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

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==Solution 1==
 
==Solution 1==
  
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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>.
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This is a time period of <math>2030 - 1980 = 50</math> years, so we can expect the ppm to increase by <math>50*1.515=75.75\approx 76</math> ppm.
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<math>76+338=\boxed{\textbf{(B)\ 414}}</math>.
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Since we started with <math>338</math> ppm, we have <math>76+338=\boxed{\textbf{(B)\ 414}}</math>.
  
 
-ILoveMath31415926535
 
-ILoveMath31415926535
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==Solution 2==
 
==Solution 2==
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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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-Benedict T (countmath1)
 
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==Solution 3==
 
 
<math>2030 - 1980 = 50</math> years.
 
<math>2030 - 1980 = 50</math> years.
 
The ppm level in 2030 is <math>338 + 50 * 1.515 = 338 + 75.75 = 413.75</math> <math>
 
The ppm level in 2030 is <math>338 + 50 * 1.515 = 338 + 75.75 = 413.75</math> <math>
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=\boxed{\textbf{(B)\ 414}}</math>.
+
\approx \boxed{\textbf{(B)\ 414}}</math>.
 
   
 
   
 
~thebanker88
 
~thebanker88
 +
 +
==Video Solution (A Clever Explanation You’ll Get Instantly)==
 +
https://youtu.be/5ZIFnqymdDQ?si=ATKTcOdaVPGWU8_V&t=1089
 +
 +
~hsnacademy
  
 
==Video Solution 1 (easy to digest) by Power Solve==
 
==Video Solution 1 (easy to digest) by Power Solve==
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==Video Solution by Interstigation==
 
==Video Solution by Interstigation==
 
https://youtu.be/ktzijuZtDas&t=938
 
https://youtu.be/ktzijuZtDas&t=938
 +
 +
==Video Solution by Daily Dose of Math (Certified, Simple, and Logical)==
 +
 +
https://youtu.be/8GHuS5HEoWc
 +
 +
~Thesmartgreekmathdude
 +
 +
==Video Solution by WhyMath==
 +
https://youtu.be/l7umQHZljeI
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2024|num-b=9|num-a=11}}
 
{{AMC8 box|year=2024|num-b=9|num-a=11}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 14:59, 9 November 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 $2030 - 1980 = 50$ years, so we can expect the ppm to increase by $50*1.515=75.75\approx 76$ ppm. Since we started with $338$ ppm, we have $76+338=\boxed{\textbf{(B)\ 414}}$.

-ILoveMath31415926535

Solution 2

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

~thebanker88

Video Solution (A Clever Explanation You’ll Get Instantly)

https://youtu.be/5ZIFnqymdDQ?si=ATKTcOdaVPGWU8_V&t=1089

~hsnacademy

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

Video Solution by Daily Dose of Math (Certified, Simple, and Logical)

https://youtu.be/8GHuS5HEoWc

~Thesmartgreekmathdude

Video Solution by WhyMath

https://youtu.be/l7umQHZljeI

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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