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Difference between revisions of "2019 AMC 10A Problems/Problem 4"

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==Solution==
 
==Solution==
  
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Before having picked <math>15</math> balls of one color, we could have chosen <math>14</math> red balls, <math>14</math> green balls, <math>14</math> yellow balls, <math>13</math> blue balls, <math>11</math> white balls, and <math>9</math> black balls, for a total of <math>75</math> balls. After this, we have to pick one more ball to guarantee that <math>15</math> balls of a single color have been drawn. Thus, our answer is <math>75 + 1 = 76 \implies \boxed{\textbf{(B)}}.</math>
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By choosing the maximum amount of balls while getting <math><15</math> of each color, we could have chosen <math>14</math> red balls, <math>14</math> green balls, <math>14</math> yellow balls, <math>13</math> blue balls, <math>11</math> white balls, and <math>9</math> black balls, for a total of <math>75</math> balls. Picking one more ball guarantees that we will get 15 balls of a color -- either red, green, or yellow. Thus, our answer is <math>75 + 1 = 76 \implies \boxed{\textbf{(B)}}.</math>
  
 
(Also explained by Rearrangement Theorem)
 
(Also explained by Rearrangement Theorem)

Revision as of 16:04, 13 February 2019

The following problem is from both the 2019 AMC 10A #4 and 2019 AMC 12A #3, so both problems redirect to this page.

Problem

A box contains $28$ red balls, $20$ green balls, $19$ yellow balls, $13$ blue balls, $11$ white balls, and $9$ black balls. What is the minimum number of balls that must be drawn from the box without replacement to guarantee that at least $15$ balls of a single color will be drawn$?$

$\textbf{(A) } 75 \qquad\textbf{(B) } 76 \qquad\textbf{(C) } 79 \qquad\textbf{(D) } 84 \qquad\textbf{(E) } 91$

Solution

By choosing the maximum amount of balls while getting $<15$ of each color, we could have chosen $14$ red balls, $14$ green balls, $14$ yellow balls, $13$ blue balls, $11$ white balls, and $9$ black balls, for a total of $75$ balls. Picking one more ball guarantees that we will get 15 balls of a color -- either red, green, or yellow. Thus, our answer is $75 + 1 = 76 \implies \boxed{\textbf{(B)}}.$

(Also explained by Rearrangement Theorem)

See Also

2019 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 3 		Followed by
Problem 5
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions
2019 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 2 		Followed by
Problem 4
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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