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

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First we find the ratio of trout: total fish using the <math>\frac{30}{180} = \frac {1}{6}</math>then since we know there is a <math>250</math> in the numerator as there are <math>250</math> trout in the whole lake we get <math>\frac{250}{250*6} =\boxed{\text{(B)}1500}</math> fish total in the lake.
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==Problem==
  
~apex304, SohumUttamchandani, wuwang2002, TaeKim, Cxrupptedpat
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A lake contains <math>250</math> trout, along with a variety of other fish. When a marine biologist catches and releases a sample of <math>180</math> fish from the lake, <math>30</math> 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?
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<math>\textbf{(A)}\ 1250 \qquad \textbf{(B)}\ 1500 \qquad \textbf{(C)}\ 1750 \qquad \textbf{(D)}\ 1800 \qquad \textbf{(E)}\ 2000</math>
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==Solution==
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~apex304, SohumUttamchandani, wuwang2002, TaeKim, Cxrupptedpat, MRENTHUSIASM
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==See Also==
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{{AMC8 box|year=2023|num-b=4|num-a=6}}
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{{MAA Notice}}

Revision as of 23:11, 24 January 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

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

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

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