Difference between revisions of "2024 AMC 8 Problems/Problem 10"
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==Solution 1== | ==Solution 1== | ||
| − | 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> | + | 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> |
| Line 22: | Line 22: | ||
==Solution 3== | ==Solution 3== | ||
2030 - 1980 = 50 years. | 2030 - 1980 = 50 years. | ||
| − | 338 + 50 * 1.515 = 338 + 75.75 = 413.75 | + | 338 + 50 * 1.515 = 338 + 75.75 = 413.75 for 2030 ppm level <math> |
=\boxed{\textbf{(B)\ 414}}</math>. | =\boxed{\textbf{(B)\ 414}}</math>. | ||
Revision as of 20:15, 28 January 2024
Contents
Problem
In January 
the Mauna Loa Observatory recorded carbon dioxide (CO2) levels of 
ppm (parts per million). Over the years the average CO2 reading has increased by about 
ppm each year. What is the expected CO2 level in ppm in January 
? Round your answer to the nearest integer.
Solution 1
This is a time period of 
years, so we can expect the ppm to increase by 
.
.
-ILoveMath31415926535
Solution 1
For each year that has passed, the ppm will increase by . In , the CO2 would have increased by 
so the total ppm of CO2 will be 
-Benedict T (countmath1)
Solution 3
2030 - 1980 = 50 years.
338 + 50 * 1.515 = 338 + 75.75 = 413.75 for 2030 ppm level 
.
-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. 
