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# Set Theory

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Set Theory [#permalink]  31 Aug 2007, 11:00
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Good question that I came across. Will post OA later.

In a class of 120 students numbered 1 to 120, all even numbered students opt for Physics, whose numbers are divisible by 5 opt for Chemistry and those whose numbers are divisible by 7 opt for Math. How many opt for none of the three subjects?

A. 19
B. 41
C. 21
D. 57
E. 26
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remove even numbers ---> 120/2 = 60

remove numbers that divisable by five and are not even ---> 5,15,25,35,45,55,65,75,85,95,105,115 = 12

remove numbers that divisable by seven and are not even ---> 7,21,35,49,63,77,91,105,119 = 9

add numbers divisable by five and seven and are not even ---> 35,105 = 2

120 - 60 - 12 - 9 + 2 = 41

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Re: Set Theory [#permalink]  31 Aug 2007, 12:05
abhi_ferrari wrote:
Good question that I came across. Will post OA later.

In a class of 120 students numbered 1 to 120, all even numbered students opt for Physics, whose numbers are divisible by 5 opt for Chemistry and those whose numbers are divisible by 7 opt for Math. How many opt for none of the three subjects?

A. 19
B. 41
C. 21
D. 57
E. 26

Bear with me, I last took the test in nov-06

Let's see, there are 60 evens and 60 odds. So let's start with 60. So 57 is out (too close to 60). There are 24 numbers that are multiples of 5, but half are even, so take away another 12. Left with 47.

Blahh, time running out!! I'm taking an educated guess at 41.

If you have some time left:
7 goes into 120 17 times. Take away about half (not sure if it should up or down; not very important right now) for even numbers. So your left with 8.5ish. how many multiples does 7 and 5 share? 35, 70, and 105. 70 is even, so eliminate that. We're left with 2

So, 47 - (8.5 - 2) = 40.5

Close enough to B)41
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Re: Set Theory [#permalink]  31 Aug 2007, 12:26
abhi_ferrari wrote:
Good question that I came across. Will post OA later.

In a class of 120 students numbered 1 to 120, all even numbered students opt for Physics, whose numbers are divisible by 5 opt for Chemistry and those whose numbers are divisible by 7 opt for Math. How many opt for none of the three subjects?

A. 19
B. 41
C. 21
D. 57
E. 26

Here is another approach, should not run out of time with this one:

I got B

the student numbers can be divisible by either

1,2,3,4,5,6,7,8, 9

the questions says that the student with numbers divisible by 3 and 9, plus 1 (the 1st student) took none of the subjects.
So,
we want to find # of numbers divisible by 3------(120-1)/3+1=40
and
# of numbers divisible by 9----(120-1)/3=13,
So that not to double count, since those divis by 9 are also divis by 3..

(40-13)+13=total number of student with numbers divisible by 3 and 9

makes it 40+1st student=41

Good one
Re: Set Theory   [#permalink] 31 Aug 2007, 12:26
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