2000 AMC 10 Problem 22

Attempt Problem 22 of the 2000 AMC 10 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2000 AMC 10 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

22.

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?

33

44

55

66

77

Answer: C
Concepts:weighted meanbounding to limit casesmixtureratio and proportion
Difficulty rating: 1900
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.

← Problem 21#21
Full Exam

Problem 22 in Other Years