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Difference between revisions of "2021 Fall AMC 10B Problems/Problem 4"

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==Solution==
 
==Solution==
  
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Let the temperature of Minneapolis at noon be <math>M</math> and let the temperature of St. Louis at noon be <math>S</math>. Then <math>M = N + S</math> and <math>(M-5)-(S+3) = |2| \implies M-S-8 = |2|</math>.
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Let the temperature of Minneapolis at noon be <math>M</math> and let the temperature of St. Louis at noon be <math>S</math>. Then <math>M = N + S</math> and <math>|(M-5)-(S+3)| = 2 \implies |M-S-8| = |2|</math>.
  
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Substituting <math>M</math> into the second equation we have <math>N - 8 = |2| \implies N = 10, 6</math>.
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Substituting <math>M</math> into the second equation we have <math>|N - 8| = 2 \implies N = 10, 6</math>.
  
 
The product of all possible values of N is therefore <math>10\cdot6=60=\boxed{C}</math>
 
The product of all possible values of N is therefore <math>10\cdot6=60=\boxed{C}</math>
  
 
~KingRavi
 
~KingRavi

Revision as of 00:50, 23 November 2021

At noon on a certain day, Minneapolis is $N$ degrees warmer than St. Louis. At $4{:}00$ the temperature in Minneapolis has fallen by $5$ degrees while the temperature in St. Louis has risen by $3$ degrees, at which time the temperatures in the two cities differ by $2$ degrees. What is the product of all possible values of $N?$

$(\textbf{A})\: 10\qquad(\textbf{B}) \: 30\qquad(\textbf{C}) \: 60\qquad(\textbf{D}) \: 100\qquad(\textbf{E}) \: 120$

Solution

Let the temperature of Minneapolis at noon be $M$ and let the temperature of St. Louis at noon be $S$. Then $M = N + S$ and $|(M-5)-(S+3)| = 2 \implies |M-S-8| = |2|$.

Substituting $M$ into the second equation we have $|N - 8| = 2 \implies N = 10, 6$.

The product of all possible values of N is therefore $10\cdot6=60=\boxed{C}$

~KingRavi

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