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

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==Problem==
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At noon on a certain day, Minneapolis is <math>N</math> degrees warmer than St. Louis. At <math>4{:}00</math> the temperature in Minneapolis has fallen by <math>5</math> degrees while the temperature in St. Louis has risen by <math>3</math> degrees, at which time the temperatures in the two cities differ by <math>2</math> degrees. What is the product of all possible values of <math>N?</math>
 
At noon on a certain day, Minneapolis is <math>N</math> degrees warmer than St. Louis. At <math>4{:}00</math> the temperature in Minneapolis has fallen by <math>5</math> degrees while the temperature in St. Louis has risen by <math>3</math> degrees, at which time the temperatures in the two cities differ by <math>2</math> degrees. What is the product of all possible values of <math>N?</math>
  
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~KingRavi
 
~KingRavi
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==See Also==
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{{AMC10 box|year=2021 Fall|ab=B|num-a=4|num-b=2}}
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{{MAA Notice}}

Revision as of 23:53, 22 November 2021

Problem

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

See Also

2021 Fall AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 2 		Followed by
Problem 4
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 AMC 10 Problems and Solutions

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