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

m (Problem)
Line 7: Line 7:
  
 
==Solution 1==
 
==Solution 1==
Let the temperature of Minneapolis be <math>M</math>, and that of St. Louis be <math>L</math>. We have <math>M=L+N</math>.
+
At noon on a certain day, let <math>M</math> be the temperature of Minneapolis and <math>L</math> be the temperature of St. Louis. It follows that <math>M=L+N.</math>
 
 
At <math>4{:}00</math>, either
 
 
 
<math>(M-5)=2+(L+3)</math>
 
 
 
or
 
 
 
<math>(M-5)+2=(L+3)</math>
 
 
 
Plug in <math>M=L+N</math> to solve the two equations respectively to get <math>N=10</math> or <math>N=6</math>. Hence the answer is <math>60 \Rightarrow \boxed{(\textbf{C})  }.</math>
 
  
 
~Wilhelm Z ~KingRavi ~MRENTHUSIASM
 
~Wilhelm Z ~KingRavi ~MRENTHUSIASM

Revision as of 03:21, 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

At noon on a certain day, let $M$ be the temperature of Minneapolis and $L$ be the temperature of St. Louis. It follows that $M=L+N.$

~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

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