Difference between revisions of "2021 Fall AMC 12B Problems/Problem 3"
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<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== | + | ==Solution 1 (Two Variables)== |
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 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> 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 | ~Wilhelm Z ~KingRavi ~MRENTHUSIASM | ||
− | == Solution 2 == | + | == Solution 2 (One Variable) == |
<cmath> | <cmath> | ||
\begin{align*} | \begin{align*} |
Revision as of 03:51, 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 degrees warmer than St. Louis. At the temperature in Minneapolis has fallen by degrees while the temperature in St. Louis has risen by degrees, at which time the temperatures in the two cities differ by degrees. What is the product of all possible values of
Solution 1 (Two Variables)
At noon on a certain day, let be the temperature of Minneapolis and be the temperature of St. Louis. It follows that
At we get We have two cases:
- If then
- If then
Together, the product of all possible values of is
~Wilhelm Z ~KingRavi ~MRENTHUSIASM
Solution 2 (One Variable)
Hence, or 6.
Therefore, the answer is .
~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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.