2022 AMC 8 Problems/Problem 9

Revision as of 17:11, 28 January 2022 by Mathfun1000 (talk | contribs)

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 1

Initially, the difference between the water temperature and the room temperature is $212-68=144^{\circ}\text{F}$. After $15$ minutes, the difference would be decreased by a factor of $2^{\tfrac{15}{3}}=8$. Now, the difference is $144\div8=18^{\circ}\text{F}$. Finally, the water temperature is $68+18=\boxed{\textbf{(B) } 86}$ degrees Fahrenheit.

~MRENTHUSIASM

~MathFun1000 (Conciseness)

See Also

2022 AMC 8 (ProblemsAnswer KeyResources)
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. AMC logo.png