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Difference between revisions of "2024 AMC 10A Problems/Problem 8"

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== Video Solution 8 by TheNeuralMathAcademy ==
 
== Video Solution 8 by TheNeuralMathAcademy ==
 
https://www.youtube.com/watch?v=4b_YLnyegtw&t=1176s
 
https://www.youtube.com/watch?v=4b_YLnyegtw&t=1176s
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== Video solution by TheNeuralMathAcademy ==
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https://www.youtube.com/watch?v=4b_YLnyegtw&t=1547s
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==See Also==
 +
{{AMC10 box|year=2024|ab=A|before=[[2023 AMC 10B Problems]]|after=[[2024 AMC 10B Problems]]}}
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* [[AMC 10]]
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* [[AMC 10 Problems and Solutions]]
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* [[Mathematics competitions]]
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* [[Mathematics competition resources]]
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{{MAA Notice}}

Revision as of 17:11, 18 August 2025

Problem

Amy, Bomani, Charlie, and Daria work in a chocolate factory. On Monday Amy, Bomani, and Charlie started working at $1:00 \ \mathrm{PM}$ and were able to pack $4$, $3$, and $3$ packages, respectively, every $3$ minutes. At some later time, Daria joined the group, and Daria was able to pack $5$ packages every $4$ minutes. Together, they finished packing $450$ packages at exactly $2:45\ \mathrm{PM}$. At what time did Daria join the group?

$\textbf{(A) }1:25\text{ PM}\qquad\textbf{(B) }1:35\text{ PM}\qquad\textbf{(C) }1:45\text{ PM}\qquad\textbf{(D) }1:55\text{ PM}\qquad\textbf{(E) }2:05\text{ PM}$

Solution 1

Note that Amy, Bomani, and Charlie pack a total of $4+3+3=10$ packages every $3$ minutes.

The total amount of time worked is $1$ hour and $45$ minutes, which when converted to minutes, is $105$ minutes. This means that since Amy, Bomani, and Charlie worked for the entire $105$ minutes, they in total packed $\dfrac{105}{3}\cdot10=350$ packages.

Since $450$ packages were packed in total, then Daria must have packed $450-350=100$ packages in total, and since she packs at a rate of $5$ packages per $4$ minutes, then Daria worked for $\dfrac{100}{5}\cdot4=80$ minutes, therefore Daria joined $80$ minutes before $2:45$ PM, which was at $\boxed{\textbf{(A) }1:25\text{ PM}}$

~Tacos_are_yummy_1, andliu766

Solution 2

Let the time, in minutes, elapsed between $1:00$ and the time Daria joined the packaging be $x$. Since Amy packages $4$ packages every $3$ minutes, she packages $\frac{4}{3}$ packages per minute. Similarly, we can see that both Bomani and Charlie package $1$ package per minute, and Daria packages $\frac{5}{4}$ packages every minute.

Before Daria arrives, we can write the total packages packaged as $x(\frac{4}{3} + 1 + 1) = x(\frac{10}{3})$. Since there are $105$ minutes between $1:00$ and $2:45$, Daria works with the other three for $105-x$ minutes, meaning for that time there are $(105-x)(\frac{4}{3} + 1 + 1 + \frac{5}{4}) = (105-x)(\frac{55}{12})$ packages packaged.

Adding the two, we get $x(\frac{10}{3}) + (105-x)(\frac{55}{12}) = 450$ (The total packaged in the entire time is $450$). Solving this equation, we get $x=25$, meaning Daria arrived $25$ minutes after $1:00$, meaning the answer is $\boxed{\textbf{(A) }1:25\text{ PM}}$.

~i_am_suk_at_math_2

Video Solution 1

https://youtu.be/l3VrUsZkv8I

Video Solution 2 by Power Solve

https://youtu.be/j-37jvqzhrg?si=bf4iiXH4E9NM65v8&t=996

Video Solution 3 by Daily Dose of Math

~Thesmartgreekmathdude

Video Solution 4 by SpreadTheMathLove

https://www.youtube.com/watch?v=_o5zagJVe1U

Video Solution 5 by Pi Academy

https://youtu.be/6qYaJsgqkbs?si=K2Ebwqg-Ro8Yqoiv

Video Solution 6 by Dr. David

https://youtu.be/Bmjli0Iq5I4

Video Solution 7 by Easy Way

https://youtu.be/BNtwJ0B2BoY

~KENJAKURA

Video Solution 8 by TheNeuralMathAcademy

https://www.youtube.com/watch?v=4b_YLnyegtw&t=1176s

Video solution by TheNeuralMathAcademy

https://www.youtube.com/watch?v=4b_YLnyegtw&t=1547s

See Also

2024 AMC 10A (Problems • Answer Key • Resources)
Preceded by
2023 AMC 10B Problems 		Followed by
2024 AMC 10B Problems
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 AMC 10 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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