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

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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?
 
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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[[File:2024-amc-8-q19.png|center|500px]]
  
 
<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>
 
<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==
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==Solution 1==
 
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 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>
  
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-Multpi12
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-ermwhatthesigma
  
 
==Solution 3==
 
==Solution 3==
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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>
 
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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-hola
  
 
==Solution 4==
 
==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>
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15 times <math>\tfrac{3}{5}</math> is <math>9</math>, which is the number of red pairs of sneakers. Then, <math>\tfrac{2}{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
 
Answer by AliceDubbleYou
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-Minor edit by angieeverfree
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==Video Solution by Central Valley Math Circle (Goes through full thought process)==
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https://youtu.be/OgWv-nRpfJA
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~mr_mathman
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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
  
 
==Video Solution (A Clever Explanation You’ll Get Instantly)==
 
==Video Solution (A Clever Explanation You’ll Get Instantly)==
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==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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https://youtu.be/BaE00H2SHQM?si=ZnK2pJGftec8ywRO&t=5589
 
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~Math-X
 
  
 
==Video Solution by NiuniuMaths (Easy to understand!)==
 
==Video Solution by NiuniuMaths (Easy to understand!)==
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https://www.youtube.com/watch?v=Svibu3nKB7E
 
https://www.youtube.com/watch?v=Svibu3nKB7E
  
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== Video Solution by CosineMethod [🔥Fast and Easy🔥]==
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== Video Solution by CosineMethod [Very Slow and Hard to Follow]==
  
 
https://www.youtube.com/watch?v=qaOkkExm57U
 
https://www.youtube.com/watch?v=qaOkkExm57U
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==Video Solution by Dr. David==
 
==Video Solution by Dr. David==
 
https://youtu.be/EbTG0F7jEqE
 
https://youtu.be/EbTG0F7jEqE
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==Video Solution by WhyMath==
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https://youtu.be/cJln3sSnkbk
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==Video Solution by Daily Dose of Math (Simple, Certified, and Logical)==
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https://youtu.be/OwJvuq6F7sQ
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~Thesmartgreekmathdude
  
 
==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}}
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[[Category:Introductory Algebra Problems]]

Latest revision as of 18:11, 1 June 2025

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?

2024-amc-8-q19.png

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

Solution 1

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 $\boxed{\dfrac{4}{15}}.$

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 $\boxed{\dfrac{4}{15}}.$

-ermwhatthesigma

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 $\boxed{\textbf{(C)}\ \frac{4}{15}}.$

-hola

Solution 4

15 times $\tfrac{3}{5}$ is $9$, which is the number of red pairs of sneakers. Then, $\tfrac{2}{3}$ 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

-Minor edit by angieeverfree

Video Solution by Central Valley Math Circle (Goes through full thought process)

https://youtu.be/OgWv-nRpfJA

~mr_mathman

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 [Very Slow and Hard to Follow]

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

Video Solution by Daily Dose of Math (Simple, Certified, and Logical)

https://youtu.be/OwJvuq6F7sQ

~Thesmartgreekmathdude

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

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