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Three piles (A, B, C) of 7 beans each are to be made from 12 red, 5 ye

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Three piles (A, B, C) of 7 beans each are to be made from 12 red, 5 ye  [#permalink]

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New post 04 Oct 2018, 12:19
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A
B
C
D
E

Difficulty:

  65% (hard)

Question Stats:

45% (01:05) correct 55% (01:31) wrong based on 29 sessions

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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!)

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Course release PROMO : finish our test drive till 31/Oct with (at least) 60 correct answers out of 92 (12-questions Mock included) to gain a 60% discount!

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Three piles (A, B, C) of 7 beans each are to be made from 12 red, 5 ye  [#permalink]

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New post 04 Oct 2018, 19:13
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!)


Restrictions:-
1) one bean of each colour
2) 7 beans in each
3) A must get the max possible

The maximum of RED beans A can be 5 out of 7 as other two will be of two other colours.
Remaining 12-5 or 7 RED beans have to be divided now among two other piles.
Since we are looking for minimum in B, let us put max in C. Here too 5 beans can be put.
So minimum possible RED in B are 7-5=2

B

Final..
A :- 5R,1G,1Y
B:- 2R,3Y,2G
C:- 5R,1G,1Y

Of course other way is to concentrate on just B..
We have 5 yellow.. give one each to other two bags, so remaining 3 in B
4 green, 1 each in other two and remaining 2 in B
So 3+2=5 already filled in B and remaining 2 have to be red
But do ensure that you do not flaw in restrictions
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Three piles (A, B, C) of 7 beans each are to be made from 12 red, 5 ye  [#permalink]

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New post 04 Oct 2018, 19:20
B to have minimum Red, should have maximum of the other two colors.
Ok let's distribute green,first.
As each pile should have at least one green, the maximum green B can have is two.
So, green done.
Let's move to yellow .

To distribute 5 yellow , first distribute 1 each to A B and C.

We have two yellow remaining, let's give it to B so that it has maximum of yellow.

B now has 3 yellow and two green. Two slots left, which will be filled by red and will be minimum.
Answer : 2

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Re: Three piles (A, B, C) of 7 beans each are to be made from 12 red, 5 ye  [#permalink]

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New post 05 Oct 2018, 08:09
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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Course release PROMO : finish our test drive till 31/Oct with (at least) 60 correct answers out of 92 (12-questions Mock included) to gain a 60% discount!

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Re: Three piles (A, B, C) of 7 beans each are to be made from 12 red, 5 ye &nbs [#permalink] 05 Oct 2018, 08:09
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