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

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==Problem==
 
==Problem==
COPY AND PASTE THE PROBLEM HERE.
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A cup of boiling water (<math>212^{\circ}\text{F}</math>) is placed to cool in a room whose temperature remains constant at <math>68^{\circ}\text{F}</math>. Suppose the difference between the water temperature and the room temperature is halved every <math>5</math> minutes. What is the water temperature, in degrees Fahrenheit, after <math>15</math> minutes?
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<math>\textbf{(A) } 77 \qquad \textbf{(B) } 86 \qquad \textbf{(C) } 92 \qquad \textbf{(D) } 98 \qquad \textbf{(E) } 104</math>
  
 
==Solution==
 
==Solution==
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Initially, the difference between the water temperature and the room temperature is <math>212-68=144</math> degrees Fahrenheit.
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After <math>5</math> minutes, the difference between the temperatures is <math>144\div2=72</math> degrees Fahrenheit.
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After <math>10</math> minutes, the difference between the temperatures is <math>72\div2=36</math> degrees Fahrenheit.
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After <math>15</math> minutes, the difference between the temperatures is <math>36\div2=18</math> degrees Fahrenheit. At this point, the water temperature is <math>68+18=\boxed{\textbf{(B) } 86}</math> degrees Fahrenheit.
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<u><b>Remark</b></u>
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Alternatively, we can condense the solution above into the following equation: <cmath>68+(212-68)\cdot\left(\frac12\right)^{\tfrac{15}{5}}=86.</cmath>
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~MRENTHUSIASM ~MathFun1000
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==Video Solution by Math-X (First understand the problem!!!)==
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https://youtu.be/oUEa7AjMF2A?si=l1o2Qg5grdDcniVg&t=1179
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~Math-X
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==Video Solution (CREATIVE THINKING!!!)==
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https://youtu.be/U6L9QiKrEW0
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~Education, the Study of Everything
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==Video Solution==
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https://www.youtube.com/watch?v=Ij9pAy6tQSg&t=627
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~Interstigation
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==Video Solution==
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https://youtu.be/1xspUFoKDnU?t=253
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~STEMbreezy
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==Video Solution==
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https://youtu.be/XQ4nVqMrgm0
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~savannahsolver
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==Video Solution==
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https://youtu.be/0YKUSrG9G-Q
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~harungurcan
  
 
==See Also==  
 
==See Also==  
 
{{AMC8 box|year=2022|num-b=8|num-a=10}}
 
{{AMC8 box|year=2022|num-b=8|num-a=10}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 21:05, 2 January 2024

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 by Math-X (First understand the problem!!!)

https://youtu.be/oUEa7AjMF2A?si=l1o2Qg5grdDcniVg&t=1179

~Math-X

Video Solution (CREATIVE THINKING!!!)

https://youtu.be/U6L9QiKrEW0

~Education, the Study of Everything

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

Video Solution

https://youtu.be/0YKUSrG9G-Q

~harungurcan

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

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