300 students participated in a symposium with 3 conferences in sequence.

Half of the students that attended the first conference, attended neither the other two.

One-third of the students that attended the second conference, attended neither the other two.

And one-fourth of the students that attended the third conference, attended neither the other two.

Knowing that the three conferences were attended by the same number of students, and that each of the 300 students attended at least one conference :

a) how many students attended each conference ?
b) how many students attended only one conference ?
c) how many students attended only two conferences ?
d) how many students attended all 3 conferences ?

One unique solution !

Since pcbouhid insists on an analytical solution...

Let x= the number of students that attended each conference.
Let y= the number of students that attended the first and second conferences, but not the third.

On my scratch paper, I have this drawn as a venn diagram, but here, I can't so easily.

x/2 students attended only the first conference
x/3 students attended only the second conference
x/4 students attended only the third conference
y students attended both the first and second, but not the third

Consider the second and third conferences.  The number that attended each is the same, so therefore the number that attended the second but not the third is equal to the number that attended the third but not the second.  x/3+y=x/4+?
Therefore, x/12+y students attended the first and third but not the second.

Consider the first and third conferences.  By the same reasoning as of that above, the number that attended the first but not the third is equal to the number that attended the third but not the first.  x/2+y = x/4+?
Therefore, x/4+y students attended the second and third but not the first.

Now, we can consider any one of the conferences.  If the total who attend a conference is to equal x, the number who attended all three conferences must be 5x/12 - 2y.

Of course, the total of all the students must be 300, so 11x/6+y=300.

Now, there are a few constraints on x and y:
x and y are both greater than 0.
5x/12 - 2y is greater than 0.
x is divisible by 12.

For y to be greater than 0, 11x/6 must be less than 300, so x must be less than or equal to 163.  For 5x/12-2y to be greater than 0, paired with the equation 11x/6+y=300, x must be greater than or equal to 147.  The only multiple of 12 between 147 and 163 is 156.

So using x=156 the answers to a) through d) are:
a) 156
b) 169
c) 94
d) 37

Posted by Tristan on 2005-05-12 23:28:58