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Revision as of 03:18, 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

Let the temperature of Minneapolis be $M$, and that of St. Louis be $L$. We have $M=L+N$.

At $4{:}00$, either

$(M-5)=2+(L+3)$

or

$(M-5)+2=(L+3)$

Plug in $M=L+N$ to solve the two equations respectively to get $N=10$ or $N=6$. Hence the answer is $60 \Rightarrow \boxed{(\textbf{C})  }.$

~Wilhelm Z ~KingRavi ~MRENTHUSIASM

Solution 2

\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 (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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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