Difference between revisions of "2024 AMC 8 Problems/Problem 9"
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+ | == Video Solution by CosineMethod [🔥Fast and Easy🔥]== | ||
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+ | https://www.youtube.com/watch?v=zqQTfBWr9T0 | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2024|num-b=8|num-a=10}} | {{AMC8 box|year=2024|num-b=8|num-a=10}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 14:41, 26 January 2024
Contents
Problem
All the marbles in Maria's collection are red, green, or blue. Maria has half as many red marbles as green marbles and twice as many blue marbles as green marbles. Which of the following could be the total number of marbles in Maria's collection?
Solution 1
Since she has half as many red marbles as green, we can call the number of red marbles , and the number of green marbles . Since she has half as many green marbles as blue, we can call the number of blue marbles . Adding them up, we have marbles. The number of marbles therefore must be a multiple of . The only possible answer is
Solution 2
Suppose Maria has green marbles and let be the total number of marbles. She then has red marbles and blue marbles. Altogether, Maria has marbles, implying that so must be a multiple of . The only multiple of is
-Benedict T (countmath1)
Video Solution 1 (easy to digest) by Power Solve
https://youtu.be/16YYti_pDUg?si=FAyIwmaHxE6UdSwP&t=238
Video Solution 2 by SpreadTheMathLove
https://www.youtube.com/watch?v=L83DxusGkSY
Video Solution by CosineMethod [🔥Fast and Easy🔥]
https://www.youtube.com/watch?v=zqQTfBWr9T0
See Also
2024 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 8 |
Followed by Problem 10 | |
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.