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2000 AMC 10 Problem 22

Problem 22 of 25HarderAlgebra

One morning each member of Angela’s family drank an 88-ounce mixture of coffee with milk. The amounts of coffee and milk varied from cup to cup, but were never zero. Angela drank a quarter of the total amount of milk and a sixth of the total amount of coffee. How many people are in the family?

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Solution

Let there be nn people, drinking 8n8n ounces total, split into milk MM and coffee C.C. Angela drank one cup, so 14M+16C=1n(M+C).\tfrac14 M + \tfrac16 C = \tfrac1n (M + C). The left side is a weighted average of 14\tfrac14 and 16,\tfrac16, so 1n\tfrac1n lies strictly between 16\tfrac16 and 14.\tfrac14. That forces 4<n<6,4 \lt n \lt 6, so n=5.n = 5. Thus, the correct answer is C.

More practice

Concepts: weighted mean · bounding to limit cases · mixture · ratio and proportion

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