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In a group of 68 students, each student is registered for at [#permalink]

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25 Aug 2007, 23:35

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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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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? 13 10 9 8 7

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? 13 10 9 8 7

please show the working.

68 = 25 +25 +34 - (ab + bc + ac) - 2(3)
ab + bc + ac = 1o

Bkk/fistail, could you tell me why did you do "2 * 3 "

bkk145 wrote:

empty_spaces wrote:

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? 13 10 9 8 7

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? 13 10 9 8 7

Basically, you count "all three" twice because question ask for "exactly two". If the question ask "two and more", then you do not have to multiply 2 since it is already count "all three" once.

The formula usually given for this type of problem use "Two or more" as the principle.

A U B U C = A + B + C - (A intersection B) - (B intersection C) - (C intersection A) + ( A intersection B intersection C)

bkk145 wrote:

empty_spaces wrote:

Bkk, could you tell me why did you do "2 * 3 "

bkk145 wrote:

empty_spaces wrote:

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? 13 10 9 8 7

Basically, you count "all three" twice because question ask for "exactly two". If the question ask "two and more", then you do not have to multiply 2 since it is already count "all three" once.

The formula usually given for this type of problem use "Two or more" as the principle.

A U B U C = A + B + C - (A intersection B) - (B intersection C) - (C intersection A) + ( A intersection B intersection C)

This is exactly what I mean. Let's use class here, so we have students taking classes A, B, and C.

Take this part as an example: (A intersection B)
This part includes students who take both "A and B" AND "A and B and C"

Same thing for (B intersection C)
This part includes students who take both "B and C" AND "A and B and C"

Which is same thing for (C intersection A)

So, you can see that (X intersection Y) counts students who take both classes as well as all three classes. Therefore, when the question ask for "exactly two", you have to multiply "all three" by 2 because "exactly two" doesn't include "A and B and C". Hope this helps.

Oh , also, I think this is the correct formula:
A U B U C = A + B + C - (A intersection B) - (B intersection C) - (C intersection A) - ( A intersection B intersection C)

thanks Bkk for your effort although am still confused, maybe my brain isnt really working at 3 in the night.

bkk145 wrote:

empty_spaces wrote:

A U B U C = A + B + C - (A intersection B) - (B intersection C) - (C intersection A) + ( A intersection B intersection C)

This is exactly what I mean. Let's use class here, so we have students taking classes A, B, and C.

Take this part as an example: (A intersection B) This part includes students who take both "A and B" AND "A and B and C"

Same thing for (B intersection C) This part includes students who take both "B and C" AND "A and B and C"

Which is same thing for (C intersection A)

So, you can see that (X intersection Y) counts students who take both classes as well as all three classes. Therefore, when the question ask for "exactly two", you have to multiply "all three" by 2 because "exactly two" doesn't include "A and B and C". Hope this helps.

Oh , also, I think this is the correct formula: A U B U C = A + B + C - (A intersection B) - (B intersection C) - (C intersection A) - ( A intersection B intersection C)

gmatclubot

Re: PS # Set theory
[#permalink]
26 Aug 2007, 00:50