fskilnik wrote:
Three piles (A, B, C) of 7 beans each are to be made from 12 red, 5 yellow and 4 green beans. If each stack must contain at least one bean of each color and stack A must get the maximum number of red beans possible, what is the minimum number of red beans to be put in stack B?
(A) 1
(B) 2
(C) 3
(D) 4
(E) 5
Source:
http://www.GMATH.net (Exercise 9 of the Quant Class #03 included in our free test drive!)
Hi, chetan2u and ShankSouljaBoi. Thank you both for your contributions!
My "official solution" is the following:
\(?\,\, = \,\,{\left( {{\rm{\# }}\,\,{\rm{red}}\,\,{\rm{in}}\,\,{\rm{B}}} \right)_{\,\,{\rm{MIN}}}}\)
Let´s focus on stack A at first. It must contain 7 beans, and 3 of them must be of different colors, exactly one (of these 3 beans) red.
Now we must choose the other 4 beans of pile A and, in order to maximize the number of red beans (why?) , we will take all 4 of them red.
(Conclusion: stack A will contain 5 red beans.)
It´s time to focus on pile C, because we are also interested in maximizing its number of red beans... repeating the same reasoning above, from the 12-5 = 7 red beans left, we will (again) take 5 of them to be put on this stack (C), remaining (the minimum of) 2 red beans to be chosen for the remainder pile (B)!
Important: all 21 beans must be allocated, otherwise we would not have 3 piles with 7 beans in each one of them!
This solution follows the notations and rationale taught in the GMATH method.
Regards,
Fabio.
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Fabio Skilnik :: GMATH method creator (Math for the GMAT)
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