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

 
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==Problem==
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== Problem ==
In how many ways can <math>6</math> juniors and <math>6</math> seniors form <math>3</math> disjoint teams of <math>4</math> people so that each team has <math>2</math> juniors and <math>2</math> seniors?
 
  
<math>\textbf{(A)}~720\qquad\textbf{(B)}~1350\qquad\textbf{(C)}~2700\qquad\textbf{(D)}~3280\qquad\textbf{(E)}~8100</math>
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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?
  
== Solution 1==
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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>
The number of ways in which we can choose the juniors for the team are <math>{6\choose2}{4\choose2}{2\choose2}=15\cdot6\cdot1=90</math>. Similarly, the number of ways to choose the seniors are the same, so the total is <math>90\cdot90=8100</math>. But we must divide the number of permutations of three teams, since the order in which the teams were chosen never mattered, which is <math>3!</math>. Thus the answer is <math>\frac{8100}{3!}=\frac{8100}{6}=\boxed{\textbf{(B) }1350}</math>!
 
  
~eevee9406
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== Solution 1 ==
~small edits by NSAoPS
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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
  
 
== Solution 2 ==
 
== Solution 2 ==
Let's first suppose that the three teams are distinguishable. The number of ways to choose the first team of two juniors and two seniors is <math>{6\choose2}{6\choose2} = 15\cdot 15 = 225</math>. Now, only four juniors and four seniors are left to choose the second team. Thus, the second team can be formed in <math>{4\choose2}{4\choose2} = 6\cdot 6 = 36</math> ways. There are now only two juniors and two seniors left, so the third team can only be formed in one way. Thus, the total number of ways in which we can choose two juniors and two teams for three distinguishable teams is <math>225\cdot 36 = 8100</math> ways. However, the problem does not require the teams to be distinguishable. Therefore, we must divide by <math>3!=6</math> to find the answer for three indistinguishable teams.
 
  
The answer is <math>\frac{8100}{6} =\boxed{\textbf{(B) }1350}</math>
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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.
  
~jjjxi
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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.
  
~edit by yanes04 (wrong number)
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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>.
  
== Video Solution by Pi Academy ==
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~i_am_suk_at_math_2
  
https://youtu.be/6qYaJsgqkbs?si=K2Ebwqg-Ro8Yqoiv
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==Video Solution==
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https://youtu.be/l3VrUsZkv8I
  
 
==Video Solution 1 by Power Solve ==
 
==Video Solution 1 by Power Solve ==
  
https://youtu.be/j-37jvqzhrg?si=IBSPzNSvdIodGvZ7&t=1145
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https://youtu.be/j-37jvqzhrg?si=bf4iiXH4E9NM65v8&t=996
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== Video Solution 2 by Daily Dose of Math ==
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[//youtu.be/W5hES6aNXAk ~Thesmartgreekmathdude]
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==Video Solution 3 by SpreadTheMathLove==
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https://www.youtube.com/watch?v=_o5zagJVe1U
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== Video Solution 4 by Pi Academy ==
  
== Video Solution by Daily Dose of Math ==
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https://youtu.be/6qYaJsgqkbs?si=K2Ebwqg-Ro8Yqoiv
  
https://youtu.be/AEd5tf1PJxk
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== Video Solution 5 by Dr. David==
  
~Thesmartgreekmathdude
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https://youtu.be/Bmjli0Iq5I4
  
==Video Solution by SpreadTheMathLove==
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== See Also ==
https://www.youtube.com/watch?v=_o5zagJVe1U
 
  
==See also==
 
 
{{AMC10 box|year=2024|ab=A|num-b=7|num-a=9}}
 
{{AMC10 box|year=2024|ab=A|num-b=7|num-a=9}}
 
{{MAA Notice}}
 
{{MAA Notice}}
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[[Category:Introductory Algebra Problems]]

Latest revision as of 19:22, 26 May 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

https://youtu.be/l3VrUsZkv8I

Video Solution 1 by Power Solve

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

Video Solution 2 by Daily Dose of Math

~Thesmartgreekmathdude

Video Solution 3 by SpreadTheMathLove

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

Video Solution 4 by Pi Academy

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

Video Solution 5 by Dr. David

https://youtu.be/Bmjli0Iq5I4

See Also

2024 AMC 10A (ProblemsAnswer KeyResources)
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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC logo.png