AMC 10 · 2019 · #4
Grade 3 arithmeticPick an answer.
AMC 10 2019 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Tool #16 (Change Focus): instead of asking 'how many draws guarantees 15 of one color', flip the question — 'what is the largest number of draws where we still avoid having 15 of any color?' Add one more to that worst case and the next ball MUST push some color to 15. Tool #2 (List): write out the maximum we can take of each color without hitting 15 — capped at 14 for the three big colors, and capped at the whole supply for the three small ones. Tool #3 (Eliminate): the choices 75, 76, 79, 84, 91 differ by only a few — once we compute the worst-case total 75, the answer is 75 + 1 = 76, picking (B).
Only red, green, and yellow have at least 15 balls, so only they can win; blue, white, and black can never reach 15.
Only colors with ≥ 15 balls can ever 'cross the line'.
3.OA.A.3Count The ComplementTake 14 of each color that could win (red, green, yellow) and every ball of blue, white, black — the biggest draw with no color at 15.
Pick the most balls you can without any single color reaching 15.
1.OA.A.2Make A Systematic ListWorst case: 14 + 14 + 14 = 42 plus 13 + 11 + 9 = 33 makes 75 balls drawn, and still no color has reached 15.
75 is the largest 'unlucky' draw — still no winning color.
2.NBT.B.5Make A Systematic ListAfter 75 draws only red, green, yellow remain, so the 76th ball must be one of them — pushing that color from 14 to 15.
One more draw past the worst case forces success.
1.OA.A.1Count The Complement75 still fails, so not (A); 79, 84, 91 work but overshoot the minimum — only (B) 76 is both smallest and sufficient.
76 is the smallest that always works — the others are too small or too big.
1.NBT.B.3Eliminate PossibilitiesThis AMC 10 problem only needs Grade 3 "word-problem reasoning" you already know — count the worst-case 'unlucky' pulls (14 + 14 + 14 + 13 + 11 + 9 = 75), then add one more so a color is forced to 15: 76.