Difference between revisions of "2024 AMC 8 Problems/Problem 19"
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− | ==Solution | + | ==Solution 2== |
We first start by finding the amount of red and white sneakers. 3/5 * 15=9 red sneakers, so 6 are white sneakers. Then 2/3 * 15=10 are high top sneakers, so 5 are low top sneakers. Now think about 15 slots and the first 10 are labeled high top sneakers. if we insert the last 5 sneakers as red sneakers there are 4 leftover over red sneakers. Putting those four sneakers as high top sneakers we have are answer as C or <math>\boxed{\dfrac{4}{15}}.</math> | We first start by finding the amount of red and white sneakers. 3/5 * 15=9 red sneakers, so 6 are white sneakers. Then 2/3 * 15=10 are high top sneakers, so 5 are low top sneakers. Now think about 15 slots and the first 10 are labeled high top sneakers. if we insert the last 5 sneakers as red sneakers there are 4 leftover over red sneakers. Putting those four sneakers as high top sneakers we have are answer as C or <math>\boxed{\dfrac{4}{15}}.</math> | ||
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-Multpi12 | -Multpi12 | ||
− | ==Video Solution by Power Solve== | + | ==Solution 3== |
+ | There are <math>\dfrac{3}{5}\cdot 15 = 9</math> red pairs of sneakers and <math>6</math> white pairs. There are also <math>\dfrac{2}{3}\cdot 15 = 10</math> high-top pairs of sneakers and <math>5</math> low-top pairs. Let <math>r</math> be the number of red high-top sneakers and let <math>w</math> be the number of white high-top sneakers. It follows that there are <math>9-r</math> red pairs of low-top sneakers and <math>6-r</math> white pairs. \\\\ | ||
+ | We must have <math>9-r \leq 5,</math> in order to have a valid amount of white sneakers. Solving this inequality gives <math>r\geq 4</math>, so the smallest possible value for <math>r</math> is <math>4</math>. This means that there would be <math>9-4=5</math> pairs of low-top red sneakers, so there are <math>0</math> pairs of low-top white sneakers and <math>6</math> pairs of high top white sneakers. This checks out perfectly, so the smallest fraction is <math>\boxed{\textbf{(C)}\ \frac{4}{15}}.</math> | ||
+ | |||
+ | -Benedict T (countmath1) | ||
+ | |||
+ | ==Video Solution (A Clever Explanation You’ll Get Instantly)== | ||
+ | https://youtu.be/5ZIFnqymdDQ?si=xdZQCcwVWhElo7Lw&t=2741 | ||
+ | |||
+ | ~hsnacademy | ||
+ | |||
+ | ==Video Solution by Power Solve (crystal clear)== | ||
https://www.youtube.com/watch?v=jmaLPhTmCeM | https://www.youtube.com/watch?v=jmaLPhTmCeM | ||
==Video Solution 1 by Math-X (First fully understand the problem!!!)== | ==Video Solution 1 by Math-X (First fully understand the problem!!!)== | ||
− | https:// | + | https://youtu.be/BaE00H2SHQM?si=ZnK2pJGftec8ywRO&t=5589 |
~Math-X | ~Math-X | ||
+ | |||
+ | ==Video Solution by NiuniuMaths (Easy to understand!)== | ||
+ | https://www.youtube.com/watch?v=V-xN8Njd_Lc | ||
+ | |||
+ | ~NiuniuMaths | ||
==Video Solution 2 by OmegaLearn.org== | ==Video Solution 2 by OmegaLearn.org== | ||
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https://www.youtube.com/watch?v=qaOkkExm57U | https://www.youtube.com/watch?v=qaOkkExm57U | ||
+ | ==Video Solution by Interstigation== | ||
+ | https://youtu.be/ktzijuZtDas&t=2211 | ||
+ | |||
+ | ==See Also== | ||
+ | {{AMC8 box|year=2024|num-b=18|num-a=20}} | ||
+ | {{MAA Notice}} |
Latest revision as of 19:59, 28 August 2024
Contents
- 1 Problem
- 2 Solution
- 3 Solution 2
- 4 Solution 3
- 5 Video Solution (A Clever Explanation You’ll Get Instantly)
- 6 Video Solution by Power Solve (crystal clear)
- 7 Video Solution 1 by Math-X (First fully understand the problem!!!)
- 8 Video Solution by NiuniuMaths (Easy to understand!)
- 9 Video Solution 2 by OmegaLearn.org
- 10 Video Solution 3 by SpreadTheMathLove
- 11 Video Solution by CosineMethod [🔥Fast and Easy🔥]
- 12 Video Solution by Interstigation
- 13 See Also
Problem
Jordan owns 15 pairs of sneakers. Three fifths of the pairs are red and the rest are white. Two thirds of the pairs are high-top and the rest are low-top. The red high-top sneakers make up a fraction of the collection. What is the least possible value of this fraction?
Solution
Jordan has high top sneakers, and
white sneakers. We would want as many white high-top sneakers as possible, so we set
high-top sneakers to be white. Then, we have
red high-top sneakers, so the answer is
~andliu766
Solution 2
We first start by finding the amount of red and white sneakers. 3/5 * 15=9 red sneakers, so 6 are white sneakers. Then 2/3 * 15=10 are high top sneakers, so 5 are low top sneakers. Now think about 15 slots and the first 10 are labeled high top sneakers. if we insert the last 5 sneakers as red sneakers there are 4 leftover over red sneakers. Putting those four sneakers as high top sneakers we have are answer as C or
-Multpi12
Solution 3
There are red pairs of sneakers and
white pairs. There are also
high-top pairs of sneakers and
low-top pairs. Let
be the number of red high-top sneakers and let
be the number of white high-top sneakers. It follows that there are
red pairs of low-top sneakers and
white pairs. \\\\
We must have
in order to have a valid amount of white sneakers. Solving this inequality gives
, so the smallest possible value for
is
. This means that there would be
pairs of low-top red sneakers, so there are
pairs of low-top white sneakers and
pairs of high top white sneakers. This checks out perfectly, so the smallest fraction is
-Benedict T (countmath1)
Video Solution (A Clever Explanation You’ll Get Instantly)
https://youtu.be/5ZIFnqymdDQ?si=xdZQCcwVWhElo7Lw&t=2741
~hsnacademy
Video Solution by Power Solve (crystal clear)
https://www.youtube.com/watch?v=jmaLPhTmCeM
Video Solution 1 by Math-X (First fully understand the problem!!!)
https://youtu.be/BaE00H2SHQM?si=ZnK2pJGftec8ywRO&t=5589
~Math-X
Video Solution by NiuniuMaths (Easy to understand!)
https://www.youtube.com/watch?v=V-xN8Njd_Lc
~NiuniuMaths
Video Solution 2 by OmegaLearn.org
Video Solution 3 by SpreadTheMathLove
https://www.youtube.com/watch?v=Svibu3nKB7E
Video Solution by CosineMethod [🔥Fast and Easy🔥]
https://www.youtube.com/watch?v=qaOkkExm57U
Video Solution by Interstigation
https://youtu.be/ktzijuZtDas&t=2211
See Also
2024 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 18 |
Followed by Problem 20 | |
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.