Difference between revisions of "2024 AMC 8 Problems/Problem 10"
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==Solution 3== | ==Solution 3== | ||
<math>2030 - 1980 = 50</math> years. | <math>2030 - 1980 = 50</math> years. | ||
| − | 338 + 50 * 1.515 = 338 + 75.75 = 413.75 for 2030 ppm level <math> | + | <math>338 + 50 * 1.515 = 338 + 75.75 = 413.75</math> for 2030 ppm level <math> |
=\boxed{\textbf{(B)\ 414}}</math>. | =\boxed{\textbf{(B)\ 414}}</math>. | ||
Revision as of 11:15, 13 July 2024
Contents
- 1 Problem
- 2 Solution 1
- 3 Solution 2
- 4 Solution 3
- 5 Video Solution 1 (easy to digest) by Power Solve
- 6 Video Solution by Math-X (First fully understand the problem!!!)
- 7 Video Solution by NiuniuMaths (Easy to understand!)
- 8 Video Solution 2 by SpreadTheMathLove
- 9 Video Solution by CosineMethod [🔥Fast and Easy🔥]
- 10 Video Solution by Interstigation
- 11 See Also
Problem
In January 
the Mauna Loa Observatory recorded carbon dioxide 
levels of 
ppm (parts per million). Over the years the average 
reading has increased by about 
ppm each year. What is the expected 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 2
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
years.
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 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. 
