Difference between revisions of "2019 AMC 10A Problems/Problem 4"
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By choosing the maximum number 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 <math>15</math> balls of a color -- either red, green, or yellow. Thus the answer is <math>75 + 1 = \boxed{\textbf{(B) } 76}</math>. | By choosing the maximum number 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 <math>15</math> balls of a color -- either red, green, or yellow. Thus the answer is <math>75 + 1 = \boxed{\textbf{(B) } 76}</math>. | ||
− | ==Video Solution== | + | ==Video Solution 1== |
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+ | https://youtu.be/givTTqH8Cqo | ||
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+ | Education, The Study of Everything | ||
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+ | ==Video Solution 2== | ||
https://youtu.be/2HmS3n1b4SI | https://youtu.be/2HmS3n1b4SI | ||
Revision as of 18:33, 20 November 2020
- 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
By choosing the maximum number of balls while getting of each color, we could have chosen red balls, green balls, yellow balls, blue balls, white balls, and black balls, for a total of balls. Picking one more ball guarantees that we will get balls of a color -- either red, green, or yellow. Thus the answer is .
Video Solution 1
Education, The Study of Everything
Video Solution 2
~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.