2019 AMC 10A Problems/Problem 4

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 number 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 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

Solution 3

Pigeon Hole Principle. You want to assume that you have very bad luck. The worst the choosing balls could go would be choosing 14 red balls, 14 green, 14 yellow, 13 blue, 11 white, and 9 black. These add up to 75, now no matter what color you choose it makes 15. (There are no more blue, white, and black socks left) = $75+1=\boxed{\textbf{(B) } 76}$

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

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