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The members of a club were asked whether they speak [#permalink]
27 Nov 2007, 09:52

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?

Re: tricky - counting and sets [#permalink]
27 Nov 2007, 10:04

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

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.

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, 09:11, edited 1 time in total.

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.

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!

>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)

Re: tricky - counting and sets [#permalink]
04 Dec 2007, 22:04

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?

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.

gmatclubot

Re: tricky - counting and sets
[#permalink]
04 Dec 2007, 22:04

I couldn’t help myself but stay impressed. young leader who can now basically speak Chinese and handle things alone (I’m Korean Canadian by the way, so...