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Difference between revisions of "2021 Fall AMC 12B Problems/Problem 3"

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==Solution 1 (Two Variables)==
 
==Solution 1 (Two Variables)==
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At noon on a certain day, let <math>M</math> be the temperature in Minneapolis and <math>L</math> be the temperature in St. Louis. It follows that <math>M=L+N.</math>
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At noon on a certain day, let <math>M</math> and <math>L</math> be the temperatures (in degrees) in Minneapolis and St. Louis, respectively. It follows that <math>M=L+N.</math>
  
 
At <math>4{:}00,</math> we get
 
At <math>4{:}00,</math> we get

Revision as of 04:44, 4 January 2022

The following problem is from both the 2021 Fall AMC 10B #4 and 2021 Fall AMC 12B #3, so both problems redirect to this page.

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 1 (Two Variables)

At noon on a certain day, let $M$ and $L$ be the temperatures (in degrees) in Minneapolis and St. Louis, respectively. It follows that $M=L+N.$

At $4{:}00,$ we get \begin{align*} |(M-5)-(L+3)| &= 2 \\ |M-L-8| &= 2 \\ |N-8| &= 2. \end{align*} We have two cases:

  1. If $N-8=2,$ then $N=10.$

  2. If $N-8=-2,$ then $N=6.$

Together, the product of all possible values of $N$ is $10\cdot6=\boxed{\textbf{(C)} \: 60}.$

~Wilhelm Z ~KingRavi ~MRENTHUSIASM

Solution 2 (One Variable)

\begin{align*} | N - 5 - 3 | = 2 . \end{align*}

Hence, $N = 10$ or 6.

Therefore, the answer is $\boxed{\textbf{(C) }60}$.

~Steven Chen (www.professorchenedu.com)

See Also

2021 Fall AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 3 		Followed by
Problem 5
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
2021 Fall AMC 12B (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 12 Problems and Solutions

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