2011 AMC 12B Problem 7 — Solution | LIVE by Po-Shen Loh

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2011 AMC 12B Problem 7

Below is the professionally curated solution for Problem 7 of the 2011 AMC 12B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2011 AMC 12B solutions, or check the answer key.

All of the real AMC 8, AMC 10, AMC 12, and AIME problems in our complete solution collection are used with official legal permission of the Mathematical Association of America (MAA).

Concepts:optimization bounding to limit cases

Difficulty rating: 1200

7.

Let $x$ and $y$ be two-digit positive integers with mean $60.$ What is the maximum value of the ratio $\dfrac{x}{y}?$

$3$

$\dfrac{33}{7}$

$\dfrac{39}{7}$

$9$

$\dfrac{99}{10}$

Solution:

Since $\dfrac{x+y}{2}=60,$ we have $x+y=120.$ To maximize $\dfrac{x}{y}$ we make $y$ small.

Because $x\le99,$ it follows that $y=120-x\ge21.$ Taking $x=99$ and $y=21$ gives the maximum $\dfrac{99}{21}=\dfrac{33}{7}.$

Thus, the correct answer is B.

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