Difference between revisions of "2024 AMC 8 Problems/Problem 10"

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This is a time period of <math>50</math> years, so we can expect the ppm to increase by <math>50*1.515=75.75~76</math>.
 
This is a time period of <math>50</math> years, so we can expect the ppm to increase by <math>50*1.515=75.75~76</math>.
<math>76+338=(B) \boxed{414}</math>.
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<math>76+338=\boxed{\textbf{(B)\ 414}}</math>.
  
 
-ILoveMath31415926535
 
-ILoveMath31415926535
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==Solution 1==
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For each year that has passed, the ppm will increase by <math>50</math>. In <math>2030</math>, the CO2 would have increased by <math>50\cdot 1.515 \approx 76,</math> so the total ppm of CO2 will be <math>76 + 338 = \boxed{\textbf{(B)\ 414}}.</math>
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-Benedict T (countmath1)
  
 
==Video Solution 1 (easy to digest) by Power Solve==
 
==Video Solution 1 (easy to digest) by Power Solve==

Revision as of 14:56, 26 January 2024

Problem

In January $1980$ the Mauna Loa Observatory recorded carbon dioxide (CO2) levels of $338$ ppm (parts per million). Over the years the average CO2 reading has increased by about $1.515$ ppm each year. What is the expected CO2 level in ppm in January $2030$? Round your answer to the nearest integer.

$\textbf{(A)}\ 24\qquad \textbf{(B)}\ 25\qquad \textbf{(C)}\ 26\qquad \textbf{(D)}\ 27\qquad \textbf{(E)}\ 28$

Solution 1

This is a time period of $50$ years, so we can expect the ppm to increase by $50*1.515=75.75~76$. $76+338=\boxed{\textbf{(B)\ 414}}$.

-ILoveMath31415926535

Solution 1

For each year that has passed, the ppm will increase by $50$. In $2030$, the CO2 would have increased by $50\cdot 1.515 \approx 76,$ so the total ppm of CO2 will be $76 + 338 = \boxed{\textbf{(B)\ 414}}.$


-Benedict T (countmath1)

Video Solution 1 (easy to digest) by Power Solve

https://youtu.be/16YYti_pDUg?si=T3FZAZoeeL5NP3yR&t=411

Video Solution 2 by SpreadTheMathLove

https://www.youtube.com/watch?v=L83DxusGkSY