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

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== Problem ==
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Amy, Bomani, Charlie, and Daria work in a chocolate factory. On Monday Amy, Bomani, and Charlie started working at <math>1:00 \ \mathrm{PM}</math> and were able to pack <math>4</math>, <math>3</math>, and <math>3</math> packages, respectively, every <math>3</math> minutes. At some later time, Daria joined the group, and Daria was able to pack <math>5</math> packages every <math>4</math> minutes. Together, they finished packing <math>450</math> packages at exactly <math>2:45\ \mathrm{PM}</math>. At what time did Daria join the group?
  
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<math>\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}</math>
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== Solution 1 ==
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Note that Amy, Bomani, and Charlie pack a total of <math>4+3+3=10</math> packages every <math>3</math> minutes.
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The total amount of time worked is <math>1</math> hour and <math>45</math> minutes, which when converted to minutes, is <math>105</math> minutes. This means that since Amy, Bomani, and Charlie worked for the entire <math>105</math> minutes, they in total packed <math>\dfrac{105}{3}\cdot10=350</math> packages.
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Since <math>450</math> packages were packed in total, then Daria must have packed <math>450-350=100</math> packages in total, and since she packs at a rate of <math>5</math> packages per <math>4</math> minutes, then Daria worked for <math>\dfrac{100}{5}\cdot4=80</math> minutes, therefore Daria joined <math>80</math> minutes before <math>2:45</math> PM, which was at <math>\boxed{\textbf{(A) }1:25\text{ PM}}</math>
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~Tacos_are_yummy_1 ~andliu766
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== Solution 2 ==
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Let the time, in minutes, elapsed between <math>1:00</math> and the time Daria joined the packaging be <math>x</math>. Since Amy packages <math>4</math> packages every <math>3</math> minutes, she packages <math>\frac{4}{3}</math> packages per minute. Similarly, we can see that both Bomani and Charlie package <math>1</math> package per minute, and Daria packages <math>\frac{5}{4}</math> packages every minute.
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Before Daria arrives, we can write the total packages packaged as <math>x(\frac{4}{3} + 1 + 1) = x(\frac{10}{3})</math>. Since there are <math>105</math> minutes between <math>1:00</math> and <math>2:45</math>, Daria works with the other three for <math>105-x</math> minutes, meaning for that time there are <math>(105-x)(\frac{4}{3} + 1 + 1 + \frac{5}{4}) = (105-x)(\frac{55}{12})</math> packages packaged.
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Adding the two, we get <math>x(\frac{10}{3}) + (105-x)(\frac{55}{12}) = 450</math> (The total packaged in the entire time is <math>450</math>). Solving this equation, we get <math>x=25</math>, meaning Daria arrived <math>25</math> minutes after <math>1:00</math>, meaning the answer is <math>\boxed{\textbf{(A) }1:25\text{ PM}}</math>.
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~i_am_suk_at_math_2
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== Video Solution by Pi Academy ==
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https://youtu.be/6qYaJsgqkbs?si=K2Ebwqg-Ro8Yqoiv
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==Video Solution 1 by Power Solve ==
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https://youtu.be/j-37jvqzhrg?si=bf4iiXH4E9NM65v8&t=996
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== Video Solution by Daily Dose of Math ==
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https://youtu.be/W5hES6aNXAk
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~Thesmartgreekmathdude
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==Video Solution by SpreadTheMathLove==
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https://www.youtube.com/watch?v=_o5zagJVe1U
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==See also==
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{{AMC10 box|year=2024|ab=A|num-b=7|num-a=9}}
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{{MAA Notice}}

Latest revision as of 19:40, 17 November 2024

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 by Pi Academy

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

Video Solution 1 by Power Solve

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

Video Solution by Daily Dose of Math

https://youtu.be/W5hES6aNXAk

~Thesmartgreekmathdude

Video Solution by SpreadTheMathLove

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

See also

2024 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 7 		Followed by
Problem 9
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

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