2022 AMC 8 Problems/Problem 9

Revision as of 17:08, 8 June 2022 by MRENTHUSIASM (talk | contribs) (Again, I like the Solution-Remark version better. It sounds more formal. :))
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

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

~Interstigation

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

Invalid username
Login to AoPS