2004 AMC 10A Problem 15 — Solution | LIVE by Po-Shen Loh

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2004 AMC 10A Problem 15

Below is the professionally curated solution for Problem 15 of the 2004 AMC 10A, 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.

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:inequality optimization algebraic manipulation

Difficulty rating: 1420

15.

Given that $-4 \le x \le -2$ and $2 \le y \le 4,$ what is the largest possible value of $\dfrac{x + y}{x}?$

$-1$

$-\dfrac{1}{2}$

$0$

$\dfrac{1}{2}$

$1$

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.

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