1971 AMC 12 Problem 27

Attempt Problem 27 of the 1971 AMC 12 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 1971 AMC 12 solutions, or check the answer key.

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27.

A box contains chips, each of which is red, white, or blue. The number of blue chips is at least half the number of white chips and at most one-third the number of red chips. The number which are white or blue is at least 55.55. The minimum number of red chips is:

2424

3333

4545

5454

5757

Answer: E
Concepts:inequalityoptimizationbounding to limit cases
Difficulty rating: 1880
Small Hint:

Let the counts be r,w,br,w,b and translate every condition into an inequality

Big Hint:

From w2bw\le2b and w+b55w+b\ge55, find the least possible integer bb

Solution:

Let the counts be r,w,b.r,w,b. The conditions give w2b,r3b,w+b55. \begin{gathered} w\le2b,\\ r\ge3b,\\ w+b\ge55. \end{gathered} Hence 3bw+b55,3b\ge w+b\ge55, so b19b\ge19 and r57.r\ge57. Equality is possible with (w,b,r)=(36,19,57),(w,b,r)=(36,19,57), so the minimum is 57.57.

Therefore, the correct answer is E.

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