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

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==Problem==
 
==Problem==
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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?
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<math>\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}</math>
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==Solution==
 
==Solution==
 
Jordan has <math>10</math> high top sneakers, and <math>6</math> white sneakers. We would want as many white high-top sneakers as possible, so we set <math>6</math> high-top sneakers to be white. Then, we have <math>10-6=4</math> red high-top sneakers, so the answer is <math>\boxed{\dfrac{4}{15}}.</math>
 
Jordan has <math>10</math> high top sneakers, and <math>6</math> white sneakers. We would want as many white high-top sneakers as possible, so we set <math>6</math> high-top sneakers to be white. Then, we have <math>10-6=4</math> red high-top sneakers, so the answer is <math>\boxed{\dfrac{4}{15}}.</math>
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~andliu766
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==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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-Multpi12
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==Solution 3==
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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. \\\\
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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>
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-Benedict T (countmath1)
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==Video Solution by Power Solve (crystal clear)==
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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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https://youtu.be/BaE00H2SHQM?si=ZnK2pJGftec8ywRO&t=5589
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~Math-X
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==Video Solution by NiuniuMaths (Easy to understand!)==
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https://www.youtube.com/watch?v=V-xN8Njd_Lc
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~NiuniuMaths
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==Video Solution 2 by OmegaLearn.org==
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https://youtu.be/W_DyNSmRSLI
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==Video Solution 3 by SpreadTheMathLove==
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https://www.youtube.com/watch?v=Svibu3nKB7E
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== Video Solution by CosineMethod [🔥Fast and Easy🔥]==
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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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==See Also==
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{{AMC8 box|year=2024|num-b=18|num-a=20}}
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{{MAA Notice}}

Revision as of 15:50, 18 June 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 are 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)

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

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

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

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