A, B and C went bird watching. Each of them saw one bird that none of the others did. Each pair saw one bird that the third did not. And one bird was seen by all three. Of the birds A saw, two were yellow. Of the birds B saw, three were yellow. Of the birds C saw, four were yellow.
(a) How many yellow birds were seen in all?
(b) How many non-yellow birds were seen in all?
Let us number the bird that only A, B and C saw separately as 1, 2 and 3.
Let us denote the bird seen by A&B be 4 B&C be 5 and, A&C be 6.
The bird seen by the three individuals A,B and C together is denoted by 7.
Accordingly, C saw 3,5,6 and 7 and it is given that all these re coloured yellow.
A saw 2 yellow birds, so the other two must be non-yellow.....(i)
Also, the birds A saw with B and ABC saw together were yellow, and accordingly fro (i), we have 6,7 are yellow and 1,4 are non-yellow birds.
B saw three yellow birds and, so the one he saw alone, that is 2 must be non-yellow. Denoting a yellow bird by Y and a non-yellow bird by NY, we then have this table as follows:
A => 1(NY) 4(NY) 6(Y) 7(Y)
B=> 2(Y) 4(NY) 5(Y) 7(Y)
C=> 3(Y) 5(Y) 6(Y) 7(Y)
Therefore, charting the # yellow birds(Y) and # non-yellow birds (NY), we now have:
Y => 2,3,5,6,7
NY => 1,4
Consequently, there are precisely 5 yellow and 2 non-yellow birds.
Edited on January 6, 2022, 10:21 pm
Puzzle Solution
