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> | ||
| − | 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 \frac34 | + | 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. |
~apex304, SohumUttamchandani, wuwang2002, TaeKim, Cxrupptedpat, MRENTHUSIASM, Nivaar | ~apex304, SohumUttamchandani, wuwang2002, TaeKim, Cxrupptedpat, MRENTHUSIASM, Nivaar | ||
Revision as of 10:33, 23 January 2024
Contents
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)
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.
~apex304, SohumUttamchandani, wuwang2002, TaeKim, Cxrupptedpat, MRENTHUSIASM, 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
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. 
