Difference between revisions of "2023 AMC 8 Problems/Problem 10"
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− | == | + | ==Problem== |
− | Harold made a plum pie. | + | Harold made a plum pie to take on a picnic. He was able to eat only <math>\frac{1}{4}</math> of the pie, and he left the rest for his friends. A moose came by and ate <math>\frac{1}{3}</math> of what Harold left behind. After that, a porcupine ate <math>\frac{1}{3}</math> of what the moose left behind. How much of the original pie still remained after the porcupine left? |
+ | <math>\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}</math> | ||
− | ~apex304, SohumUttamchandani, wuwang2002, TaeKim, Cxrupptedpat, lpieleanu | + | ==Solution== |
+ | |||
+ | Note that: | ||
+ | |||
+ | |||
+ | |||
+ | Alternatively, 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, lpieleanu, MRENTHUSIASM | ||
+ | |||
+ | ==See Also== | ||
+ | {{AMC8 box|year=2023|num-b=9|num-a=11}} | ||
+ | {{MAA Notice}} |
Revision as of 23:25, 24 January 2023
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:
Alternatively, we can condense the solution above into the following equation:
~apex304, SohumUttamchandani, wuwang2002, TaeKim, Cxrupptedpat, lpieleanu, MRENTHUSIASM
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.