Question 200548: There are 15 blue jays and 14 orioles perched in 3 trees . each tree has at least 4 blue jays, and 2 orioles, if no tree has more orioles than blue jays, then the largest number of birds that can be in one tree is
a)11 b) 12 c) 13 d) 14 e) 15
Answer by Edwin McCravy(20056)   (Show Source):
You can put this solution on YOUR website! There are 15 blue jays and 14 orioles perched in 3 trees . each tree has at least 4 blue jays, and 2 orioles, if no tree has more orioles than blue jays, then the largest number of birds that can be in one tree is
a)11 b) 12 c) 13 d) 14 e) 15
Pretend we have three trees, TREES 1, 2, and 3 and
we are going to place the birds in the trees so that
TREE 1 will have as many birds as possible, under the
given conditions.
Make this chart
TREE 1 TREE 2 TREE 3
No. of Blue Jays
No. of Orioles
We start by placing 4 bluejays and 2 orioles in each tree,
since each tree must have that:
TREE 1 TREE 2 TREE 3
No. of Blue Jays 4 4 4
No. of Orioles 2 2 2
So
1. Now we have accounted for 12 of the 15 blue jays, leaving
3 more blue jays to place.
and
2. We have accounted for 6 of the 14 orioles, leaving 8 more orioles
to place.
Now since we're letting Tree 1 have the most birds, let's put all
three of the remaining blue jays in TREE 1, making there be 7 Blue
Jays there.
TREE 1 TREE 2 TREE 3
No. of Blue Jays 7 4 4
No. of Orioles 2 2 2
Now we have placed all the Blue Jays. We still have 8 more orioles
to place. We can only put 5 more orioles in TREE 1, because TREE 1
can't have more orioles than blue jays. So we have
TREE 1 TREE 2 TREE 3
No. of Blue Jays 7 4 4
No. of Orioles 7 2 2
That leaves 3 more orioles and so we can put two of them in in TREE 2
and 1 in TREE 2, or vice-versa. So we either end up with THIS:
TREE 1 TREE 2 TREE 3
No. of Blue Jays 7 4 4
No. of Orioles 7 4 3
OR with THIS:
TREE 1 TREE 2 TREE 3
No. of Blue Jays 7 4 4
No. of Orioles 7 3 4
Either way TREE 1 has 14 birds. So that's the answer d).
Edwin
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