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| The three conferences (Posted on 2005-05-11) |
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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 !
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Submitted by pcbouhid
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Rating: 4.0000 (1 votes)
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Solution:
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We'll use the "Venn's diagram" to solve this problem.
Drawing three circles that intersect one another, we identify seven independent areas called A, B, C, D, E, F and G.
The area A is the portion of the first circle that is outside of the other two, so represents the number of students that attended only the first conference, and attended neither of the other two.
The areas B and C represent the same as above, respectively to the second and the third conferences.
The areas D, E and F represent the number of students that attended, respectively, both the first and second conferences (but not the third), both the first and the third conferences (but not the second) and both the second and third conferences (but not the first).
Finally, the area G (the common area of the three circles) represents the number of students that attended all the three conferences.
Once this is made, let's start the solution.
The first conference was attended by (A + D + E + G) students, the second by (B + D + F + G) and the third by (C + E + F + G) students.
Since the three conferences was attended by the same number of students (let's call it N), we have :
N = (A + D + E + G) = (B + D + F + G) = (C + E + F + G)
A is half of the students that attended only the first conference, so :
A = (A + D + E + G)/2
2A = (A + D + E + G)......(1)
For the second and third conferences, we have :
3B = (B + D + F + G)..........(2)
and
4C = (C + E + F + G)...........(3)
And also : N = 2A = 3B = 4C ..........(4)
From (1)...... A = D + E + G .........(5)
From (2)......2B = D + F + G .........(6)
From (3)......3C = E + F + G..........(7)
Summing up (5), (6) and (7) :
A + 2B + 3C = 2(D + E + F) + 3G.......(8)
We know that (A + B + C + D + E + F + G) = 300.....(9)
(4) and (8) give us :
A + 4A/3 + 3A/2 = 2(D + E + F) + 3G........(10)
(4) and (9) give us :
A + 2A/3 + A/2 + (D + E + F) + G = 300........(11)
Eliminating (D + E + F) in (10) and (11) we'll find :
A = (3600 + 6G)/49
or
A = 73 + (6G + 23)/49
Since A is an integer, (6G + 23) must be a multiple of 49.
Making (6G + 23) = 49K (where K=1,2,3,4,...) we obtain :
G = (49K - 23)/6 = (8K - 4) + (K + 1)/6
and A = 73 + K.
By (5), A is greater than G, since A = (D + E + G), so :
73 + K > (49K - 23)/6,
wich gives K < 10,....
But (K + 1) is a multiple of 6 (see value of G, above).
THEN, K = 5 (unique solution), that gives G = 37 and A = 78.
A few calculations give us :
B = 52, C = 39, D = 14, E = 27 and F = 53.
Answering the questions :
a) how many students attended each conference ?
answer : N = 2A (or 3B or 4C) = 2 x 78 = 156
b) how many students attended only one conference :
answer : A + B + C = 78 + 52 + 39 = 169
c) how many students attended only two conferences :
answer : D + E + F = 14 + 27 + 53 = 94
d) how many students attended all the three conferences:
answer : G = 37
Solution : a) 156; b) 169; c) 94; d) 37
Not one single trial.
One (curious) fact that call my attention looking at the results, is that N, A, B and C, are all multiples of 13.
And more : A = 6 x 13; B = 4 x 13 and C = 3 x 13....and 6 + 4 + 3 = 13.
Since we know, right from the start, that N is a multiple of 12, if we could prove that it is also a multiple of 13, the answer would be obtained in seconds : N = 12 x 13 = 156.
It's all a great coincidence (multiples of 13) ? In math ? I don't think so. But I ever succeeded in proving that N is a multiple of 13, also.
GOLDEN MEDAL FOR TRISTAN AND SILVER MEDAL FOR CORY (because of the two trials !)
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