2023 AMC 8 Problems/Problem 5

Revision as of 23:20, 26 January 2023 by MRENTHUSIASM (talk | contribs) (Undo revision 188142 by Megaboy6679 (talk) I don't really think the last edit is necessary. We want to use as few lines as possible ...)

Problem

A lake contains $250$ trout, along with a variety of other fish. When a marine biologist catches and releases a sample of $180$ fish from the lake, $30$ are identified as trout. Assume that the ratio of trout to the total number of fish is the same in both the sample and the lake. How many fish are there in the lake?

$\textbf{(A)}\ 1250 \qquad \textbf{(B)}\ 1500 \qquad \textbf{(C)}\ 1750 \qquad \textbf{(D)}\ 1800 \qquad \textbf{(E)}\ 2000$

Solution

Note that \[\frac{\text{number of trout}}{\text{total number of fish}} = \frac{30}{180} = \frac16.\] So, the total number of fish is $6$ times the number of trout. Since the lake contains $250$ trout, there are $250\cdot6=\boxed{\textbf{(B)}\ 1500}$ fish in the lake.

~apex304, SohumUttamchandani, wuwang2002, TaeKim, Cxrupptedpat, lpieleanu, MRENTHUSIASM

Video Solution by Magic Square

https://youtu.be/-N46BeEKaCQ?t=5308

Video Solution by SpreadTheMathLove

https://www.youtube.com/watch?v=EcrktBc8zrM

See Also

2023 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 4
Followed by
Problem 6
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 AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png