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2017 AMC 12A Problem 1

Problem 1 of 25EasierAlgebra

Pablo buys popsicles for his friends. The store sells single popsicles for $1\$1 each, 33-popsicle boxes for $2,\$2, and 55-popsicle boxes for $3.\$3. What is the greatest number of popsicles that Pablo can buy with $8?\$8?

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Solution

The cheapest popsicles come from the 55-popsicle box, at $35=$0.60\dfrac{\$3}{5}=\$0.60 each. Even at that rate, 1414 popsicles would cost 14$0.60=$8.40,14\cdot\$0.60=\$8.40, more than $8.\$8. So Pablo can buy at most 13,13, and he achieves this with two 55-boxes for $6\$6 and one 33-box for $2,\$2, giving 25+3=132\cdot5+3=13 popsicles. Thus, the correct answer is D.

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Concepts: optimization · rate

Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.