M01-14

Question

Rachel wants to divide 21 beans into three piles of seven beans each. She has 10 red beans, 5 yellow beans, and 6 green beans. If each pile must contain at least one bean of each color, what is the maximum number of red beans that can she put in one of the piles?

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

Show Answer

Official Answer

A

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

Bunuel · Accepted Answer

Official Solution:

Rachel wants to divide 21 beans into three piles of seven beans each. She has 10 red beans, 5 yellow beans, and 6 green beans. If each pile must contain at least one bean of each color, what is the maximum number of red beans that can she put in one of the piles?

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

Show Answer

Official Answer

A

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

To maximize the number of red beans in a single pile, it is necessary to minimize the number of other colored beans in that same pile. As each pile must contain at least one bean of every color, the minimum number of yellow and green beans in a pile is one each. Given the rule that each pile must contain exactly 7 beans, the highest possible number of red beans in one pile is, therefore, 7 - 1 - 1 = 5. This can be accomplished by having one pile with 1 yellow, 1 green, and 5 red beans, while still distributing the remaining 14 beans (5 red, 4 yellow, and 5 green) across the other two piles so that each of them also includes at least one bean of each color.

Answer: A

Bunuel · Answer

I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.

BottomJee · Answer

I think this is a  high-quality  question and I agree with explanation.

ShreeyaV · Answer

For Red to be maximum in a pile, 3 out of 7 spots of the beans will be filled by each color.
R, G, Y. 
With this, we will left with 4 slots out of 7. Since we have total of 10 red beans and 3 piles where each pile must have at least 1 color bean, we need to add another 4 R to fill up the capacity of 7. Hence, 5R + 1Y + 1G = 7 Beans in total.

Answer is  A , 5 red beans

CockySaint · Answer

out of the options in the question,
only 5 seems possible.
if you out 6 red beans, there's only one other bean possible.
all other options are greater than 7 so by default out.

M01-14

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