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

Revision as of 15:50, 18 June 2024

1 Problem
2 Solution
3 Solution 2
4 Solution 3
5 Video Solution by Power Solve (crystal clear)
6 Video Solution 1 by Math-X (First fully understand the problem!!!)
7 Video Solution by NiuniuMaths (Easy to understand!)
8 Video Solution 2 by OmegaLearn.org
9 Video Solution 3 by SpreadTheMathLove
10 Video Solution by CosineMethod [🔥Fast and Easy🔥]
11 Video Solution by Interstigation
12 See Also

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

@@ Line 43: / Line 43: @@
 https://www.youtube.com/watch?v=qaOkkExm57U
+==Video Solution by Interstigation==
+https://youtu.be/ktzijuZtDas&t=2211
 ==See Also==
 {{AMC8 box|year=2024|num-b=18|num-a=20}}
 {{MAA Notice}}

2024 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 18	Followed by Problem 20
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