Difference between revisions of "2022 AMC 8 Problems/Problem 9"
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<u><b>Remark</b></u> | <u><b>Remark</b></u> | ||
| − | Alternatively, we can condense the solution above into the following equation: <cmath>68+(212-68)\cdot\left(\frac12\right)^{15 | + | Alternatively, we can condense the solution above into the following equation: <cmath>68+(212-68)\cdot\left(\frac12\right)^{\tfrac{15}{5}}=86.</cmath> |
~MRENTHUSIASM ~Mathfun1000 | ~MRENTHUSIASM ~Mathfun1000 | ||
Revision as of 21:48, 28 January 2022
Problem
A cup of boiling water (
) is placed to cool in a room whose temperature remains constant at 
. Suppose the difference between the water temperature and the room temperature is halved every 
minutes. What is the water temperature, in degrees Fahrenheit, after 
minutes?
Solution
Initially, the difference between the water temperature and the room temperature is 
degrees Fahrenheit.
After minutes, the difference between the temperatures is 
degrees Fahrenheit.
After 
minutes, the difference between the temperatures is 
degrees Fahrenheit.
After minutes, the difference between the temperatures is 
degrees Fahrenheit. At this point, the water temperature is 
degrees Fahrenheit.
Remark
Alternatively, we can condense the solution above into the following equation: ![\[68+(212-68)\cdot\left(\frac12\right)^{\tfrac{15}{5}}=86.\]](http://latex.artofproblemsolving.com/5/a/9/5a909744444c81a91c9dfe019ba59b1e1a1fa52f.png)
~MRENTHUSIASM ~Mathfun1000
See Also
| 2022 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 8 |
Followed by Problem 10 | |
| 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. 
