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# A school has 3 classes, math class has 14 students

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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

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

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01 Apr 2013, 12:01
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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)
Answer $$=x-3*3$$
$$12-9=3$$
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Re: A school has 3 classes, math class has 14 students [#permalink]

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01 Apr 2013, 12:52
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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!!

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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

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

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

Comparing I & II we get Y = 3
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Re: A school has 3 classes, math class has 14 students [#permalink]

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02 Apr 2013, 08:02
1
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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.
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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 2378 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.

Hope that helped
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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 .

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

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02 Apr 2013, 21:26
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Expert's post
guerrero25 wrote:
mridulparashar1 wrote:

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 .

thanks again !

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

SetsThree_1_23Sept.jpg [ 20.19 KiB | Viewed 2331 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.

To avoid confusion, it is a good idea to use venn diagrams instead.
Check out my post: http://www.veritasprep.com/blog/2012/09 ... ping-sets/
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Current Student Joined: 24 Nov 2012 Posts: 177 Concentration: Sustainability, Entrepreneurship GMAT 1: 770 Q50 V44 WE: Business Development (Internet and New Media) Followers: 36 Kudos [?]: 205 [0], given: 73 Re: A school has 3 classes, math class has 14 students [#permalink] ### Show Tags 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 ... nprep-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 - from-the-ashes-thou-shall-rise-770-q-50-v-44-awa-5-ir-162299.html#p1284542 Verbal Forum Moderator Joined: 10 Oct 2012 Posts: 630 Followers: 72 Kudos [?]: 919 [0], given: 136 Re: A school has 3 classes, math class has 14 students [#permalink] ### Show Tags 16 Apr 2013, 09:53 Expert's post 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". _________________ Current Student Joined: 24 Nov 2012 Posts: 177 Concentration: Sustainability, Entrepreneurship GMAT 1: 770 Q50 V44 WE: Business Development (Internet and New Media) Followers: 36 Kudos [?]: 205 [0], given: 73 Re: A school has 3 classes, math class has 14 students [#permalink] ### Show Tags 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 ... nprep-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 - from-the-ashes-thou-shall-rise-770-q-50-v-44-awa-5-ir-162299.html#p1284542 Verbal Forum Moderator Joined: 10 Oct 2012 Posts: 630 Followers: 72 Kudos [?]: 919 [0], given: 136 Re: A school has 3 classes, math class has 14 students [#permalink] ### Show Tags 16 Apr 2013, 10:23 Expert's post 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. _________________ Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 6473 Location: Pune, India Followers: 1754 Kudos [?]: 10458 [0], given: 205 Re: A school has 3 classes, math class has 14 students [#permalink] ### Show Tags 16 Apr 2013, 22:40 Expert's post 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... 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. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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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.
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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

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