Difference between revisions of "2019 AMC 10A Problems/Problem 4"
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==Problem== | ==Problem== | ||
| − | A box contains <math>28</math> red balls, <math>20</math> green balls, <math>19</math> yellow balls, <math>13</math> blue balls, <math>11</math> white balls, and <math>9</math> black balls. What is the minimum number of balls that must be drawn from the box without replacement to guarantee that at least <math>15</math> balls of a single color will be drawn | + | A box contains <math>28</math> red balls, <math>20</math> green balls, <math>19</math> yellow balls, <math>13</math> blue balls, <math>11</math> white balls, and <math>9</math> black balls. What is the minimum number of balls that must be drawn from the box without replacement to guarantee that at least <math>15</math> balls of a single color will be drawn? |
<math>\textbf{(A) } 75 \qquad\textbf{(B) } 76 \qquad\textbf{(C) } 79 \qquad\textbf{(D) } 84 \qquad\textbf{(E) } 91</math> | <math>\textbf{(A) } 75 \qquad\textbf{(B) } 76 \qquad\textbf{(C) } 79 \qquad\textbf{(D) } 84 \qquad\textbf{(E) } 91</math> | ||
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==Solution== | ==Solution== | ||
| − | + | We try to find the worst case scenario where we can find the maximum number of balls that can be drawn while getting <math><15</math> of each color by applying the [[pigeonhole principle]] and through this we get a perfect guarantee. | |
| + | Namely, we can draw up to <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, without drawing <math>15</math> balls of any one color. Drawing one more ball guarantees that we will get <math>15</math> balls of one color — either red, green, or yellow. Thus, the answer is <math>75 + 1 = \boxed{\textbf{(B) } 76}</math>. | ||
==Video Solution 1== | ==Video Solution 1== | ||
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~savannahsolver | ~savannahsolver | ||
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==See Also== | ==See Also== | ||
Latest revision as of 22:04, 15 March 2022
- 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 
red balls, 
green balls, 
yellow balls, 
blue balls, 
white balls, and 
black balls. What is the minimum number of balls that must be drawn from the box without replacement to guarantee that at least 
balls of a single color will be drawn?
Solution
We try to find the worst case scenario where we can find the maximum number of balls that can be drawn while getting 
of each color by applying the pigeonhole principle and through this we get a perfect guarantee.
Namely, we can draw up to 
red balls, green balls, yellow balls, blue balls, white balls, and black balls, for a total of 
balls, without drawing balls of any one color. Drawing one more ball guarantees that we will get balls of one color — either red, green, or yellow. Thus, the answer is 
.
Video Solution 1
Education, The Study of Everything
Video Solution 2
https://youtu.be/8WrdYLw9_ns?t=23
~ pi_is_3.14
Video Solution 3
~savannahsolver
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. 
