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"

m (→‎Solution)
(Undo revision 188142 by Megaboy6679 (talk) I don't really think the last edit is necessary. We want to use as few lines as possible ...)
(Tag: Undo)
Line 7: Line 7:
 
==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.
+
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
 
~apex304, SohumUttamchandani, wuwang2002, TaeKim, Cxrupptedpat, lpieleanu, MRENTHUSIASM

Revision as of 00:20, 27 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

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, lpieleanu, 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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2023_AMC_8_Problems/Problem_5&oldid=188145"