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M01-14

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M01-14 [#permalink]

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New post 16 Sep 2014, 00:15
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Three piles of 7 beans each are to be made from 10 red, 5 yellow, and 6 green beans. If all of the beans must be used and each stack must contain at least one bean of each color, then what is the maximum number of red beans that can be put in one of the stacks?

A. 5
B. 6
C. 8
D. 9
E. 10

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Re M01-14 [#permalink]

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New post 16 Sep 2014, 00:15
Official Solution:

Three piles of 7 beans each are to be made from 10 red, 5 yellow, and 6 green beans. If all of the beans must be used and each stack must contain at least one bean of each color, then what is the maximum number of red beans that can be put in one of the stacks?

A. 5
B. 6
C. 8
D. 9
E. 10


All piles should contain 7 beans. If there must be 1 yellow and 1 green bean in the pile, then there is room for 5 red beans in the pile. Therefore the answer is 5.


Answer: A
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Collection of Questions:
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Re M01-14 [#permalink]

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New post 28 Nov 2016, 21:32
I think this is a poor-quality question and I agree with explanation. I don't think the question was clear that there was a restriction in the total number of beans. The question says that all the beans must be used only but doesn't mention anything about amount.
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M01-14 [#permalink]

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New post 06 Mar 2017, 05:09
If all of the beans must be used and each stack must contain at least one bean of each color, and three piles of 7 beans, then the solution is not complete because according to the solution there are only two piles of 7 beans and not three as the question stem mentions:
red yellow green
5 1 1
5 1 1

What am I missing? Brunel, could you please chime in?
Thanks in advance
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New post 06 Mar 2017, 05:33
racingip wrote:
If all of the beans must be used and each stack must contain at least one bean of each color, and three piles of 7 beans, then the solution is not complete because according to the solution there are only two piles of 7 beans and not three as the question stem mentions:
red yellow green
5 1 1
5 1 1

What am I missing? Brunel, could you please chime in?
Thanks in advance


I don't understand what you mean.

The question asks: what is the maximum number of red beans that can be put in one of the stacks? And the answer is - maximum of 5 red beans can be in one of the stacks.
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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M01-14 [#permalink]

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New post 06 Mar 2017, 05:43
Thank you for your prompt response Brunel.

My doubt is as follows:

The question says that we must form three piles of 7 beans (but if we select 5 (red), 1 (yellow) and 1 (green) we can only form two piles of 7 beans. Plus, the question says that all the beans must be used but again if we select 5 (red), 1 (yellow) and 1 (green) two times, then there are yellow and green beans left. I am not sure what I am missing. Thanks
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Re: M01-14 [#permalink]

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New post 06 Mar 2017, 06:18
1
racingip wrote:
Thank you for your prompt response Brunel.

My doubt is as follows:

The question says that we must form three piles of 7 beans (but if we select 5 (red), 1 (yellow) and 1 (green) we can only form two piles of 7 beans. Plus, the question says that all the beans must be used but again if we select 5 (red), 1 (yellow) and 1 (green) two times, then there are yellow and green beans left. I am not sure what I am missing. Thanks


We have 21 beans. We want to divide all of them equally in 3 stacks so that each stack must contain at least one bean of each color. The question asks: what is the maximum number of red beans that can be put in one of the stacks? The answer is 5. You cannot put more than 5 red beans in any of the stacks. One of the distributions is shown below:

R - Y - G
5 - 1 - 1
3 - 3 - 3
2 - 1 - 2

Hope it's clear.
_________________

New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: M01-14   [#permalink] 06 Mar 2017, 06:18
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