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Difference between revisions of "2021 Fall AMC 10A Problems/Problem 2"

(→‎Solution)
(→‎Solution)
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== Solution ==
 
== Solution ==
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If Menkara instead shortens the length of the other side by <math>1</math> inch, then the dimensions of the index card will be <math>4\times5,</math> with area <math>\boxed{\textbf{(E) } 20}.</math>
+
We construct the following table:
 +
<cmath>\begin{array}{c||c|c||c}
 +
& & & \\ [-2.5ex]
 +
\textbf{Scenario} & \textbf{Length} & \textbf{Width} & \textbf{Area} \\ [0.5ex]
 +
\hline
 +
& & & \\ [-2ex]
 +
\textbf{Initial} & 4 & 6 & 24 \\
 +
\textbf{Menkara shortens one side.} & 3 & 6 & 18 \\
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\textbf{Menkara shortens other side instead.} & 4 & 5 & 20
 +
\end{array}</cmath>
 +
Therefore, the answer is <math>\boxed{\textbf{(E) } 20}.</math>
  
 
~MRENTHUSIASM
 
~MRENTHUSIASM

Revision as of 18:06, 22 November 2021

Problem

Menkara has a $4 \times 6$ index card. If she shortens the length of one side of this card by $1$ inch, the card would have area $18$ square inches. What would the area of the card be in square inches if instead she shortens the length of the other side by $1$ inch?

$\textbf{(A) } 16 \qquad\textbf{(B) } 17 \qquad\textbf{(C) } 18 \qquad\textbf{(D) } 19 \qquad\textbf{(E) } 20$

Solution

We construct the following table: \[\begin{array}{c||c|c||c} & & & \\ [-2.5ex] \textbf{Scenario} & \textbf{Length} & \textbf{Width} & \textbf{Area} \\ [0.5ex] \hline & & & \\ [-2ex] \textbf{Initial} & 4 & 6 & 24 \\ \textbf{Menkara shortens one side.} & 3 & 6 & 18 \\ \textbf{Menkara shortens other side instead.} & 4 & 5 & 20 \end{array}\] Therefore, the answer is $\boxed{\textbf{(E) } 20}.$

~MRENTHUSIASM

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