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Difference between revisions of "2023 AMC 8 Problems/Problem 5"

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Revision as of 21:17, 29 December 2023

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, MRENTHUSIASM



*Simple Solution by MathTalks_Now*




Video Solution (How to CREATIVELY THINK!!!)

https://youtu.be/Rhg5mu7pdNU

~Education the Study of everything

Video Solution by Math-X (Let's first Understand the question)

https://youtu.be/Ku_c1YHnLt0?si=daQltLOjgTuUiFTM&t=593 ~Math-X


Video Solution by Magic Square

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

Video Solution by SpreadTheMathLove

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

Video Solution by Interstigation

https://youtu.be/DBqko2xATxs&t=345

Video Solution by WhyMath

https://youtu.be/J23Ljt3uV-8

~savannahsolver

Video Solution by harungurcan

https://www.youtube.com/watch?v=35BW7bsm_Cg&t=547s

~harungurcan

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. AMC logo.png

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