Difference between revisions of "2021 Fall AMC 12B Problems/Problem 3"
(→Solution 1) |
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~Wilhelm Z | ~Wilhelm Z | ||
| + | |||
| + | == 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) | ||
{{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}} | ||
Revision as of 21:17, 25 November 2021
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
Let the temperature of Minneapolis be 
, and that of St. Louis be 
. We have 
.
At , either
or
Plug in to solve the two equations respectively to get 
or 
. Hence the answer is 
~Wilhelm Z
Solution 2
Hence, 
or 6.
Therefore, the answer is 
.
~Steven Chen (www.professorchenedu.com)
| 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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