Difference between revisions of "2021 Fall AMC 12A Problems/Problem 2"
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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>. | 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>. | ||
| − | Therefore, the answer is <math>\boxed{\textbf{(E )} 20}</math>. | + | Therefore, the answer is <math>\boxed{\textbf{(E) } 20}</math>. |
~Steven Chen (www.professorchenedu.com) | ~Steven Chen (www.professorchenedu.com) | ||
Revision as of 20:54, 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.
Contents
Problem
Menkara has a 
index card. If she shortens the length of one side of this card by 
inch, the card would have area 
square inches. What would the area of the card be in square inches if instead she shortens the length of the other side by inch?
Solution 1
We construct the following table:
![\[\begin{array}{c||c|c||c} & & & \\ [-2.5ex] \textbf{Scenario} & \textbf{Length} & \textbf{Width} & \textbf{Area} \\ [0.5ex] \hline & & & \\ [-2ex] \text{Initial} & 4 & 6 & 24 \\ \text{Menkara shortens one side.} & 3 & 6 & 18 \\ \text{Menkara shortens other side instead.} & 4 & 5 & 20 \end{array}\]](http://latex.artofproblemsolving.com/c/6/7/c675a041e6c76adb56e8213ad259239344a1b750.png)
Therefore, the answer is 
~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 
.
Therefore, the answer is 
.
~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. 
