2021 Fall AMC 12B Problems/Problem 3

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:

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

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)

At noon on a certain day, the difference of temperatures in Minneapolis and St. Louis is $N$ degrees.

At $4{:}00,$ the difference of temperatures in Minneapolis and St. Louis is $N-8$ degrees.

It follows that $|N-8|=2.$ We continue with the casework in Solution 1 to get the answer $\boxed{\textbf{(C)} \: 60}.$

~Steven Chen (www.professorchenedu.com) ~MRENTHUSIASM

Video Solution by Interstigation

https://youtu.be/p9_RH4s-kBA?t=291

~Interstigation

Video Solution

https://youtu.be/480KnrVnbOc

~Education, the Study of Everything

Video Solution by WhyMath

https://youtu.be/y4U3wv2euH8

~savannahsolver

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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