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A school has 3 classes, math class has 14 students [#permalink]

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01 Apr 2013, 11:43

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A school has 3 classes, math class has 14 students. English class has 10 students, PE class has 11 students. There are 20 students taking only one class, 3 students are taking all three classes. How many students are taking two classes?

A) 3 B) 6 C) 9 D) 18 E)20

could any one show me the solution using 3 overlapping groups formula .

T = A + B + C - (AB + AC + BC) - 2(ABC) ... I am getting a wrong answer

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 \(=x-3*3\) \(12-9=3\) _________________

It is beyond a doubt that all our knowledge that begins with experience.

Re: A school has 3 classes, math class has 14 students [#permalink]

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01 Apr 2013, 21:54

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guerrero25 wrote:

A school has 3 classes, math class has 14 students. English class has 10 students, PE class has 11 students. There are 20 students taking only one class, 3 students are taking all three classes. How many students are taking two classes?

A) 3 B) 6 C) 9 D) 18 E)20

could any one show me the solution using 3 overlapping groups formula .

T = A + B + C - (AB + AC + BC) - 2(ABC) ... I am getting a wrong answer

thanks !

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

Re: A school has 3 classes, math class has 14 students [#permalink]

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02 Apr 2013, 08:02

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guerrero25 wrote:

A school has 3 classes, math class has 14 students. English class has 10 students, PE class has 11 students. There are 20 students taking only one class, 3 students are taking all three classes. How many students are taking two classes?

A) 3 B) 6 C) 9 D) 18 E)20

could any one show me the solution using 3 overlapping groups formula .

T = A + B + C - (AB + AC + BC) - 2(ABC) ... I am getting a wrong answer

thanks !

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. _________________

“If you can't fly then run, if you can't run then walk, if you can't walk then crawl, but whatever you do you have to keep moving forward.”

Re: A school has 3 classes, math class has 14 students [#permalink]

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02 Apr 2013, 10:11

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

Attachment:

venn.jpg [ 55.63 KiB | Viewed 2523 times ]

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.

Re: A school has 3 classes, math class has 14 students [#permalink]

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02 Apr 2013, 15:10

mridulparashar1 wrote:

guerrero25 wrote:

A school has 3 classes, math class has 14 students. English class has 10 students, PE class has 11 students. There are 20 students taking only one class, 3 students are taking all three classes. How many students are taking two classes?

A) 3 B) 6 C) 9 D) 18 E)20

could any one show me the solution using 3 overlapping groups formula .

T = A + B + C - (AB + AC + BC) - 2(ABC) ... I am getting a wrong answer

thanks !

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.

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 .

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.

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:

Attachment:

SetsThree_1_23Sept.jpg [ 20.19 KiB | Viewed 2475 times ]

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.

Re: A school has 3 classes, math class has 14 students [#permalink]

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16 Apr 2013, 09:34

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... _________________

You've been walking the ocean's edge, holding up your robes to keep them dry. You must dive naked under, and deeper under, a thousand times deeper! - Rumi

http://www.manhattangmat.com/blog/index.php/author/cbermanmanhattanprep-com/ - This is worth its weight in gold

Economist GMAT Test - 730, Q50, V41 Aug 9th, 2013 Manhattan GMAT Test - 670, Q45, V36 Aug 11th, 2013 Manhattan GMAT Test - 680, Q47, V36 Aug 17th, 2013 GmatPrep CAT 1 - 770, Q50, V44 Aug 24th, 2013 Manhattan GMAT Test - 690, Q45, V39 Aug 30th, 2013 Manhattan GMAT Test - 710, Q48, V39 Sep 13th, 2013 GmatPrep CAT 2 - 740, Q49, V41 Oct 6th, 2013

GMAT - 770, Q50, V44, Oct 7th, 2013 My Debrief - http://gmatclub.com/forum/from-the-ashes-thou-shall-rise-770-q-50-v-44-awa-5-ir-162299.html#p1284542

Re: A school has 3 classes, math class has 14 students [#permalink]

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16 Apr 2013, 09:53

Transcendentalist wrote:

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". _________________

Re: A school has 3 classes, math class has 14 students [#permalink]

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16 Apr 2013, 09:57

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've been walking the ocean's edge, holding up your robes to keep them dry. You must dive naked under, and deeper under, a thousand times deeper! - Rumi

http://www.manhattangmat.com/blog/index.php/author/cbermanmanhattanprep-com/ - This is worth its weight in gold

Economist GMAT Test - 730, Q50, V41 Aug 9th, 2013 Manhattan GMAT Test - 670, Q45, V36 Aug 11th, 2013 Manhattan GMAT Test - 680, Q47, V36 Aug 17th, 2013 GmatPrep CAT 1 - 770, Q50, V44 Aug 24th, 2013 Manhattan GMAT Test - 690, Q45, V39 Aug 30th, 2013 Manhattan GMAT Test - 710, Q48, V39 Sep 13th, 2013 GmatPrep CAT 2 - 740, Q49, V41 Oct 6th, 2013

GMAT - 770, Q50, V44, Oct 7th, 2013 My Debrief - http://gmatclub.com/forum/from-the-ashes-thou-shall-rise-770-q-50-v-44-awa-5-ir-162299.html#p1284542

Re: A school has 3 classes, math class has 14 students [#permalink]

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16 Apr 2013, 10:23

Transcendentalist wrote:

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. _________________

Re: A school has 3 classes, math class has 14 students [#permalink]

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25 Jan 2014, 03:41

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. _________________

Re: A school has 3 classes, math class has 14 students [#permalink]

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Re: A school has 3 classes, math class has 14 students [#permalink]

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11 Oct 2015, 09:25

guerrero25 wrote:

A school has 3 classes, math class has 14 students. English class has 10 students, PE class has 11 students. There are 20 students taking only one class, 3 students are taking all three classes. How many students are taking two classes?

A) 3 B) 6 C) 9 D) 18 E)20

could any one show me the solution using 3 overlapping groups formula .

T = A + B + C - (AB + AC + BC) - 2(ABC) ... I am getting a wrong answer

thanks !

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.

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Re: A school has 3 classes, math class has 14 students
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11 Oct 2015, 09:25

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