Difference between revisions of "2021 Fall AMC 10B Problems/Problem 4"

(Since this problem is a duplicate from the AMC 12 Problem, I redirected accordingly.)
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==Problem==
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#REDIRECT [[2021_Fall_AMC_12B_Problems/Problem_3]]
 
 
At noon on a certain day, Minneapolis is <math>N</math> degrees warmer than St. Louis. At <math>4{:}00</math> the temperature in Minneapolis has fallen by <math>5</math> degrees while the temperature in St. Louis has risen by <math>3</math> degrees, at which time the temperatures in the two cities differ by <math>2</math> degrees. What is the product of all possible values of <math>N?</math>
 
 
 
<math>(\textbf{A})\: 10\qquad(\textbf{B}) \: 30\qquad(\textbf{C}) \: 60\qquad(\textbf{D}) \: 100\qquad(\textbf{E}) \: 120</math>
 
 
 
==Solution 1==
 
 
 
Let the temperature of Minneapolis at noon be <math>M</math> and let the temperature of St. Louis at noon be <math>S</math>. Then <math>M = N + S</math> and <math>|(M-5)-(S+3)| = 2 \implies |M-S-8| = |2|</math>.
 
 
 
Substituting <math>M</math> into the second equation we have <math>|N - 8| = 2 \implies N = 10, 6</math>.
 
 
 
The product of all possible values of N is therefore <math>10\cdot6=60=\boxed{C}</math>
 
 
 
~KingRavi
 
 
 
== Solution 2 ==
 
<cmath>
 
\begin{align*}
 
| N - 5 - 3 | = 2 .
 
\end{align*}
 
</cmath>
 
 
 
Hence, <math>N = 10</math> or 6.
 
 
 
Therefore, the answer is <math>\boxed{\textbf{(C) }60}</math>.
 
 
 
~Steven Chen (www.professorchenedu.com)
 
 
 
==Video Solution by Interstigation==
 
https://youtu.be/p9_RH4s-kBA?t=291
 
 
 
==See Also==
 
{{AMC10 box|year=2021 Fall|ab=B|num-a=5|num-b=3}}
 
{{MAA Notice}}
 

Latest revision as of 17:46, 3 January 2022