Difference between revisions of "2024 AMC 10A Problems/Problem 8"
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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. | 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. | ||
− | Since <math>450</math> packages were packed in total, then Daria must have packed <math>450-350=100</math> packages in total, and since he 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{\ | + | Since <math>450</math> packages were packed in total, then Daria must have packed <math>450-350=100</math> packages in total, and since he 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 | ||
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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> hours 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. | 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> hours 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. | ||
− | 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{\ | + | 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>. |
~i_am_suk_at_math_2 | ~i_am_suk_at_math_2 | ||
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== Solution 3 == | == Solution 3 == | ||
− | We notice that every <math>3</math> minutes, Amy, Bomani, Charlie pack <math>4 + 3 + 3 = 10</math> packages in total. Thus, over the entire course, they pack <math>105 \cdot \frac{10}{3} = 350</math>. Thus, Daria packaged <math>450 - 350 = 100</math> packages. Thus, Daria worked for <math>100 \cdot \frac{4}{5} = 80</math> minutes, so Daria starting working <math>80</math> minutes before <math>2:45</math> pm, or <math>\boxed{\ | + | We notice that every <math>3</math> minutes, Amy, Bomani, Charlie pack <math>4 + 3 + 3 = 10</math> packages in total. Thus, over the entire course, they pack <math>105 \cdot \frac{10}{3} = 350</math>. Thus, Daria packaged <math>450 - 350 = 100</math> packages. Thus, Daria worked for <math>100 \cdot \frac{4}{5} = 80</math> minutes, so Daria starting working <math>80</math> minutes before <math>2:45</math> pm, or <math>\boxed{\textbf{(A) }1:25\text{ PM}}</math>. |
~andliu766 | ~andliu766 |
Revision as of 02:09, 9 November 2024
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 he 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
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
hours 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
Solution 3
We notice that every minutes, Amy, Bomani, Charlie pack
packages in total. Thus, over the entire course, they pack
. Thus, Daria packaged
packages. Thus, Daria worked for
minutes, so Daria starting working
minutes before
pm, or
.
~andliu766
Video Solution 1 by Power Solve
https://youtu.be/j-37jvqzhrg?si=bf4iiXH4E9NM65v8&t=996
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 |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.