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2024 AMC 10A Problem 8

2024 AMC 10A Problem 8AMC 10 Step by Step

Name: Date: October 28, 2026

Linear Equations & Word Problems · difficulty 2 of 5 · about 3 min · position P1-10

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?

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

Answer

Key insight

Amy, Bomani and Charlie work the whole 105 minutes and pack 350 packages. Daria packs the other 100, which takes her 80 minutes, so she started 80 minutes before 2:45.

Solution

The first three workers are there the whole time, so their work is easy to count. Whatever is left must be Daria's.

How long is the work time? From 1:00 PM to 2:45 PM is $1$ hour $45$ minutes, which is $105$ minutes.
The first three workers. Together they pack $4 + 3 + 3 = 10$ packages every $3$ minutes. In $105$ minutes there are $105 \div 3 = 35$ blocks of $3$ minutes, so they pack $35 \cdot 10 = 350$ packages.
Daria's share. The total is $450$ , so Daria packed $450 - 350 = 100$ packages.
Daria's time. She packs $5$ packages every $4$ minutes. $100$ packages is $100 \div 5 = 20$ groups of $5$ , which takes $20 \cdot 4 = 80$ minutes.
Work backwards from the end. She finished at 2:45 PM after working $80$ minutes ( $1$ hour $20$ minutes). So she started at 1:25 PM.

The answer is $\boxed{\textbf{(A)}\ 1{:}25\text{ PM}}$ .

Why this works

Split the work by who did it. The people who worked the whole time produce a fixed amount, and the leftover belongs to the person who joined late. Dividing that leftover by her speed gives how long she worked, and counting back from the known finishing time gives when she started, without ever needing an unknown start time.

Alternative approach

Let Daria work for $t$ minutes. The first three pack $\frac{10}{3}$ packages per minute for $105$ minutes, and Daria packs $\frac{5}{4}$ per minute for $t$ minutes:
$\frac{10}{3} \cdot 105 + \frac{5}{4}\, t = 450.$
The first term is $350$ , so $\frac{5}{4}\, t = 100$ and $t = 80$ minutes. She started $80$ minutes before 2:45 PM, which is 1:25 PM.

The trap

Counting 1:00 to 2:45 as 145 minutes instead of 105, or giving Daria the whole time period.

Concise solution

Amy, Bomani and Charlie together pack $4 + 3 + 3 = 10$ packages every $3$ minutes. From 1:00 PM to 2:45 PM is $105$ minutes, which is $35$ blocks of $3$ minutes, so the three of them pack $35\cdot 10 = 350$ packages.

That leaves $450 - 350 = 100$ packages for Daria. She packs $5$ every $4$ minutes, so $100$ packages take $\frac{100}{5}\cdot 4 = 80$ minutes.

Daria worked the last $80$ minutes, so she joined $80$ minutes before 2:45 PM, at 1:25 PM.

The answer is $\boxed{\textbf{(A)}\ 1{:}25\text{ PM}}$ .

Review in the AoPS books

Introduction to Algebra, Chapter 3: One-Variable Linear Equations (§3.3, §3.4)
Prealgebra, Chapter 15: Problem-Solving Strategies (§15.4)
Introduction to Algebra, Chapter 7: Proportion (§7.4)

Techniques

Set up the equation/formula and compute; no special trick needed · Start from the end state / desired conclusion and reverse