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Difference between revisions of "2022 AMC 8 Problems/Problem 9"

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Alternatively, we can condense the solution above into the following equation: <math>68+(212-68)\cdot\left(\frac12\right)^{\tfrac{15}{5}}=86.</math>
 
Alternatively, we can condense the solution above into the following equation: <math>68+(212-68)\cdot\left(\frac12\right)^{\tfrac{15}{5}}=86.</math>
  

Revision as of 16:03, 7 June 2022

Problem

A cup of boiling water ($212^{\circ}\text{F}$) is placed to cool in a room whose temperature remains constant at $68^{\circ}\text{F}$. Suppose the difference between the water temperature and the room temperature is halved every $5$ minutes. What is the water temperature, in degrees Fahrenheit, after $15$ minutes?

$\textbf{(A) } 77 \qquad \textbf{(B) } 86 \qquad \textbf{(C) } 92 \qquad \textbf{(D) } 98 \qquad \textbf{(E) } 104$

Solution

Initially, the difference between the water temperature and the room temperature is $212-68=144$ degrees Fahrenheit.

After $5$ minutes, the difference between the temperatures is $144\div2=72$ degrees Fahrenheit.

After $10$ minutes, the difference between the temperatures is $72\div2=36$ degrees Fahrenheit.

After $15$ minutes, the difference between the temperatures is $36\div2=18$ degrees Fahrenheit. At this point, the water temperature is $68+18=\boxed{\textbf{(B) } 86}$ degrees Fahrenheit.

Shorter Version

Alternatively, we can condense the solution above into the following equation: $68+(212-68)\cdot\left(\frac12\right)^{\tfrac{15}{5}}=86.$

~MRENTHUSIASM ~Mathfun1000

Video Solution

https://youtu.be/Ij9pAy6tQSg?t=627

~Interstigation

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

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