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2024 AMC 10A Problems/Problem 8

Revision as of 16:16, 8 November 2024 by Andliu766 (talk | contribs)

Problem

Amy, Bomani, Charlie, and Daria work in a chocolate factory. On Monday Amy, Bomani, and Charlie started working at $1:00 PM$ 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 $2:45 PM$. At what time did Daria join the group?

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

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 he 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 $\boxed{\text{(A) }1:25\text{ PM}}$

~Tacos_are_yummy_1

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 $105$ hours 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 $\boxed{\text{(A) }1:25\text{ PM}}$.

~i_am_suk_at_math_2

Solution 3

We notice that every $3$ minutes, Amy, Bomani, Charlie pack $4 + 3 + 3 = 10$ packages in total. Thus, over the entire course, they pack $105 \cdot \frac{10}{3} = 350$. Thus, Daria packaged $450 - 350 = 100$ packages. Thus, Daria worked for $100 \cdot \frac{4}{5} = 80$ minutes, so Daria starting working $80$ minutes before $2:45$ pm, or $\boxed{\text{(A) }1:25\text{ PM}}$.

~andliu766


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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