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

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-Benedict T (countmath1)
 
-Benedict T (countmath1)
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==Solution 3==
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2030 - 1980 = 50 years.
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338 + 50 * 1.515 = 338 + 75.75 = 413.75 \approx. 414 for 2030<math> ppm</math> level <math>
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=\boxed{\textbf{(B)\ 414}}</math>.
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-thebanker88
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==Video Solution 1 (easy to digest) by Power Solve==
 
==Video Solution 1 (easy to digest) by Power Solve==

Revision as of 20:07, 28 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.

$\textbf{(A)}\ 399\qquad \textbf{(B)}\ 414\qquad \textbf{(C)}\ 420\qquad \textbf{(D)}\ 444\qquad \textbf{(E)}\ 459$

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 $1.515$. 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)


Solution 3

2030 - 1980 = 50 years. 338 + 50 * 1.515 = 338 + 75.75 = 413.75 \approx. 414 for 2030$ppm$ level $=\boxed{\textbf{(B)\ 414}}$.

-thebanker88


Video Solution 1 (easy to digest) by Power Solve

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

Video Solution by NiuniuMaths (Easy to understand!)

https://www.youtube.com/watch?v=V-xN8Njd_Lc

~NiuniuMaths

Video Solution 2 by SpreadTheMathLove

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

Video Solution by CosineMethod [🔥Fast and Easy🔥]

https://www.youtube.com/watch?v=uceM9Gek944

See Also

2024 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 9 		Followed by
Problem 11
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

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