2011 AMC 12B Problem 7

Attempt Problem 7 of the 2011 AMC 12B 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 2011 AMC 12B solutions, or check the answer key.

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

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}

The mean condition gives

x+y=120

To maximize

\dfrac{x}{y},

make

x

as large and

y

as small as allowed

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.

Problem 7 in other years