Difference between revisions of "2009 AMC 12A Problems/Problem 1"
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== Solution == | == Solution == | ||
| − | There is <math>1</math> hour and <math>60-34 = 26</math> minutes between 10:34 AM and noon; and there is <math>1</math> hour and <math>18</math> minutes between noon and 1:18 PM. Hence the flight took <math>2</math> hours and <math>26 + 18 = 44</math> minutes, and <math>h+m= 46\ \textbf{(A)}</math>. | + | There is <math>1</math> hour and <math>60-34 = 26</math> minutes between 10:34 AM and noon; and there is <math>1</math> hour and <math>18</math> minutes between noon and 1:18 PM. Hence the flight took <math>2</math> hours and <math>26 + 18 = 44</math> minutes,and 2+44=46s0 <math>h+m= 46\ \textbf{(A)}</math>. |
== See Also == | == See Also == | ||
Latest revision as of 15:24, 1 July 2025
Problem
Kim's flight took off from Newark at 10:34 AM and landed in Miami at 1:18 PM. Both cities are in the same time zone. If her flight took 
hours and 
minutes, with 
, what is 
?
Solution
There is 
hour and 
minutes between 10:34 AM and noon; and there is hour and 
minutes between noon and 1:18 PM. Hence the flight took 
hours and 
minutes,and 2+44=46s0 
.
See Also
| 2009 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by First question |
Followed by Problem 2 |
| 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 | |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
