Difference between revisions of "2021 Fall AMC 12A Problems/Problem 2"
MRENTHUSIASM (talk | contribs) (→Solution) |
MRENTHUSIASM (talk | contribs) (→Solution) |
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\hline | \hline | ||
& & & \\ [-2ex] | & & & \\ [-2ex] | ||
| − | Initial & 4 & 6 & 24 \\ | + | \text{Initial} & 4 & 6 & 24 \\ |
| − | Menkara shortens one side. & 3 & 6 & 18 \\ | + | \text{Menkara shortens one side.} & 3 & 6 & 18 \\ |
| − | Menkara shortens other side instead. & 4 & 5 & 20 | + | \text{Menkara shortens other side instead.} & 4 & 5 & 20 |
\end{array}</cmath> | \end{array}</cmath> | ||
Therefore, the answer is <math>\boxed{\textbf{(E) } 20}.</math> | Therefore, the answer is <math>\boxed{\textbf{(E) } 20}.</math> | ||
Revision as of 01:09, 24 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 
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
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
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. 
