Problem 19 · 2024 AMC 8
Hard
Fractions, Decimals & Percents
percent-multipliercareful-counting
Jordan owns 15 pairs of sneakers. Three fifths of the pairs are red and the rest are white. Two thirds of the pairs are high-top and the rest are low-top. The red high-top sneakers make up a fraction of the collection. What is the least possible value of this fraction?
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Answer: C — 4/15.
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Hint 1 of 2
There are a FIXED 10 high-tops. To make as few of them red as possible, give as many high-tops as you can to the white pairs instead — crowd red out.
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Hint 2 of 2
Technique (minimize an overlap by pushing to the extreme): 9 red, 6 white; 10 high-top, 5 low-top. Whites can absorb at most 6 high-tops, so red is forced to take whatever high-tops are left.
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Approach: push white pairs into high-top to crowd out red
- First nail the four counts. Red = 35×15 = 9, white = 6; high-top = 23×15 = 10, low-top = 5. The total of 10 high-tops is fixed; we only choose WHO gets them.
- To minimize red high-tops, hand high-tops to white first. There are only 6 white pairs, so at most 6 high-tops can be white — do exactly that.
- That leaves 10 − 6 = 4 high-top spots with nowhere to go but red. So the smallest red-high-top fraction is 415. This transfers: to minimize the overlap of two groups, shove the limited "other" group as full as it goes — the forced leftover is your minimum (it's the same logic as the Pigeonhole/inclusion bound).
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