In a contest of intelligence, three problems A, B and C were posed.

Among the contestants there were 25 who solved at least one problem each.

Of all the contestants who did not solve problem A, the number who solved B was twice the number who solved C.

The number of participants who solved only problem A was one more than the number who solved problem A and at least one other problem.

Of all students who solved just one problem, half did not solve problem A.

How many students solved only problem B?

Six students solved only B.

If we designate the number that got A alone as a, B alone as b, C alone as c; and that got A and B without C as d, that got B and C without A as e, that got C and A without B as f; and that got all three correct as g, then:

a+b+c+d+e+f+g = 25
b+e=2(c+e)
a=f+g+d+1
a=b+c

Then
2b+2c+e+b+c-1=25
or
3b+3c+e = 26

and since b=2c+e,

4b+c=26
or
c=26-4b

We can work backwards to deduce from b what a, c and e are, and what remains for d+f+g:

   a     b     c             e           the rest


  23     1    22           -43                  22


  20     2    18           -34                  19


  17     3    14           -25                  16


  14     4    10           -16                  13


  11     5     6            -7                  10


   8     6     2             2                   7


   5     7    -2            11                   4


   2     8    -6            20                   1


  -1     9   -10            29                  -2


  -4    10   -14            38                  -5


  -7    11   -18            47                  -8


 -10    12   -22            56                 -11

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The only line for which all the variables are positive is that on which b is 6.

We can also see that 8 people got A alone correct and 2 people got C alone correct. Two people got both B and C correct but not A, but we don't know how the remaining 7 split among the other possibilities.

Posted by Charlie on 2004-01-16 10:34:12