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

Revision as of 01:09, 9 November 2024

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?

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

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 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

Video Solution 1 by Power Solve

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

@@ Line 2: / Line 2: @@
 Amy, Bomani, Charlie, and Daria work in a chocolate factory. On Monday Amy, Bomani, and Charlie started working at <math>1:00 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 PM</math>. At what time did Daria join the group?
-<math>\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</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 ==

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