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

(Solution)
m (Solution)
 
(17 intermediate revisions by 11 users not shown)
Line 3: Line 3:
 
==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<math>?</math>
+
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>
Line 9: Line 9:
 
==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==
 +
 +
https://youtu.be/givTTqH8Cqo
 +
 +
Education, The Study of Everything
 +
 +
== Video Solution 2 ==
 +
https://youtu.be/8WrdYLw9_ns?t=23
 +
 +
~ pi_is_3.14
 +
 +
==Video Solution 3==
 +
https://youtu.be/2HmS3n1b4SI
 +
 +
~savannahsolver
  
 
==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 $28$ 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

We try to find the worst case scenario where we can find the maximum number of balls that can be drawn while getting $<15$ of each color by applying the pigeonhole principle and through this we get a perfect guarantee. Namely, we can draw up to $14$ red balls, $14$ green balls, $14$ yellow balls, $13$ blue balls, $11$ white balls, and $9$ black balls, for a total of $75$ balls, without drawing $15$ balls of any one color. Drawing one more ball guarantees that we will get $15$ balls of one color — either red, green, or yellow. Thus, the answer is $75 + 1 = \boxed{\textbf{(B) } 76}$.

Video Solution 1

https://youtu.be/givTTqH8Cqo

Education, The Study of Everything

Video Solution 2

https://youtu.be/8WrdYLw9_ns?t=23

~ pi_is_3.14

Video Solution 3

https://youtu.be/2HmS3n1b4SI

~savannahsolver

See Also

2019 AMC 10A (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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