Difference between revisions of "2024 AMC 8 Problems/Problem 9"
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Since she has half as many red marbles as green, we can call the number of red marbles <math>x</math>, and the number of green marbles <math>2x</math>. | Since she has half as many red marbles as green, we can call the number of red marbles <math>x</math>, and the number of green marbles <math>2x</math>. | ||
Since she has half as many green marbles as blue, we can call the number of blue marbles <math>4x</math>. | Since she has half as many green marbles as blue, we can call the number of blue marbles <math>4x</math>. | ||
− | Adding them up, we have <math>7x</math> marbles. The number of marbles therefore must be a multiple of <math>7</math>. The only possible answer is <math>\boxed{(E)28}</math>. | + | Adding them up, we have <math>7x</math> marbles. The number of marbles therefore must be a multiple of <math>7</math>. The only possible answer is <math>\boxed{\textbf{(E) 28}}.</math> |
+ | |||
+ | ==Solution 2== | ||
+ | |||
+ | Suppose Maria has <math>g</math> green marbles and let <math>t</math> be the total number of marbles. She then has <math>\frac{g}{2}</math> red marbles and <math>2g</math> blue marbles. Altogether, Maria has | ||
+ | <cmath>g + \frac{g}{2} + 2g = \frac{7g}{2} = t</cmath> | ||
+ | marbles, implying that <math>g = \dfrac{2t}{7},</math> so <math>t</math> must be a multiple of <math>7</math>. The only multiple of <math>7</math> is <math>\boxed{\textbf{(E) 28}}.</math> | ||
+ | |||
+ | -Benedict T (countmath1) | ||
==Video Solution 1 (easy to digest) by Power Solve== | ==Video Solution 1 (easy to digest) by Power Solve== | ||
https://youtu.be/16YYti_pDUg?si=FAyIwmaHxE6UdSwP&t=238 | https://youtu.be/16YYti_pDUg?si=FAyIwmaHxE6UdSwP&t=238 | ||
+ | |||
+ | ==Video Solution by Math-X (First fully understand the problem!!!)== | ||
+ | https://youtu.be/BaE00H2SHQM?si=ORnDevWxiB6tv1jO&t=1912 | ||
+ | |||
+ | ~Math-X | ||
+ | |||
+ | ==Video Solution by NiuniuMaths (Easy to understand!)== | ||
+ | https://www.youtube.com/watch?v=V-xN8Njd_Lc | ||
+ | |||
+ | ~NiuniuMaths | ||
+ | |||
+ | ==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 | ||
+ | ==Video Solution by Interstigation== | ||
+ | https://youtu.be/ktzijuZtDas&t=890 | ||
+ | |||
+ | ==See Also== | ||
+ | {{AMC8 box|year=2024|num-b=8|num-a=10}} | ||
+ | {{MAA Notice}} |
Revision as of 03:30, 6 February 2024
Contents
- 1 Problem
- 2 Solution 1
- 3 Solution 2
- 4 Video Solution 1 (easy to digest) by Power Solve
- 5 Video Solution by Math-X (First fully understand the problem!!!)
- 6 Video Solution by NiuniuMaths (Easy to understand!)
- 7 Video Solution 2 by SpreadTheMathLove
- 8 Video Solution by CosineMethod [🔥Fast and Easy🔥]
- 9 Video Solution by Interstigation
- 10 See Also
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 by Math-X (First fully understand the problem!!!)
https://youtu.be/BaE00H2SHQM?si=ORnDevWxiB6tv1jO&t=1912
~Math-X
Video Solution by NiuniuMaths (Easy to understand!)
https://www.youtube.com/watch?v=V-xN8Njd_Lc
~NiuniuMaths
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
Video Solution by Interstigation
https://youtu.be/ktzijuZtDas&t=890
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.