Difference between revisions of "2024 AMC 10A Problems/Problem 8"
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== Problem == | == Problem == | ||
− | 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 3 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? | + | |
+ | 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? | ||
<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> | <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> | ||
== Solution 1 == | == Solution 1 == | ||
+ | |||
Note that Amy, Bomani, and Charlie pack a total of <math>4+3+3=10</math> packages every <math>3</math> minutes. | 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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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> | 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> | ||
− | ~Tacos_are_yummy_1 | + | ~Tacos_are_yummy_1, andliu766 |
== Solution 2 == | == Solution 2 == | ||
+ | |||
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. | 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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~i_am_suk_at_math_2 | ~i_am_suk_at_math_2 | ||
− | == Video Solution by | + | ==Video Solution 1 by Power Solve == |
− | https://youtu.be/ | + | https://youtu.be/j-37jvqzhrg?si=bf4iiXH4E9NM65v8&t=996 |
− | ==Video Solution | + | == Video Solution 2 by Daily Dose of Math == |
− | + | [//youtu.be/W5hES6aNXAk ~Thesmartgreekmathdude] | |
− | == Video Solution by | + | ==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 | |
− | == | + | == 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}} | ||
+ | [[Category:Introductory Algebra Problems]] |
Latest revision as of 09:31, 22 February 2025
Contents
Problem
Amy, Bomani, Charlie, and Daria work in a chocolate factory. On Monday Amy, Bomani, and Charlie started working at and were able to pack
,
, and
packages, respectively, every
minutes. At some later time, Daria joined the group, and Daria was able to pack
packages every
minutes. Together, they finished packing
packages at exactly
. At what time did Daria join the group?
Solution 1
Note that Amy, Bomani, and Charlie pack a total of packages every
minutes.
The total amount of time worked is hour and
minutes, which when converted to minutes, is
minutes. This means that since Amy, Bomani, and Charlie worked for the entire
minutes, they in total packed
packages.
Since packages were packed in total, then Daria must have packed
packages in total, and since she packs at a rate of
packages per
minutes, then Daria worked for
minutes, therefore Daria joined
minutes before
PM, which was at
~Tacos_are_yummy_1, andliu766
Solution 2
Let the time, in minutes, elapsed between and the time Daria joined the packaging be
. Since Amy packages
packages every
minutes, she packages
packages per minute. Similarly, we can see that both Bomani and Charlie package
package per minute, and Daria packages
packages every minute.
Before Daria arrives, we can write the total packages packaged as . Since there are
minutes between
and
, Daria works with the other three for
minutes, meaning for that time there are
packages packaged.
Adding the two, we get (The total packaged in the entire time is
). Solving this equation, we get
, meaning Daria arrived
minutes after
, meaning the answer is
.
~i_am_suk_at_math_2
Video Solution 1 by Power Solve
https://youtu.be/j-37jvqzhrg?si=bf4iiXH4E9NM65v8&t=996
Video Solution 2 by Daily Dose of Math
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
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.