Difference between revisions of "2023 AMC 8 Problems/Problem 10"
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More simply, we can condense the solution above into the following equation: <cmath>\left(1-\frac14\right)\left(1-\frac13\right)\left(1-\frac13\right) = \frac34\cdot\frac23\cdot\frac23 = \frac13.</cmath> | More simply, we can condense the solution above into the following equation: <cmath>\left(1-\frac14\right)\left(1-\frac13\right)\left(1-\frac13\right) = \frac34\cdot\frac23\cdot\frac23 = \frac13.</cmath> | ||
| − | + | ~apex304, SohumUttamchandani, wuwang2002, TaeKim, Cxrupptedpat, MRENTHUSIASM | |
| − | + | ==Remark== | |
| + | This problem is a great example of the implementation of <b>remaining portions</b>. This concept, in short, refers to the method of finding out what remains of a whole when part of it is taken. If you look closely at our solution, we can see that the denominator is the same when we said Harold ate <math>\frac14</math> of the pie and we got <math>\frac34</math> of the pie left. We can also see the numerator is just the denominator minus the old denominator. | ||
| + | |||
| + | ~Nivaar | ||
==Video Solution by Math-X (Let's first Understand the question)== | ==Video Solution by Math-X (Let's first Understand the question)== | ||
| Line 39: | Line 42: | ||
==Video Solution by SpreradTheMathLove== | ==Video Solution by SpreradTheMathLove== | ||
https://www.youtube.com/watch?v=TAa6jarbATE | https://www.youtube.com/watch?v=TAa6jarbATE | ||
| + | |||
| + | ==Video Solution by Dr. David== | ||
| + | https://youtu.be/BCEMLHWsw2o | ||
| + | |||
| + | ==Video Solution by WhyMath== | ||
| + | https://youtu.be/EYCq78Fg1RY | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2023|num-b=9|num-a=11}} | {{AMC8 box|year=2023|num-b=9|num-a=11}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 10:09, 18 November 2024
Contents
- 1 Problem
- 2 Solution
- 3 Remark
- 4 Video Solution by Math-X (Let's first Understand the question)
- 5 Video Solution (HOW TO THINK CREATIVELY!!!)
- 6 Video Solution by Magic Square
- 7 Video Solution by Interstigation
- 8 Video Solution by harungurcan
- 9 Video Solution by SpreradTheMathLove
- 10 Video Solution by Dr. David
- 11 Video Solution by WhyMath
- 12 See Also
Problem
Harold made a plum pie to take on a picnic. He was able to eat only 
of the pie, and he left the rest for his friends. A moose came by and ate 
of what Harold left behind. After that, a porcupine ate of what the moose left behind. How much of the original pie still remained after the porcupine left?
Solution
Note that:
- Harold ate
of the pie. After that, 
of the pie was left behind.
of the pie was left behind.- The moose ate
of the pie. After that, 
of the pie was left behind.
of the pie. After that, 
of the pie was left behind.- The porcupine ate
of the pie. After that, 
of the pie was left behind.
of the pie. After that, 
of the pie was left behind.More simply, we can condense the solution above into the following equation: ![\[\left(1-\frac14\right)\left(1-\frac13\right)\left(1-\frac13\right) = \frac34\cdot\frac23\cdot\frac23 = \frac13.\]](http://latex.artofproblemsolving.com/b/7/9/b79c7d75d940ecb0d045b002bc7f0e20a2bafa3e.png)
~apex304, SohumUttamchandani, wuwang2002, TaeKim, Cxrupptedpat, MRENTHUSIASM
Remark
This problem is a great example of the implementation of remaining portions. This concept, in short, refers to the method of finding out what remains of a whole when part of it is taken. If you look closely at our solution, we can see that the denominator is the same when we said Harold ate 
of the pie and we got 
of the pie left. We can also see the numerator is just the denominator minus the old denominator.
~Nivaar
Video Solution by Math-X (Let's first Understand the question)
https://youtu.be/Ku_c1YHnLt0?si=N3jXnFg5Zy3GCgrB&t=1500 ~Math-X
Video Solution (HOW TO THINK CREATIVELY!!!)
https://youtu.be/WKcH7Cyeipo ~Education the Study of everything
Video Solution by Magic Square
https://youtu.be/-N46BeEKaCQ?t=4814
Video Solution by Interstigation
https://youtu.be/DBqko2xATxs&t=859
Video Solution by harungurcan
https://www.youtube.com/watch?v=oIGy79w1H8o&t=246s
Video Solution by SpreradTheMathLove
https://www.youtube.com/watch?v=TAa6jarbATE
Video Solution by Dr. David
Video Solution by WhyMath
See Also
| 2023 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 9 |
Followed by Problem 11 | |
| 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 AJHSME/AMC 8 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
