2024 AMC 8 Problems/Problem 19: Difference between revisions

Latest revision as of 06:05, 7 November 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?

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

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

Solution 5

Let the number of red high-top sneakers be $x$ . Making a two way table we get:

Two-way table
	Red	White	Total
High-top	$x$	$10 - x$	$10$
Low-top	$9-x$	$x-4$	$5$
Total	$9$	$6$	$15$

From the table we can see that $4\leq x\leq 9$ . Therefore the minimum value is when $x=4$ .

So the answer is $\boxed{\textbf{(C)}\ \frac{4}{15}}.$

~Seafowl23

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

@@ Line 3: / Line 3: @@
 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?
-<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>
+[[File:2024-amc-8-q19.png|center|500px]]
-==Solution==
+<imath>\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}</imath>
-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 1==
+Jordan has <imath>10</imath> high top sneakers, and <imath>6</imath> white sneakers. We would want as many white high-top sneakers as possible, so we set <imath>6</imath> high-top sneakers to be white. Then, we have <imath>10-6=4</imath> red high-top sneakers, so the answer is <imath>\boxed{\dfrac{4}{15}}.</imath>
 ==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. <imath>\frac{3}{5} \times 15=9</imath> red sneakers, so 6 are white. Then <imath>\frac{2}{3} \times 15=10</imath> are high top sneakers, so <imath>5</imath> are low top sneakers. Now think about <imath>15</imath> slots, and the first <imath>10</imath> are labeled high-top sneakers. If we insert the last <imath>5</imath> sneakers as red sneakers, there are <imath>4</imath> leftover red sneakers. Putting those <imath>4</imath> sneakers as high top sneakers, we have our answer as C, or <imath>\boxed{\dfrac{4}{15}}.</imath>
--Multpi12
+-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 <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 <imath>\dfrac{3}{5}\cdot 15 = 9</imath> red pairs of sneakers and <imath>6</imath> white pairs. There are also <imath>\dfrac{2}{3}\cdot 15 = 10</imath> high-top pairs of sneakers and <imath>5</imath> low-top pairs. Let <imath>r</imath> be the number of red high-top sneakers and let <imath>w</imath> be the number of white high-top sneakers. It follows that there are <imath>9-r</imath> red pairs of low-top sneakers and <imath>6-r</imath> 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 <imath>9-r \leq 5,</imath> in order to have a valid amount of white sneakers. Solving this inequality gives <imath>r\geq 4</imath>, so the smallest possible value for <imath>r</imath> is <imath>4</imath>. This means that there would be <imath>9-4=5</imath> pairs of low-top red sneakers, so there are <imath>0</imath> pairs of low-top white sneakers and <imath>6</imath> pairs of high top white sneakers. This checks out perfectly, so the smallest fraction is <imath>\boxed{\textbf{(C)}\ \frac{4}{15}}.</imath>
+-hola
+==Solution 4==
+<imath>15 \times \frac{3}{5}=9</imath>, which is the number of red pairs of sneakers. Then, <imath>\frac{2}{3} \times 15=10</imath>, 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 <imath>\boxed{\textbf{(C)}\ \frac{4}{15}}.</imath>
+Answer by AliceDubbleYou
+-Minor edit by angieeverfree
+-Minor edit by SmartyDigits
+==Solution 5==
+Let the number of red high-top sneakers be <imath>x</imath>. Making a two way table we get:
--Benedict T (countmath1)
+{| border="1" cellpadding="4" cellspacing="0" style="border-collapse:collapse;"
+|+ Two-way table
+!
+! Red
+! White
+! Total
+|-
+! High-top
+| <imath>x</imath> || <imath>10 - x</imath> || <imath>10</imath>
+|-
+! Low-top
+| <imath>9-x</imath> || <imath>x-4</imath> || <imath>5</imath>
+|-
+! Total
+| <imath>9</imath> || <imath>6</imath> || <imath>15</imath>
+|}
+From the table we can see that <imath> 4\leq x\leq 9 </imath>. Therefore the minimum value is when <imath>x=4</imath>.
+So the answer is <imath>\boxed{\textbf{(C)}\ \frac{4}{15}}.</imath>
+~Seafowl23
+==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 1 by Math-X (First fully understand the problem!!!)==
-https://www.youtube.com/watch?v=eYnLh_SGy7c
-~Math-X
 ==Video Solution 2 by OmegaLearn.org==
@@ Line 39: / Line 91: @@
 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
+==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==
 {{AMC8 box|year=2024|num-b=18|num-a=20}}
 {{MAA Notice}}
+[[Category:Introductory Algebra Problems]]

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
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