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

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{{AMC10 box|year=2019|ab=A|num-b=3|num-a=5}}
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{{AMC12 box|year=2019|ab=A|num-b=2|num-a=4}}
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{{MAA Notice}}

Revision as of 17:06, 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 $28$ 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

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