A school has 3 classes, math class has 14 students

This could be solved easier:

(1) There are 14+10+11 = 35 seats to fill (some students occupy only 1 seat, others 2, others 3).

(2) 20 students occupy only 1 seat: 35-20 = 15... now we have 15 seats to fill

(3) 3 students occupy 3 seats: 15-9 = 6 seats left

(4) We have 6 seats for students that occupy 2 seats ---> solution is 3 students!!

Answer A

If we put up a Venn diagram of the situation, we get something like this :

With :

A : the number of students only attending math class ;
B : the number of students only attending english class ;
C : the number of students only attending PE class ;
x : the number of students attending both math and english ;
y : the number of students attending both math and PE ;
z : the number of students attending both english and PE.

If we use the data we're given, we get the following system :

14 students attend math classes => \(x + y + 3 + A = 14\) (1)
10 students attend english classes => \(x + z + 3 + B = 10\) (2)
11 students attend PE classes =>\(y + z + 3 + C = 11\) (3)

We also know that there are only 20 students attending only one class, which means that : \(A + B + C = 20\) (4)

If we add equations (1), (2) and (3) and take into account equation (4) we get : \(2*(x + y + z) + 9 + 20 =35\)

The quantity \((x + y + z)\) represents the total number of students attending 2 classes, which is what we're looking for. So completing the computation, we get :
\(2*(x + y + z) + 9 + 20 =35\) => \(2*(x + y + z) = 6\) => \((x + y + z) = 3\) which is answer choice A.

Hope that helped

The right formula is:

P(A u B u C) : P(A) + P(B) + P(C) – P(A n B) – P(A n C) – P(B n C) + P(A n B n C)
formulae-for-3-overlapping-sets-69014.html

Here I would use those:
2. To determine the No of persons in exactly one set : P(A) + P(B) + P(C) – 2P(A n B) – 2P(A n C) – 2P(B n C) + 3P(A n B n C)
\(x\)=number in 2 classes or P(A n B) + P(A n C) + P(B n C)
\(20=14+10+11-2x+9*3\)
\(x=12\)
3. To determine the No of persons in exactly two of the sets : P(A n B) + P(A n C) + P(B n C) – 3P(A n B n C)
That's the answer:
Answer \(=x-3*3\)
\(12-9=3\)

T = 14 + 10 + 11 - 2[P(A U B) + P(A U C) + P(C U B)] + 3P(A U B U C)
Let the sum of students attending 2 classes be Y

T = 35 - 2Y + 9 = 44 - 2Y ------- I

Also,
T = 20 + Y + 3
T = 23 + Y --------- II

Comparing I & II we get Y = 3

Hi guerrero25,

There are 2 forumulaes
T = A+B+C - (Sum of 2 groups) + ( Sum of all three groups) + Neither

and T = A+B+C - (Sum of exactly 2 groups) - 2( sum of all three groups) + Neither

In this Q, Neither = 0

T = 14+10+11

35= 14+10+11 - x + 3 +0
x =3

Note the 2nd Forumlae will be applied when we are asked about exactly 2 classes where as 1st formulae will be applied for group attending 2 classes.
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Hello Mridul ,Thanks for the explanation . Could you elaborate more when you say "Note the 2nd Forumlae will be applied when we are asked about exactly 2 classes where as 1st formulae will be applied for group attending 2 classes"..I could not understand the difference .

thanks again !

To understand the two different formulas, look at the diagram given here:



A - No of people in set A = a + d + e + g
B - No of people in set B = b + d + g + f
C - No of people in set C = c + e + g + f

Notice that Total = a + b+ c + d + e + f + g
But depending on the given data, we often don't have values for a, b, c etc separately. When the question says, Math class has 14 students, it means a + d + e + g = 14. Similarly, if English class has 10 students, it means b + d + g + f = 10 and so on...
So we need to subtract the things we have counted twice/thrice. Note that if we write 14 + 10, we have already counted d and g twice here.

When we add A + B, we have counted (d +g) twice so we need to subtract it out once.
When we add C to it as well, we have counted (e + g) and (f + g) twice too so we need to subtract them out as well.
But when we do this, we have subtracted the g region 3 times and hence, it's not accounted for now. So we add it back.
This is how you get
T = A+B+C - (Sum of 2 groups) + ( Sum of all three groups) + Neither

On the other hand,
If after adding A + B + C, you subtract the region which is common to ONLY two sets i.e. d, e and f, then you still need to subtract the g region twice since it is present in A, B and C.
That is how you get the formula:
T = A+B+C - (Sum of ONLY 2 groups) - 2*( Sum of all three groups) + Neither

The formula to use depends on what data is given in the question.

To avoid confusion, it is a good idea to use venn diagrams instead.

Isn't the question ambiguous?

A is answer is the question is only two classes

B is the answer if it is at least two classes

Correct me if i am wrong...

I don't think there is any ambiguity. If they were asking for "atleast two classes", the question would have mentioned that specifically.Also, the fact that they have given the number of students who take all three classes, doesn't really add anything new to the scenario of "atleast two classes".
_________________

Maybe i am confusing my verbal with quant

But the right answer to the question "How many students are taking two classes? " is 6 as there are 6 students taking two classes

You can think of it as this : number of students taking two classes : M and E OR E and PE OR M and PE.
Question asks how many take either M and E OR E and PE OR M and PE? The students who take all three will not fall under this category as they take all three.
Again, the answer will be 6 , only for number of students taking atleast two classes,i.e. two or more.
I think the word "only" is inherent for the given question.
_________________

Yes, I would think twice too. The official questions will say either 'how many take EXACTLY two classes' or 'How many take AT LEAST two classes'. GMAC digs potholes for you but it is not unfair. You will never wonder what they want to know - you may ignore what they want to know or you may wonder how to get it.

Yup I agree with Transcendentalist. The question asks how many students take 2 classes. Those 3 people who take 3 classes ALSO take two classes. Hence 6 seems correct choice.

Here we have to find (AB + AC + BC), and given (A + B + C) = 20, (ABC) =3

so we use formula
Sum of individual Total of A , B and C = (A + B + C) +2(AB + AC + BC) +3(ABC)

=> 14 + 10 + 11 = 20 + 2(AB + AC + BC) +3(3)

=> (AB + AC + BC) = 3 Answer


If explanation is of help, give kudos to motivate me.

Here,

Let T be the total no of students,

then T = 20 + x +3,
where, x be the no of students taking exactly 2 classes.

So,
20+x+3 = 14+10+11 -x -2 *3. This should give you the answer.

here x is :
x = AB+BC+AC - 3(ABC), where ABC means students taking all classes and ABC = 3 (GIVEN), AB,BC,AC means students taking 2 classes (and not the only two classes).