Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution 1)
m (→‎Video Solution by WhyMath)
 
(21 intermediate revisions by 7 users not shown)
Line 1: Line 1:
 +
{{duplicate|[[2021 Fall AMC 10B Problems/Problem 4|2021 Fall AMC 10B #4]] and [[2021 Fall AMC 12B Problems/Problem 3|2021 Fall AMC 12B #3]]}}
 +
 
==Problem==
 
==Problem==
 
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>
 
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>
+
<math>\textbf{(A)}\: 10\qquad\textbf{(B)} \: 30\qquad\textbf{(C)} \: 60\qquad\textbf{(D)} \: 100\qquad\textbf{(E)} \: 120</math>
 +
 
 +
==Solution 1 (Two Variables)==
 +
At noon on a certain day, let <math>M</math> and <math>L</math> be the temperatures (in degrees) in Minneapolis and St. Louis, respectively. It follows that <math>M=L+N.</math>
 +
 
 +
At <math>4{:}00,</math> we get
 +
<cmath>\begin{align*}
 +
|(M-5)-(L+3)| &= 2 \\
 +
|M-L-8| &= 2 \\
 +
|N-8| &= 2.
 +
\end{align*}</cmath>
 +
We have two cases:
 +
<ol style="margin-left: 1.5em;">
 +
  <li>If <math>N-8=2,</math> then <math>N=10.</math></li><p>
 +
  <li>If <math>N-8=-2,</math> then <math>N=6.</math></li><p>
 +
</ol>
 +
Together, the product of all possible values of <math>N</math> is <math>10\cdot6=\boxed{\textbf{(C)} \: 60}.</math>
 +
 
 +
~Wilhelm Z ~KingRavi ~MRENTHUSIASM
 +
 
 +
== Solution 2 (One Variable) ==
 +
At noon on a certain day, the difference of temperatures in Minneapolis and St. Louis is <math>N</math> degrees.
 +
 
 +
At <math>4{:}00,</math> the difference of temperatures in Minneapolis and St. Louis is <math>N-8</math> degrees.
 +
 
 +
It follows that <cmath>|N-8|=2.</cmath>
 +
We continue with the casework in Solution 1 to get the answer <math>\boxed{\textbf{(C)} \: 60}.</math>
 +
 
 +
~Steven Chen (www.professorchenedu.com) ~MRENTHUSIASM
 +
==Video Solution by Interstigation==
 +
https://youtu.be/p9_RH4s-kBA?t=291
 +
 
 +
~Interstigation
  
−
==Solution 1==
+
==Video Solution==
−
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>.
+
https://youtu.be/480KnrVnbOc
  
−
At <math>4{:}00</math>, either
+
~Education, the Study of Everything
  
−
<math>(M-5)=2+(L+3)</math>
+
==Video Solution by WhyMath==
 +
https://youtu.be/y4U3wv2euH8
  
−
or
+
~savannahsolver
  
−
<math>(M-5)+2=(L+3)</math>
+
==Video Solution by TheBeautyofMath==
 +
For AMC 10: https://youtu.be/lC7naDZ1Eu4?t=670
  
−
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>
+
For AMC 12: https://youtu.be/yaE5aAmeesc?t=379
  
−
~Wilhelm Z
+
~IceMatrix
  
 +
==See Also==
 +
{{AMC10 box|year=2021 Fall|ab=B|num-a=5|num-b=3}}
 
{{AMC12 box|year=2021 Fall|ab=B|num-a=4|num-b=2}}
 
{{AMC12 box|year=2021 Fall|ab=B|num-a=4|num-b=2}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 12:51, 30 December 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 (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:

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

  2. 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

Video Solution by TheBeautyofMath

For AMC 10: https://youtu.be/lC7naDZ1Eu4?t=670

For AMC 12: https://youtu.be/yaE5aAmeesc?t=379

~IceMatrix

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

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2021_Fall_AMC_12B_Problems/Problem_3&oldid=185257"