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

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==Problem==
 
==Problem==
  
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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.  
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In January <math>1980</math> the Mauna Loa Observatory recorded carbon dioxide <math>(CO2)</math> levels of <math>338</math> ppm (parts per million). Over the years the average <math>CO2</math> reading has increased by about <math>1.515</math> ppm each year. What is the expected <math>CO2</math> level in ppm in January <math>2030</math>? Round your answer to the nearest integer.  
  
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(A) <math>399</math>  (B) <math>414</math>  (C) <math>420</math>  (D) <math>444</math>  (E) <math>459</math>
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<math>\textbf{(A)}\ 399\qquad \textbf{(B)}\ 414\qquad \textbf{(C)}\ 420\qquad \textbf{(D)}\ 444\qquad \textbf{(E)}\ 459</math>
  
 
==Solution 1==
 
==Solution 1==
  
 
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>.
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<math>76+338=</math>(B) \boxed{414}$.
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<math>76+338=\boxed{\textbf{(B)\ 414}}</math>.
  
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==Video Solution 1(easy to digest) by Power Solve==
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-ILoveMath31415926535
 +
 
 +
==Solution 2==
 +
 
 +
For each year that has passed, the ppm will increase by <math>1.515</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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 +
 
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-Benedict T (countmath1)
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==Solution 3==
 +
2030 - 1980 = 50 years.
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338 + 50 * 1.515 = 338 + 75.75 = 413.75 for 2030 ppm level <math>
 +
=\boxed{\textbf{(B)\ 414}}</math>.
 +
 +
-thebanker88
 +
 
 +
==Video Solution 1 (easy to digest) by Power Solve==
 
https://youtu.be/16YYti_pDUg?si=T3FZAZoeeL5NP3yR&t=411
 
https://youtu.be/16YYti_pDUg?si=T3FZAZoeeL5NP3yR&t=411
 +
 +
==Video Solution by Math-X (First fully understand the problem!!!)==
 +
https://youtu.be/BaE00H2SHQM?si=kk52okhz__aC8g9l&t=2078
 +
 +
~Math-X
 +
 +
==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
 +
==Video Solution by Interstigation==
 +
https://youtu.be/ktzijuZtDas&t=938
 +
 +
==See Also==
 +
{{AMC8 box|year=2024|num-b=9|num-a=11}}
 +
{{MAA Notice}}

Latest revision as of 10:36, 20 March 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 2

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 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 Math-X (First fully understand the problem!!!)

https://youtu.be/BaE00H2SHQM?si=kk52okhz__aC8g9l&t=2078

~Math-X

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

Video Solution by Interstigation

https://youtu.be/ktzijuZtDas&t=938

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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