Difference between revisions of "2021 Fall AMC 12A Problems/Problem 2"
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| + | {{duplicate|[[2021 Fall AMC 10A Problems/Problem 2|2021 Fall AMC 10A #2]] and [[2021 Fall AMC 12A Problems/Problem 2|2021 Fall AMC 12A #2]]}} | ||
| + | |||
| + | == Problem == | ||
| + | |||
Menkara has a <math>4 \times 6</math> index card. If she shortens the length of one side of this card by <math>1</math> inch, the card would have area <math>18</math> square inches. What would the area of the card be in square inches if instead she shortens the length of the other side by <math>1</math> inch? | Menkara has a <math>4 \times 6</math> index card. If she shortens the length of one side of this card by <math>1</math> inch, the card would have area <math>18</math> square inches. What would the area of the card be in square inches if instead she shortens the length of the other side by <math>1</math> inch? | ||
| − | <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> |
| + | |||
| + | == Solution == | ||
| + | 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] | ||
| + | \text{Initial} & 4 & 6 & 24 \\ | ||
| + | \text{Menkara shortens one side.} & 3 & 6 & 18 \\ | ||
| + | \text{Menkara shortens other side instead.} & 4 & 5 & 20 | ||
| + | \end{array}</cmath> | ||
| + | Therefore, the answer is <math>\boxed{\textbf{(E) } 20}.</math> | ||
| + | |||
| + | ~MRENTHUSIASM | ||
| + | |||
| + | ==Video Solution== | ||
| + | https://youtu.be/8I2nzPmscTc | ||
| + | |||
| + | ~Charles3829 | ||
| + | |||
| + | ==Video Solution (Simple and Quick)== | ||
| + | https://youtu.be/Q3kkg-z0CXo | ||
| + | |||
| + | ~Education, the Study of Everything | ||
| + | |||
| + | ==Video Solution== | ||
| + | https://youtu.be/158dTGn0zhI | ||
| + | |||
| + | ~savannahsolver | ||
| + | |||
| + | ==Video Solution by TheBeautyofMath== | ||
| + | for AMC 10: https://youtu.be/o98vGHAUYjM?t=100 | ||
| + | |||
| + | for AMC 12: https://youtu.be/jY-17W6dA3c?t=101 | ||
| + | |||
| + | ~IceMatrix | ||
| + | |||
| + | ==Video Solution== | ||
| + | https://youtu.be/aMkfa8VHjB8 | ||
| + | |||
| + | ~Lucas | ||
| + | |||
| + | ==See Also== | ||
| + | {{AMC12 box|year=2021 Fall|ab=A|num-b=1|num-a=3}} | ||
| + | {{AMC10 box|year=2021 Fall|ab=A|num-b=1|num-a=3}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 09:53, 18 September 2024
- 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
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
Video Solution
~Charles3829
Video Solution (Simple and Quick)
~Education, the Study of Everything
Video Solution
~savannahsolver
Video Solution by TheBeautyofMath
for AMC 10: https://youtu.be/o98vGHAUYjM?t=100
for AMC 12: https://youtu.be/jY-17W6dA3c?t=101
~IceMatrix
Video Solution
~Lucas
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. 
