2021 AMC Fall 10A Problem 2

The card is $4$ by $6$, so there are only two sides she could shorten. Try both and see which one gives $18$.

1. Shorten the $4$-inch side: the card becomes $3$ by $6$, with area $3 \times 6 = 18$.
2. Shorten the $6$-inch side: the card becomes $4$ by $5$, with area $4 \times 5 = 20$.
3. The problem says the area would be $18$, so the first change was to the $4$-inch side.
4. The question asks about the other side, which is the $6$-inch side. From step 2, that gives area $20$.

The answer is $\boxed{\textbf{(E)}\ 20}$.

The area $18$ is not extra information. It tells you which side was shortened first. When a problem allows only two possibilities, work out both, and let the fact the problem gives you choose between them.

Test which side was shortened. Shortening the $4$-inch side gives a $3 \times 6$ card of area $18$; shortening the $6$-inch side gives a $4 \times 5$ card of area $20$. Since the problem says the result was $18$, the first change was to the $4$-inch side.

"The other side" is therefore the $6$-inch side, and shortening it by $1$ inch produces a $4 \times 5$ card with area $4 \cdot 5 = 20$.

The answer is $\boxed{\textbf{(E)}\ 20}$.