2000 AMC 8 Problem 10

Below is the professionally curated solution for Problem 10 of the 2000 AMC 8, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2000 AMC 8 solutions, or check the answer key.

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Concepts:percentage linear equation

Difficulty rating: 1070

10.

Ara and Shea were once the same height. Since then Shea has grown $20\%$ while Ara has grown half as many inches as Shea. Shea is now $60$ inches tall. How tall, in inches, is Ara now?

$48$

$51$

$52$

$54$

$55$

Solution:

Let $x$ be Ara and Shea's initial height. Then we get that $\begin{align*} 1.2x &= 60 \\ x &= 50. \end{align*}$

This means that Shea grew $10$ inches, which means that Ara grew $10 \div 2 = 5$ inches, making her $50 + 5 = 55$ inches tall.

Thus, E is the correct answer.

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