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

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==Solution==
 
==Solution==
  
Note that the ratio between number of trout and total fish is: <cmath>\frac{\text{number of trout}}{\text{total number of fish}} = \frac{30}{180} = \frac16.</cmath> Therefore, 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.
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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.
  
~apex304, SohumUttamchandani, wuwang2002, TaeKim, Cxrupptedpat, lpieleanu, MRENTHUSIASM
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~apex304, SohumUttamchandani, wuwang2002, TaeKim, Cxrupptedpat, MRENTHUSIASM
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 +
==Video Solution by CoolMathProblems==
 +
https://youtu.be/Pf93RGtKo1I?feature=shared&t=280
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 +
==Video Solution (A Clever Explanation You’ll Get Instantly)==
 +
https://youtu.be/zntZrtsnyxc?si=nM5eWOwNU6HRdleZ&t=219
 +
~hsnacademy
 +
 
 +
==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==
 
==Video Solution by Magic Square==
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==Video Solution by SpreadTheMathLove==
 
==Video Solution by SpreadTheMathLove==
 
https://www.youtube.com/watch?v=EcrktBc8zrM
 
https://www.youtube.com/watch?v=EcrktBc8zrM
 +
==Video Solution by Interstigation==
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https://youtu.be/DBqko2xATxs&t=345
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 +
==Video Solution by WhyMath==
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https://youtu.be/J23Ljt3uV-8
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 +
~savannahsolver
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==Video Solution by harungurcan==
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https://www.youtube.com/watch?v=35BW7bsm_Cg&t=547s
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 +
~harungurcan
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==Simple Solution by MathTalks_Now==
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* https://studio.youtube.com/video/PMOeiGLkDH0/edit
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==Video Solution (How to CREATIVELY THINK!!!)==
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https://youtu.be/Rhg5mu7pdNU
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~Education the Study of Everything
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==Video Solution by Dr. David==
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https://youtu.be/n78DeBJNfjY
  
 
==See Also==  
 
==See Also==  
 
{{AMC8 box|year=2023|num-b=4|num-a=6}}
 
{{AMC8 box|year=2023|num-b=4|num-a=6}}
 
{{MAA Notice}}
 
{{MAA Notice}}
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[[Category:Introductory Algebra Problems]]

Latest revision as of 19:10, 4 June 2025

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

Video Solution by CoolMathProblems

https://youtu.be/Pf93RGtKo1I?feature=shared&t=280

Video Solution (A Clever Explanation You’ll Get Instantly)

https://youtu.be/zntZrtsnyxc?si=nM5eWOwNU6HRdleZ&t=219 ~hsnacademy

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

Simple Solution by MathTalks_Now

Video Solution (How to CREATIVELY THINK!!!)

https://youtu.be/Rhg5mu7pdNU

~Education the Study of Everything

Video Solution by Dr. David

https://youtu.be/n78DeBJNfjY

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png