I prefer to do these as a formula while other prefer to draw these out.
When you have three groups (as in this case with Eng, French and German), then the formula is:
total - GroupA + GroupsB + GroupC - AB - BC - AC - ABC - ABC + None.
You have to subtract those that are in all three groups twice since they have been counted 3 times and we need them only counted once.
Now, with the percentages, we know that:
5 speak English
10 speak German
6 speak French
1 speaks E, G, & F
In this case, we don't care which two languages they speak, so we can group together all the "speak 2 lang" together (AB, BC and AC) and get:
15 = 5 + 6 + 10 -2lang -1 -1 + None
We are missing a bit of information to really be able to solve this. Without knowing that everyone speaks at least one language, there are multiple solutions.
Solution if everyone speaks at least 1 language:
15 = 5 + 6 + 10 -2lang -1 -1 +0
15 = 21 - 2 - 2lang
15 = 19 - 2lang
4 people speak 2 languages
So the answers to your questions are:
a) exactly 2 lang = 4
b) at least 2 lang = 5
c) only one lang = 10
However, the % still work fine if 5 people speak no languages.
Then you get:
15 = 5 + 6 + 10 - 2lang - 1 -1 +5
15 = 26 - 2 - 2lang
15 = 24 - 2lang
9 = 2lang
So the answers to your questions are:
a) exactly 2 lang = 9
b) at least 2 lang = 10
c) only one lang = 0
Another way to look at this is that there are 21 languages spoken by 15 people.
1 person counts for 3 languages
If the other 14 all speak at least 1 lang, then you have 21 - 14 + 3 = 21-17 = 4 who have to speak 2 languages.
OR
21 languages spoken by 15 people.
1 person counts for 3 languages
9 people count for 2 languages
= 3 + (9*2) = 3 + 18 = 21, so there are 15-(9+1) = 15-10 = 5 people who don't speak any languages.
With the none not = to 0, I am probably looking too hard to be tricked, but since this isn't really in the form of a GMAT question (one question with 5 choices), then it lends itself to this openness.
-Carrie
