2022 AMC 8 Problems/Problem 9

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.

Remark

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://www.youtube.com/watch?v=Ij9pAy6tQSg&t=627

~Interstigation

Video Solution

https://youtu.be/1xspUFoKDnU?t=253

~STEMbreezy

Video Solution

https://youtu.be/XQ4nVqMrgm0

~savannahsolver

2022 AMC 8 (Problems • Answer Key • Resources)
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All AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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2022 AMC 8 Problems/Problem 9

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