Three piles (A, B, C) of 7 beans each are to be made from 12

fskilnik@GMATH 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

Answer: [spoiler]__(B)____ [/spoiler]

Piles - A,B,C
Since we are looking for least red in B, let us distribution max red in other two..
1) A has to get max possible..
As one each is a must remaining 5 can be red
2) similarly for C too

So total 5+5 red gone,
Remaining 12-10=2 have to go to B

Ans B

fskilnik@GMATH 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: https://www.GMATH.net
(Exercise 9 of the Quant Class #03 included in our free test drive!)

Thank you for your contribution, chetan.sharma !

My "official solution" (the one presented in GMATH´s course) 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.