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

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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>
 
<math>\textbf{(A) } 77 \qquad \textbf{(B) } 86 \qquad \textbf{(C) } 92 \qquad \textbf{(D) } 98 \qquad \textbf{(E) } 104</math>
  
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==Solution==
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==Solution 1==
  
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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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Initially, the difference between the water temperature and the room temperature is <math>212-68=144^{\circ}\text{F}</math>. After <math>15</math> minutes, the difference would be decreased by a factor of <math>2^{\tfrac{15}{3}}=8</math>. Now, the difference is <math>144\div8=18^{\circ}\text{F}</math>. Finally, the water temperature is <math>68+18=\boxed{\textbf{(B) } 86}</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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~MRENTHUSIASM
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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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~MRENTHUSIASM
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~MathFun1000 (Conciseness)
  
 
==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}}

Revision as of 18:11, 28 January 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 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 (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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