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Help requred for Sets problem

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Help requred for Sets problem [#permalink] New post 08 Apr 2005, 02:24
Can anyone help me solve this sets problem. Sets is my weakest link in PS.

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

Thanks,
:twisted:
Darth
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 [#permalink] New post 08 Apr 2005, 02:31
Total students in all 3 classes= 40+35+30=105
Students in only one class= 70
students in 3 classes= 15
students in 2 classes= 105-15-70=20
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Re: Help requred for Sets problem [#permalink] New post 08 Apr 2005, 02:52
Darth_McDaddy wrote:
Can anyone help me solve this sets problem. Sets is my weakest link in PS.

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

Thanks,
:twisted:
Darth


If R1 - R7 represent regions,

R1 + R2 + R3 + R4 + R5 + R6 + R7 = 70.

When you just add Maths + English + German, you get
(R1+R2+R7+R6) + (R4+R5+R6+R7) + (R2+R3+R4+R7) = 40 + 35 + 30

70 + R2 + R4 + R6 + R7 + R7 = 105
=> R2 + R4 + R6 + 2xR7 = 35.
Since R7 = 15,
R2 + R4 + R6 = 5.

Hope that helps.
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 [#permalink] New post 08 Apr 2005, 03:51
Godd explanation and drawing, kapslock.
Could you explain why
70=R(1->7)
and not
70=R1+R3+R5
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 [#permalink] New post 08 Apr 2005, 04:17
Good question theReach !!

Actually this has to do with the conventional way of stating the set questions.
When they say 70 people are enrolled in Maths, English or German, they just mean to imply that 70 makes up the Universal Set.

Of course the real statement should have been "70 people are enrolled in Maths, English and German, not counting the people pursuing two or more courses two or more times".

Taking this assumption, you'd agree with me that R(1-7) does specify the universal set.

Hope this helps.
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Kapslock is right [#permalink] New post 11 Apr 2005, 04:08
Thank you so much Kapslock. Your answer is correct. It is 5. I got the question from Barrons, but the explanation wasn't clear. Your venn diagram was very useful, and so were your explanations.

I also thank Thearch for helping me answer the question.
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 [#permalink] New post 11 Apr 2005, 17:57
Here is the way I think about any set problem

Total = # in set 1 + # in set 2 + # in set 3 - # in two sets - 2*(number in 3 sets) - 3 * (number in 4 sets)

So for this example:

70 = (40 + 35 + 30) - x - (2*15)
70 = 105 - 30 - x
70 = 75 - x
so x == 5
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 [#permalink] New post 14 Apr 2005, 18:50
I get 5 as well.. Same methodology as Kapslock - but slightly different way. I used x, y & z for the 2 subject intersections

70 = 25 -(x+y) + 20 -(y+z) + 15 -(x+z) + x+y+z +15
=> x+y+z = 75 -70 =5
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Re: Help requred for Sets problem [#permalink] New post 14 Apr 2005, 22:24
Darth_McDaddy wrote:
Can anyone help me solve this sets problem. Sets is my weakest link in PS.

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

Thanks,
:twisted:
Darth


I have a document on "working with sets" here. Hope it's helpful

http://www.gmatclub.com/phpbb/viewtopic.php?t=14576&start=20
Re: Help requred for Sets problem   [#permalink] 14 Apr 2005, 22:24
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