2004 AMC 10A Problem 15

Attempt Problem 15 of the 2004 AMC 10A 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 2004 AMC 10A solutions, or check the answer key.

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15.

Given that

-4 \le x \le -2

and

2 \le y \le 4,

what is the largest possible value of

\dfrac{x + y}{x}?

-1

-\dfrac{1}{2}

0

\dfrac{1}{2}

1

Answer: D

Concepts:inequality optimization algebraic manipulation

Difficulty rating: 1420

Small hint:

Rewrite

\dfrac{x + y}{x} = 1 + \dfrac{y}{x}

Big hint:

Since

\dfrac{y}{x} \lt 0,

make its absolute value as small as possible

Solution:

Write

\dfrac{x + y}{x} = 1 + \dfrac{y}{x}.

Here

\dfrac{y}{x} \lt 0,

so the expression is largest when

\left|\dfrac{y}{x}\right|

is smallest.

That happens with

y = 2

and

x = -4,

giving

1 + \dfrac{2}{-4} = 1 - \dfrac{1}{2} = \dfrac{1}{2}.

Thus, the correct answer is D.