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Difference between revisions of "2024 AMC 8 Problems/Problem 19"

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==Solution 2==
 
==Solution 2==
  
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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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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 our answer as C, or <math>\boxed{\dfrac{4}{15}}.</math>
  
 
-Multpi12
 
-Multpi12
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-Benedict T (countmath1)
 
-Benedict T (countmath1)
  
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==Video Solution by NiuniuMaths (Easy to understand!)==
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==Solution 4==
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https://www.youtube.com/watch?v=V-xN8Njd_Lc
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15 times <math>3</math> over <math>5</math> is <math>9</math>, which is the number of red pairs of sneakers. Then, <math>2</math> over <math>3</math> times <math>15</math> is <math>10</math>, so there are ten pairs of high-top sneakers.  If there are ten high-top sneakers, there are five low top sneakers. To have the lowest amount of high-top red sneakers, all five of the low-top sneakers need to be red. There are four red sneakers remaining so they have to be high-top. So the answer is <math>\boxed{\textbf{(C)}\ \frac{4}{15}}.</math>
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Answer by AliceDubbleYou
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==Video Solution 1 by Math-X (First fully understand the problem!!!)==
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https://youtu.be/BaE00H2SHQM?si=ZnK2pJGftec8ywRO&t=5589
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~Math-X
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==Video Solution (A Clever Explanation You’ll Get Instantly)==
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https://youtu.be/5ZIFnqymdDQ?si=xdZQCcwVWhElo7Lw&t=2741
  
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~NiuniuMaths
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~hsnacademy
  
 
==Video Solution by Power Solve (crystal clear)==
 
==Video Solution by Power Solve (crystal clear)==
 
https://www.youtube.com/watch?v=jmaLPhTmCeM
 
https://www.youtube.com/watch?v=jmaLPhTmCeM
  
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==Video Solution 1 by Math-X (First fully understand the problem!!!)==
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==Video Solution by NiuniuMaths (Easy to understand!)==
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https://www.youtube.com/watch?v=eYnLh_SGy7c
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https://www.youtube.com/watch?v=V-xN8Njd_Lc
  
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~Math-X
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~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
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==Video Solution by Interstigation==
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https://youtu.be/ktzijuZtDas&t=2211
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==Video Solution by Dr. David==
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https://youtu.be/EbTG0F7jEqE
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==Video Solution by WhyMath==
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https://youtu.be/cJln3sSnkbk
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2024|num-b=18|num-a=20}}
 
{{AMC8 box|year=2024|num-b=18|num-a=20}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 15:14, 22 December 2024

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?

$\textbf{(A) } 0\qquad\textbf{(B) } \dfrac{1}{5} \qquad\textbf{(C) } \dfrac{4}{15} \qquad\textbf{(D) } \dfrac{1}{3} \qquad\textbf{(E) } \dfrac{2}{5}$

Solution

Jordan has $10$ high top sneakers, and $6$ white sneakers. We would want as many white high-top sneakers as possible, so we set $6$ high-top sneakers to be white. Then, we have $10-6=4$ red high-top sneakers, so the answer is $\boxed{\dfrac{4}{15}}.$ ~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 our answer as C, or $\boxed{\dfrac{4}{15}}.$

-Multpi12

Solution 3

There are $\dfrac{3}{5}\cdot 15 = 9$ red pairs of sneakers and $6$ white pairs. There are also $\dfrac{2}{3}\cdot 15 = 10$ high-top pairs of sneakers and $5$ low-top pairs. Let $r$ be the number of red high-top sneakers and let $w$ be the number of white high-top sneakers. It follows that there are $9-r$ red pairs of low-top sneakers and $6-r$ white pairs. \\\\ We must have $9-r \leq 5,$ in order to have a valid amount of white sneakers. Solving this inequality gives $r\geq 4$, so the smallest possible value for $r$ is $4$. This means that there would be $9-4=5$ pairs of low-top red sneakers, so there are $0$ pairs of low-top white sneakers and $6$ pairs of high top white sneakers. This checks out perfectly, so the smallest fraction is $\boxed{\textbf{(C)}\ \frac{4}{15}}.$

-Benedict T (countmath1)

Solution 4

15 times $3$ over $5$ is $9$, which is the number of red pairs of sneakers. Then, $2$ over $3$ times $15$ is $10$, so there are ten pairs of high-top sneakers. If there are ten high-top sneakers, there are five low top sneakers. To have the lowest amount of high-top red sneakers, all five of the low-top sneakers need to be red. There are four red sneakers remaining so they have to be high-top. So the answer is $\boxed{\textbf{(C)}\ \frac{4}{15}}.$

Answer by AliceDubbleYou

Video Solution 1 by Math-X (First fully understand the problem!!!)

https://youtu.be/BaE00H2SHQM?si=ZnK2pJGftec8ywRO&t=5589

~Math-X

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 by NiuniuMaths (Easy to understand!)

https://www.youtube.com/watch?v=V-xN8Njd_Lc

~NiuniuMaths

Video Solution 2 by OmegaLearn.org

https://youtu.be/W_DyNSmRSLI

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

Video Solution by Dr. David

https://youtu.be/EbTG0F7jEqE

Video Solution by WhyMath

https://youtu.be/cJln3sSnkbk

See Also

2024 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 18 		Followed by
Problem 20
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