members of a club were asked whether they speak [#permalink]

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27 Nov 2007, 11:04
marcodonzelli wrote:
The members of a club were asked whether they speak Cantonese, Mandarin, and Japanese. 100 said that they spoke Cantonese, 150 said that they spoke Mandarin and 200 said that they spoke Japanese. 120 said that they spoke exactly 2 of the 3 languages. How many members does the club have?

1. there are twice as many members who speak none of the languages as there are who speak all 3 languages.

2. half of the members who speak Japanese and Cantonese also speak Mandarin.

I don't have the answer...in my opinion that's E....does smbd thinks differently?

x = 100 + 150 +200 - 120 - [all three languages *2] + neither

1. tell us relative info on the 2 variables
insuff
2. irrelvant as we already know 120 consists of all the overlap of 2 language-speaking ppl
insuff

E

thoughts?

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Re: The members of a club were asked whether they speak [#permalink]

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03 Dec 2007, 09:50
I think it's C, the two statements together are sufficient.

The total number of members has to be:

100 Cantonese
+150 Mandarin
+200 Japanese
less 120 (speak exactly 2)
less (speak all 3)*2
less speak none

So we need to know how many speak all three languages and how many speak none.

Does 1 answer this for us? Obviously not. But it does give us interesting information - twice as many speak none as speak all three.

2 tells us that half of those who speak Japanese (100) and half of those who speak Cantonese (50) also speak Mandarin. Since only 120 speak exactly 2 languages, 30 must speak all three. But still we don't know how many speak none.

If we combine 2 with 1 we get the answer.

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Re: The members of a club were asked whether they speak [#permalink]

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03 Dec 2007, 10:10
Raffie wrote:
I think it's C, the two statements together are sufficient.

The total number of members has to be:

100 Cantonese
+150 Mandarin
+200 Japanese
less 120 (speak exactly 2)
less (speak all 3)*2
less speak none

So we need to know how many speak all three languages and how many speak none.

Does 1 answer this for us? Obviously not. But it does give us interesting information - twice as many speak none as speak all three.

2 tells us that half of those who speak Japanese (100) and half of those who speak Cantonese (50) also speak Mandarin. Since only 120 speak exactly 2 languages, 30 must speak all three. But still we don't know how many speak none.

If we combine 2 with 1 we get the answer.

We are trying to find ALL members.
Why are you using for all of them MINUS non languages???
But U right, we can find answer because "exactly 2 languages" gives us necessary info.

Last edited by Blad on 03 Dec 2007, 10:11, edited 1 time in total.

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Re: The members of a club were asked whether they speak [#permalink]

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03 Dec 2007, 10:11
If you fill it up this way, it doesn't add up. 120 people who speak both Japanese and Cantonese + 30 who speak all three = 150. But only 100 speak Cantonese.

Without more information on how all the people who speak 2 languages we can't solve. At least I can't. Love to hear to OE.

Raffie wrote:
I think it's C, the two statements together are sufficient.

The total number of members has to be:

100 Cantonese
+150 Mandarin
+200 Japanese
less 120 (speak exactly 2)
less (speak all 3)*2
less speak none

So we need to know how many speak all three languages and how many speak none.

Does 1 answer this for us? Obviously not. But it does give us interesting information - twice as many speak none as speak all three.

2 tells us that half of those who speak Japanese (100) and half of those who speak Cantonese (50) also speak Mandarin. Since only 120 speak exactly 2 languages, 30 must speak all three. But still we don't know how many speak none.

If we combine 2 with 1 we get the answer.

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Re: The members of a club were asked whether they speak [#permalink]

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04 Dec 2007, 07:37
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Let those that speak C&M (but not J) = x, C&J (but not M) = y and M&J (but not C) = z

From the stimulus, we know that x + y + z = 120

Let those that speak C&M&J = w

Total members = (C - x - y - w) + (M - x - z - w) + (J - y - z - w) + x + y + z + w + None

Total members = C + M + J -(x + y + z) - 2w + None

Total members = 330 - 2w + None (as mentioned in previous post)

From Statement1: we know that: None = 2w

Total members = 330 - 2w + 2w = 330. Therefore answer is A or D

From Statement2: we know that: y + w (all those that speak Japanese & Canotense) = 2w (all those that speak Japanese, Cantonese & Mandarin). Therefore y = w

Total members = 300 - 2y + None...still have two unknown variables, therefore INSUFFICIENT!

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Re: The members of a club were asked whether they speak [#permalink]

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04 Dec 2007, 07:46
>2. half of the members who speak Japanese and Cantonese also speak Mandarin.
To me this implies
0.5 of the members how speak both Japanese and Cantonese (info we don't have)
and NOT
0.5 of the members how speak either Japanese or Cantonese (info we do have)

so i think the answer is E

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Re: The members of a club were asked whether they speak [#permalink]

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04 Dec 2007, 23:04
2
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marcodonzelli wrote:
The members of a club were asked whether they speak Cantonese, Mandarin, and Japanese. 100 said that they spoke Cantonese, 150 said that they spoke Mandarin and 200 said that they spoke Japanese. 120 said that they spoke exactly 2 of the 3 languages. How many members does the club have?

1. there are twice as many members who speak none of the languages as there are who speak all 3 languages.

2. half of the members who speak Japanese and Cantonese also speak Mandarin.

I don't have the answer...in my opinion that's E....does smbd thinks differently?

Getting A

Total = C + M + J - (2 of 3) - 2(all 3) + Niether
Total = 330 - 2(all 3) + Niether

Stat 1:
If x = all 3, then 2x = Niether
Total = 330 - 2x + 2x = 330. Sufficient

Stat 2:
Tells us that all 3 = 150 but doesn't say anything about niether. Insuff.

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Re: The members of a club were asked whether they speak [#permalink]

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22 Nov 2014, 03:57
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Re: The members of a club were asked whether they speak [#permalink]

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22 Nov 2014, 04:15
1
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mantymooney wrote:
Total members = 330 - 2w + None (as mentioned in previous post)

How are we getting 330 here? For those who didn't understand how 330 is arrived at, here it is.
Great work mantymooney!

Total members = (C - x - y - w) + (M - x - z - w) + (J - y - z - w) + x + y + z + w + None
Total members = C - x - y - w + M - x - z- w + J - y - z - w + x + y + z + w + None
Total members = C + M + J -(x + y + z) - 2w + None
OR Total members = 100 + 150 + 200 -(120) - 2w + None
OR Total members = 330 - 2w + None
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Re: The members of a club were asked whether they speak [#permalink]

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25 Nov 2015, 17:53
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Re: The members of a club were asked whether they speak [#permalink]

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13 Aug 2016, 06:45
Total - neither = C + M + J - ( (C&M) + (M&J) + ( C&J)) - 2 ( C&J&M)

Given C = 100 , M =150 , J =200
C&M) + (M&J) + ( C&J) = 120

a) A says Neither = 2 * ( C &J&M)
When we equate this is above 3 overlapping set equation , then neither and all 3 cancelled out
we left with
Total = C + M + J - 120

A is sufficient to answer

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Re: The members of a club were asked whether they speak   [#permalink] 13 Aug 2016, 06:45
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