Track / Stage 3 / 197 of 260 #197 of 1964

Problem 197

AMC 10/12, early questions

Algebra Difficulty 3.5 Find the answer

A group of clerks is assigned the task of sorting $1775$ files. Each clerk sorts at a constant rate of $30$ files per hour. At the end of the first hour, some of the clerks are reassigned to another task; at the end of the second hour, the same number of the remaining clerks are also reassigned to another task, and a similar assignment occurs at the end of the third hour. The group finishes the sorting in $3$ hours and $10$ minutes. Find the number of files sorted during the first one and a half hours of sorting.

Your answer

A number or a short expression. Spacing and $ signs are ignored.

Official solution

There are $x$ clerks at the beginning, and $t$ clerks are reassigned to another task at the end of each hour. So, $30x+30(x-t)+30(x-2t)+30\cdot\frac{10}{60} \cdot (x-3t)=1775$ , and simplify that we get $19x-21t=355$ .
Now the problem is to find a reasonable integer solution. Now we know $x= \frac{355+21t}{19}$ , so $19$ divides $355+21t$ , AND as long as $t$ is a integer, $19$ must divide $2t+355$ . Now, we suppose that $19m=2t+355$ , similarly we get $t=\frac{19m-355}{2}$ , and so in order to get a minimum integer solution for $t$ , it is obvious that $m=19$ works. So we get $t=3$ and $x=22$ . One and a half hour's work should be $30x+15(x-t)$ , so the answer is $\boxed{945}$ .

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.