Difference between revisions of "2023 AMC 8 Problems/Problem 10"

(Solution)
(Solution)
Line 17: Line 17:
 
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 to show students 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$ of the pie left. We can also see the numerator is just the denominator minus the old denominator.
+
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$ 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 10:32, 23 January 2024

Problem

Harold made a plum pie to take on a picnic. He was able to eat only $\frac{1}{4}$ of the pie, and he left the rest for his friends. A moose came by and ate $\frac{1}{3}$ of what Harold left behind. After that, a porcupine ate $\frac{1}{3}$ of what the moose left behind. How much of the original pie still remained after the porcupine left?

$\textbf{(A)}\ \frac{1}{12} \qquad \textbf{(B)}\ \frac{1}{6} \qquad \textbf{(C)}\ \frac{1}{4} \qquad \textbf{(D)}\ \frac{1}{3} \qquad \textbf{(E)}\ \frac{5}{12}$

Solution

Note that:

  • Harold ate $\frac14$ of the pie. After that, $1-\frac14=\frac34$ of the pie was left behind.
  • The moose ate $\frac13\cdot\frac34 = \frac14$ of the pie. After that, $\frac34 - \frac14 = \frac12$ of the pie was left behind.
  • The porcupine ate $\frac13\cdot\frac12 = \frac16$ of the pie. After that, $\frac12 - \frac16 = \boxed{\textbf{(D)}\ \frac{1}{3}}$ of the pie was left behind.

More simply, we can condense the solution above into the following equation: \[\left(1-\frac14\right)\left(1-\frac13\right)\left(1-\frac13\right) = \frac34\cdot\frac23\cdot\frac23 = \frac13.\]

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 $\frac14$ of the pie and we got \frac34$ 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 (ProblemsAnswer KeyResources)
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. AMC logo.png