Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2024 AMC 8 Problems/Problem 19"
m (→‎Solution 4)
 
(32 intermediate revisions by 21 users not shown)
Line 2: Line 2:
  
 
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?
 +
 +
[[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>
  
−
==Solution==
+
==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>
−
~andliu766
 
  
 
==Solution 2==
 
==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 number of red and white sneakers. <math>\frac{3}{5} \times 15=9</math> red sneakers, so 6 are white. Then <math>\frac{2}{3} \times 15=10</math> are high top sneakers, so <math>5</math> are low top sneakers. Now think about <math>15</math> slots, and the first <math>10</math> are labeled high-top sneakers. If we insert the last <math>5</math> sneakers as red sneakers, there are <math>4</math> leftover red sneakers. Putting those <math>4</math> sneakers as high top sneakers, we have our answer as C, or <math>\boxed{\dfrac{4}{15}}.</math>
 +
 
 +
-ermwhatthesigma
 +
 
  
−
-Multpi12
+
-slight change by the-guy-with-the"W"hairline-and-a-not-goofy-looking-gyat-with-so-much-recoil-it-bounces-to-the-nearest-blackhole...
  
 
==Solution 3==
 
==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. \\\\
+
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>
 
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)
+
-hola
 +
 
 +
==Solution 4==
 +
 
 +
<math>15 \times \frac{3}{5}=9</math>, which is the number of red pairs of sneakers. Then, <math>\frac{2}{3} \times 15=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
 +
 
 +
-Minor edit by angieeverfree
 +
 
 +
-Minor edit by SmartyDigits
 +
 
 +
==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!!!)==
 
==Video Solution 1 by Math-X (First fully understand the problem!!!)==
Line 25: Line 44:
  
 
~Math-X
 
~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!)==
 
==Video Solution by NiuniuMaths (Easy to understand!)==
Line 30: Line 57:
  
 
~NiuniuMaths
 
~NiuniuMaths
−
 
−
==Video Solution by Power Solve (crystal clear)==
 
−
https://www.youtube.com/watch?v=jmaLPhTmCeM
 
  
 
==Video Solution 2 by OmegaLearn.org==
 
==Video Solution 2 by OmegaLearn.org==
Line 40: Line 64:
 
https://www.youtube.com/watch?v=Svibu3nKB7E
 
https://www.youtube.com/watch?v=Svibu3nKB7E
  
−
== Video Solution by CosineMethod [🔥Fast and Easy🔥]==
+
== Video Solution by CosineMethod [Very Slow and Hard to Follow]==
  
 
https://www.youtube.com/watch?v=qaOkkExm57U
 
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
 +
 +
== Video solution by TheNeuralMathAcademy ==
 +
https://youtu.be/f63MY1T2MgI&t=2109s
  
 
==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}}
 +
 +
[[Category:Introductory Algebra Problems]]

Latest revision as of 12:31, 28 September 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 number of red and white sneakers. $\frac{3}{5} \times 15=9$ red sneakers, so 6 are white. Then $\frac{2}{3} \times 15=10$ 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 red sneakers. Putting those $4$ sneakers as high top sneakers, we have our answer as C, or $\boxed{\dfrac{4}{15}}.$

-ermwhatthesigma


-slight change by the-guy-with-the"W"hairline-and-a-not-goofy-looking-gyat-with-so-much-recoil-it-bounces-to-the-nearest-blackhole...

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

$15 \times \frac{3}{5}=9$, which is the number of red pairs of sneakers. Then, $\frac{2}{3} \times 15=10$, 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

-Minor edit by SmartyDigits

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

Video solution by TheNeuralMathAcademy

https://youtu.be/f63MY1T2MgI&t=2109s

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2024_AMC_8_Problems/Problem_19&oldid=261695"