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So I've never stumbled upon this type of problem (Vienn

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So I've never stumbled upon this type of problem (Vienn [#permalink] New post 29 Sep 2006, 07:38
So I've never stumbled upon this type of problem (Vienn diagram) until I noticed it in Rhyme's gmat-knowledge pdf.
(http://www.gmatclub.com/phpbb/viewtopic.php?t=30030&postdays=0&postorder=asc&start=40 )

So I want to know if I did it correcly, since it doesn't provide the final answer.

PROBLEM I:

20 people are on marketing team, 30 on sales,
40 on vision. 5 are on both marketing and sales, 6 on sales
and vision, 9 on both marketing and vision, 4 in all three.
How many workers are there in total?



PROBLEM II from MGMAT:


In a group of 68 students, each student is registered for at least one of three classes – History, Math and English. Twenty-five students are registered for History, twenty-five students are registered for Math, and thirty-four students are registered for English. If only three students are registered for all three classes, how many students are registered for exactly two classes?



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 [#permalink] New post 29 Sep 2006, 13:23
Problem 1: 74

Problem II : 49
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 [#permalink] New post 29 Sep 2006, 13:46
abhis0 wrote:
Problem 1: 74

Problem II : 49


Could you show your math for I, I got 74. II is incorrect.
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 [#permalink] New post 30 Sep 2006, 16:12
Nsentra wrote:
abhis0 wrote:
Problem 1: 74

Problem II : 49

quote]

bump. I thought the formula T=A+B+C-(AB+BC+AC)+ABC couple be applied to for Porlem II? but it doesn't seem to work. Please explain. 49 is not correct answer.
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Re: Overlapping sets review [#permalink] New post 30 Sep 2006, 20:39
I - I got 74

20+30+40-(5+9+6)+4

II - I got 19

68 = (25+25+34)-X+3
X=87-68 = 19

Nsentra wrote:
So I've never stumbled upon this type of problem (Vienn diagram) until I noticed it in Rhyme's gmat-knowledge pdf.
(http://www.gmatclub.com/phpbb/viewtopic.php?t=30030&postdays=0&postorder=asc&start=40 )

So I want to know if I did it correcly, since it doesn't provide the final answer.

PROBLEM I:

20 people are on marketing team, 30 on sales,
40 on vision. 5 are on both marketing and sales, 6 on sales
and vision, 9 on both marketing and vision, 4 in all three.
How many workers are there in total?



PROBLEM II from MGMAT:


In a group of 68 students, each student is registered for at least one of three classes – History, Math and English. Twenty-five students are registered for History, twenty-five students are registered for Math, and thirty-four students are registered for English. If only three students are registered for all three classes, how many students are registered for exactly two classes?



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 [#permalink] New post 30 Sep 2006, 20:47
I 74. II 10.

II is not 19. you have to sutract those who take all 3. In your 19, you added this 3 three times.
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 [#permalink] New post 01 Oct 2006, 06:46
tennis_ball wrote:
I 74. II 10.

II is not 19. you have to sutract those who take all 3. In your 19, you added this 3 three times.


So how come in first problem you don't subtract 4, but in second one you do subtract 3?
I - I got 74

20+30+40-(5+9+6)+4

II - I got 19

68 = (25+25+34)-X+3
X=87-68 = 19


10 is OA btw
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 [#permalink] New post 01 Oct 2006, 18:10
You didn't read the question clearly.

It is asking for students with exact 2 choices.
Say: E for English, M for Math and H for history.

X for only E&M
Y for only E&H
Z for only M&H

Remember these X Y Z don't include those students who take all 3 courses, and they are what the question is asking about.

so the formula is:

68 = 25 + 25 + 34 - (X+3) - (Y+3) - (Z+3) + 3.
then: X + Y + Z = 10.

you have to subtract the 3 two-overlapped areas and plus 1 three-overlapped area back.
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 [#permalink] New post 01 Oct 2006, 19:12
tennis_ball wrote:
You didn't read the question clearly.

It is asking for students with exact 2 choices.
Say: E for English, M for Math and H for history.

X for only E&M
Y for only E&H
Z for only M&H

Remember these X Y Z don't include those students who take all 3 courses, and they are what the question is asking about.

so the formula is:

68 = 25 + 25 + 34 - (X+3) - (Y+3) - (Z+3) + 3.
then: X + Y + Z = 10.

you have to subtract the 3 two-overlapped areas and plus 1 three-overlapped area back.


Thanks tennis_ball!!
  [#permalink] 01 Oct 2006, 19:12
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