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

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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>
 
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>
  
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~Wilhelm Z
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~Wilhelm Z ~KingRavi ~MRENTHUSIASM
  
 
== Solution 2 ==
 
== Solution 2 ==

Revision as of 18:07, 3 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 (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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