Difference between revisions of "2023 AMC 8 Problems/Problem 5"
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Note that <cmath>\frac{\text{number of trout}}{\text{total number of fish}} = \frac{30}{180} = \frac16.</cmath> So, the total number of fish is <math>6</math> times the number of trout. Since the lake contains <math>250</math> trout, there are <math>250\cdot6=\boxed{\textbf{(B)}\ 1500}</math> fish in the lake. | Note that <cmath>\frac{\text{number of trout}}{\text{total number of fish}} = \frac{30}{180} = \frac16.</cmath> So, the total number of fish is <math>6</math> times the number of trout. Since the lake contains <math>250</math> trout, there are <math>250\cdot6=\boxed{\textbf{(B)}\ 1500}</math> fish in the lake. | ||
| − | ~apex304, SohumUttamchandani, wuwang2002, TaeKim, Cxrupptedpat | + | ~apex304, SohumUttamchandani, wuwang2002, TaeKim, Cxrupptedpat, MRENTHUSIASM |
==Video Solution by Magic Square== | ==Video Solution by Magic Square== | ||
Revision as of 16:20, 14 February 2023
Contents
Problem
A lake contains 
trout, along with a variety of other fish. When a marine biologist catches and releases a sample of 
fish from the lake, 
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?
Solution
Note that ![\[\frac{\text{number of trout}}{\text{total number of fish}} = \frac{30}{180} = \frac16.\]](http://latex.artofproblemsolving.com/c/2/d/c2d56c81b6970c98c737fea028a7743c6d3616ec.png)
So, the total number of fish is 
times the number of trout. Since the lake contains trout, there are 
fish in the lake.
~apex304, SohumUttamchandani, wuwang2002, TaeKim, Cxrupptedpat, 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 (Problems • Answer Key • Resources) | ||
| 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. 
