Difference between revisions of "2021 Fall AMC 10B Problems/Problem 4"
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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== | + | ==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>. | 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>. | ||
| Line 14: | Line 14: | ||
~KingRavi | ~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== | ==Video Solution by Interstigation== | ||
https://youtu.be/p9_RH4s-kBA?t=291 | https://youtu.be/p9_RH4s-kBA?t=291 | ||
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 at noon be 
and let the temperature of St. Louis at noon be 
. Then 
and 
.
Substituting into the second equation we have 
.
The product of all possible values of N is therefore 
~KingRavi
Solution 2
Hence, 
or 6.
Therefore, the answer is 
.
~Steven Chen (www.professorchenedu.com)
Video Solution by Interstigation
https://youtu.be/p9_RH4s-kBA?t=291
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
