Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution: Removed contributor who did not really contribute.)
(→‎Video Solution by SpreadTheMathLove)
Line 16: Line 16:
 
==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==
 +
https://youtu.be/1bA7fD7Lg54?t=260
  
 
==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}}

Revision as of 21:39, 16 February 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

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/1bA7fD7Lg54?t=260

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2023_AMC_8_Problems/Problem_5&oldid=189783"