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Problem Solving (OG) Question 178

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by seekerk » Wed Jan 11, 2012 1:14 am
club no of students
chess 40
drama 30
maths 25

The table above shows the number of students in 3 clubs in McAuliffe School. Although no student is in all 3 clubs, 10 students are in both chess and drama, 5 students are in both chess and maths, and 6 students are in both drama and maths. How many different students are in the 3 clubs?

a) 68
b) 69
c) 74
d) 79
e) 84


The ans is C.

I understand the logic behind not double counting (25+14+14+10+5+6)=74. But what I don't understand is the question itself. "How many DIFFERENT students are in the three clubs?"
I was thinking: you have three clubs, and each club must have students that are different from each other. That would mean to say those students who are in 2 clubs shouldn't be included, which would lead to (25+14+14)=53.
I find that some of the quantitative questions can be tricky and they test your English as well.
These are the kind of mistakes that we should all avoid. Any advise on this?
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by Whitney Garner » Wed Jan 11, 2012 5:50 am
seekerk wrote:club no of students
chess 40
drama 30
maths 25

The table above shows the number of students in 3 clubs in McAuliffe School. Although no student is in all 3 clubs, 10 students are in both chess and drama, 5 students are in both chess and maths, and 6 students are in both drama and maths. How many different students are in the 3 clubs?

a) 68
b) 69
c) 74
d) 79
e) 84


The ans is C.

I understand the logic behind not double counting (25+14+14+10+5+6)=74. But what I don't understand is the question itself. "How many DIFFERENT students are in the three clubs?"
I was thinking: you have three clubs, and each club must have students that are different from each other. That would mean to say those students who are in 2 clubs shouldn't be included, which would lead to (25+14+14)=53.
I find that some of the quantitative questions can be tricky and they test your English as well.
These are the kind of mistakes that we should all avoid. Any advise on this?
Hi seekerk!

I think that the biggest thing they are trying to accomplish here with the word "different" is to explicitly tell you NOT to double-count (ie. do not count the same person 2 times because one person cannot be 2 different people). I know that the language of some of the Quant problems can be a bear, but more often than not, the test writers are just trying to be as definitive as possible to ensure that no one complains later of ambiguity.

This is the same reason that they will explicitly omit values or variables during certain problems. For example, a problem might tell you at the beginning that xy /= 0. But knowing that neither is equal to zero isn't a big part of answering the question, it just happens that at some point in the calculations, you have xy in a denominator. A recent example I saw was for a DS question that asked:

What is 1/(x-1)? The stem had to also specify that x/=1 or someone might have called them out for the possibility that the expression could be Undefined.

In sets problems, they do not always get this explicit, but there is the concern with your example that a person might say - "How many students are in the clubs, well just add them up". By saying different they are trying to help avoid that confusion (but I completely understand it might make many of us second guess what they are trying to say!!).

Check out a similar occurance in OG12th PS #135:
"The diagram below shows the various paths along which a mouse can travel from point X, where it is released, to point Y, where it is rewarded with a food pellet. How many <b>different</b> paths from X to Y can the mouse take if it goes directly from X to Y without retracing any point along its path?"

If they didn't expressly say DIFFERENT, someone might have said, well it could be infinite because it didn't say I couldn't count the same one over and over and over!

:)
Hope this helps!
Whit
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by LalaB » Wed Jan 11, 2012 11:31 am
from the q.stem u can see,that one person can be in 2 groups. the q. is asking u to find a real numbers of persons-

(40+30+25)-10-5-6=74
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by ArunangsuSahu » Thu Jan 12, 2012 7:19 am
Total= 40+30+25-10-5-6=74

(C) is the answer
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by ronnie1985 » Thu Jan 12, 2012 11:29 pm
USE VENN DIAGRAM OR THE FORMULA n(AUBUC) = N(A)+ n(B0 + n(C) - n(AintB)- n(BintC) - n(CintA)
ANS = 74 (c)
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by seekerk » Fri Jan 13, 2012 11:39 pm
Thanks. Are there any more examples where the English is not too clear and we need to take note?
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by megha kumar » Thu Dec 25, 2014 6:58 pm
Can someone explain me how to solve this using Venn Diagram? Thanks
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