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Difference between revisions of "2021 Fall AMC 12A Problems/Problem 2"

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<math>\textbf{(A) } 16 \qquad\textbf{(B) } 17 \qquad\textbf{(C) } 18 \qquad\textbf{(D) } 19 \qquad\textbf{(E) } 20</math>
 
<math>\textbf{(A) } 16 \qquad\textbf{(B) } 17 \qquad\textbf{(C) } 18 \qquad\textbf{(D) } 19 \qquad\textbf{(E) } 20</math>
  
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== Solution ==
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== Solution 1 ==
 
We construct the following table:
 
We construct the following table:
 
<cmath>\begin{array}{c||c|c||c}
 
<cmath>\begin{array}{c||c|c||c}
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~MRENTHUSIASM
 
~MRENTHUSIASM
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 +
== Solution 2 ==
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To get the area 18 square inches, Menkara shortens the side with length 4 inches.
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 +
Therefore, if Menkara shortens the other side with length 6 inches by 1 inch, the area of the card becomes <math>4 \cdot \left( 6 - 1 \right) = 20</math>.
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Therefore, the answer is <math>\boxed{\textbf{(E) 20}}</math>.
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~Steven Chen (www.professorchenedu.com)
  
 
==See Also==
 
==See Also==

Revision as of 19:53, 25 November 2021

The following problem is from both the 2021 Fall AMC 10A #2 and 2021 Fall AMC 12A #2, so both problems redirect to this page.

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 1

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

~MRENTHUSIASM

Solution 2

To get the area 18 square inches, Menkara shortens the side with length 4 inches.

Therefore, if Menkara shortens the other side with length 6 inches by 1 inch, the area of the card becomes $4 \cdot \left( 6 - 1 \right) = 20$.

Therefore, the answer is $\boxed{\textbf{(E) 20}}$.

~Steven Chen (www.professorchenedu.com)

See Also

2021 Fall AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 1 		Followed by
Problem 3
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions
2021 Fall AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 1 		Followed by
Problem 3
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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