2019 AMC 10A Problem 4

Question

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?

75

76

79

84

91

To avoid getting

15

of a color, draw at most

14

from each large color

For colors with fewer than

15

, all of that color can be drawn

Note that we can pull as many as

14

balls of each color without ensuring that

15

balls of one color are drawn.

This means that we can draw all of the black, white and blue balls, along with

14

red, green, and yellow balls.

This gives us a total of

9 + 11 + 13 + 3 \cdot 14

= 33 + 42

= 75.

We need to add one at the end, however, to ensure that we get that

15

th ball of some color,

75 + 1 = 76.

Thus, B is the correct answer.

Problem 4 in other years