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# Sets / Counting Problem

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01 Jun 2009, 12:20
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Here is the problem.

70 students are enrolled in Math, English, or German. 40 students are in Math, 35 are in English, 30 are in German. 15 students are enrolled in all 3 of the courses. How many of the students are enrolled in exactly two of the courses: Math, English, and German?

The book has very bad explanation and possibly wrong answer for this problem. Using my own logic I get a different answer than the book.

Please let me know what you guys get as answer for this problem.

thank you.
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Re: Sets / Counting Problem [#permalink]

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01 Jun 2009, 14:31
bfman wrote:
Here is the problem.

70 students are enrolled in Math, English, or German. 40 students are in Math, 35 are in English, 30 are in German. 15 students are enrolled in all 3 of the courses. How many of the students are enrolled in exactly two of the courses: Math, English, and German?

The book has very bad explanation and possibly wrong answer for this problem. Using my own logic I get a different answer than the book.

Please let me know what you guys get as answer for this problem.

thank you.

Try using a Venn diagram...
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AkamaiBrah
Former Senior Instructor, Manhattan GMAT and VeritasPrep
Vice President, Midtown NYC Investment Bank, Structured Finance IT
MFE, Haas School of Business, UC Berkeley, Class of 2005
MBA, Anderson School of Management, UCLA, Class of 1993

Intern
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Re: Sets / Counting Problem [#permalink]

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01 Jun 2009, 14:49
I did. Didn't really help. I am getting 20 as my answer. Book's answer is 5.
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Re: Sets / Counting Problem [#permalink]

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01 Jun 2009, 14:59
Obviously, Every student is taking at least one course.

M + E + G - ME - MG - GE + MGE = 70

M + E + G + MGE - 70 = ME + MG + GE = Students with two courses

40 + 35 + 30 + 15 - 70 = 50 Students are taking exactly two courses.
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Re: Sets / Counting Problem [#permalink]

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01 Jun 2009, 20:53
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I got the answer as 5.

AuBuC = A + B + C - ( AnB + BnC + CnA) - 2( AnBnC)

70 = 40 + 35 + 30 - ( x) - 2 ( 15 )

x = 5.
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Re: Sets / Counting Problem [#permalink]

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01 Jun 2009, 21:02
The answer is 5. I figured out the problem.
Thank you all for your help.
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Re: Sets / Counting Problem [#permalink]

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02 Jun 2009, 03:48
tkarthi4u wrote:
I got the answer as 5.

AuBuC = A + B + C - ( AnB + BnC + CnA) - 2( AnBnC)

70 = 40 + 35 + 30 - ( x) - 2 ( 15 )

x = 5.

i'm not able to understand why you subtracted "2( AnBnC)". please explain.

i did it by venn diagram.
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Re: Sets / Counting Problem [#permalink]

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02 Jun 2009, 08:02
Here is a better explanation.

You might want to set up a venn diagriam with 3 circles and plug in the following values:

all 3 classes: 15

M+E = x
M+G = y
E+G = z

Only M = 25 -(x+y)
Only E = 20-(x+z)
Only G = 15 -(y+z)

The question is asking for x+y+z

(25-x-y) + (20-x-z) + (15-y-z) + (x+y+z) +15 = 70

75-x-y-z = 70

x+y+z = 5
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Re: Sets / Counting Problem [#permalink]

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02 Jun 2009, 08:03
goldeneagle94 wrote:
tkarthi4u wrote:
I got the answer as 5.

AuBuC = A + B + C - ( AnB + BnC + CnA) - 2( AnBnC)

70 = 40 + 35 + 30 - ( x) - 2 ( 15 )

x = 5.

i'm not able to understand why you subtracted "2( AnBnC)". please explain.

i did it by venn diagram.

I would also be interested in knowing this as the book applies the same logic which I didn't get. Thank you.
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Re: Sets / Counting Problem [#permalink]

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02 Jun 2009, 09:17
bfman wrote:
Here is a better explanation.

You might want to set up a venn diagriam with 3 circles and plug in the following values:

all 3 classes: 15

M+E = x
M+G = y
E+G = z

Only M = 25 -(x+y)
Only E = 20-(x+z)
Only G = 15 -(y+z)

The question is asking for x+y+z

(25-x-y) + (20-x-z) + (15-y-z) + (x+y+z) +15 = 70

75-x-y-z = 70

x+y+z = 5

Thx. I got this approach now.
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Re: Sets / Counting Problem [#permalink]

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12 Jun 2009, 14:58
goldeneagle94 wrote:
tkarthi4u wrote:
I got the answer as 5.

AuBuC = A + B + C - ( AnB + BnC + CnA) - 2( AnBnC)

70 = 40 + 35 + 30 - ( x) - 2 ( 15 )

x = 5.

i'm not able to understand why you subtracted "2( AnBnC)". please explain.

i did it by venn diagram.

The formula that tkarthi4u listed is a standard formula for venn diagrams with 3 circles.

If you add A + B + C, you have overlaps that you need to remove to prevent double-counting. AnB, BnC, and CnA overlap and are double-counted, so you need to subtract each of them once. If you subtracted AnB twice, you would not count that section altogether.

(AnBnC) is subtracted twice because when you add A, B, and C, you are triple-counting that area. You need to subtract out that area twice (and thereby only counting that area once).
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Re: Sets / Counting Problem [#permalink]

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13 Jun 2009, 04:32
disc108 wrote:
If you add A + B + C, you have overlaps that you need to remove to prevent double-counting. AnB, BnC, and CnA overlap and are double-counted, so you need to subtract each of them once. If you subtracted AnB twice, you would not count that section altogether.

(AnBnC) is subtracted twice because when you add A, B, and C, you are triple-counting that area. You need to subtract out that area twice (and thereby only counting that area once).

I think the standard formula is AuBuC = A + B + C - ( AnB + BnC + CnA) + ( AnBnC)

(AnBnC) is subtracted twice while adding the third circle, that is why we need to add it Once.
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Re: Sets / Counting Problem [#permalink]

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13 Jun 2009, 05:16
Slightly other way:

1) (40+35+30) - 70 = 35 - the number of students with two courses and the double number of students with three courses.
2) 35 - 2*15 = 5 - the number of students with two courses.
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Re: Sets / Counting Problem   [#permalink] 13 Jun 2009, 05:16
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