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 "2024 AMC 8 Problems/Problem 10"
m (→‎Solution 1)
 
(18 intermediate revisions by 12 users not shown)
Line 1: Line 1:
−
==Video Solution 1(easy to digest) by Power Solve==
+
==Problem==
 +
 
 +
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.
 +
 
 +
<math>\textbf{(A)}\ 399\qquad \textbf{(B)}\ 414\qquad \textbf{(C)}\ 420\qquad \textbf{(D)}\ 444\qquad \textbf{(E)}\ 459</math>
 +
 
 +
==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>.
 +
<math>76+338=\boxed{\textbf{(B)\ 414}}</math>.
 +
 
 +
-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>
 +
 
 +
 
 +
-Benedict T (countmath1)
 +
 
 +
==Solution 3==
 +
2030 - 1980 = 50 years.
 +
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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2024_AMC_8_Problems/Problem_10&oldid=217192"