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

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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
  
−
==Solution 1==
+
==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>
  
 
-Multpi12
 
-Multpi12
 +
 +
==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. \\\\
 +
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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 +
==Solution 4==
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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>
 +
 +
Answer by AliceDubbleYou
  
 
==Video Solution 1 by Math-X (First fully understand the problem!!!)==
 
==Video Solution 1 by Math-X (First fully understand the problem!!!)==
−
https://www.youtube.com/watch?v=eYnLh_SGy7c
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https://youtu.be/BaE00H2SHQM?si=ZnK2pJGftec8ywRO&t=5589
  
 
~Math-X
 
~Math-X
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 +
==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==
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https://youtu.be/cJln3sSnkbk
 +
 +
==See Also==
 +
{{AMC8 box|year=2024|num-b=18|num-a=20}}
 +
{{MAA Notice}}

Latest revision as of 06:49, 15 November 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)

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
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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