Difference between revisions of "2019 AMC 10A Problems/Problem 4"
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==Solution== | ==Solution== | ||
| + | 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{B}.</math> | ||
==See Also== | ==See Also== | ||
Revision as of 17:17, 9 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 
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
Before having picked balls of one color, we could have chosen 
red balls, green balls, yellow balls, blue balls, white balls, and black balls, for a total of 
balls. After this, we have to pick one more ball to guarantee that balls of a single color have been drawn. Thus, our answer is 
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. 
