Difference between revisions of "2017 AMC 8 Problems/Problem 23"
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<cmath>\dfrac{60}{5}+\dfrac{60}{10}+\dfrac{60}{15}+\dfrac{60}{20}=12+6+4+3=\boxed{\textbf{(C)}\ 25}.</cmath> | <cmath>\dfrac{60}{5}+\dfrac{60}{10}+\dfrac{60}{15}+\dfrac{60}{20}=12+6+4+3=\boxed{\textbf{(C)}\ 25}.</cmath> | ||
| − | ==Video | + | == Video Solution by OmegaLearn == |
| − | https://youtu.be/ | + | https://youtu.be/XixU0JZ5FLk?t=395 |
| − | + | ~ pi_is_3.14 | |
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==See Also== | ==See Also== | ||
Latest revision as of 09:52, 24 July 2024
Problem
Each day for four days, Linda traveled for one hour at a speed that resulted in her traveling one mile in an integer number of minutes. Each day after the first, her speed decreased so that the number of minutes to travel one mile increased by 
minutes over the preceding day. Each of the four days, her distance traveled was also an integer number of miles. What was the total number of miles for the four trips?
Solution
It is well known that 
. In the question, we want distance. From the question, we have that the time is 
minutes or 
hour. By the equation derived from , we have 
, so the speed is mile per 
minutes. Because we want the distance, we multiply the time and speed together yielding 
. The minutes cancel out, so now we have 
as our distance for the first day. The distance for the following days are:
![\[\dfrac{60}{x},\dfrac{60}{x+5},\dfrac{60}{x+10},\dfrac{60}{x+15}.\]](http://latex.artofproblemsolving.com/2/1/0/2106706e668970b4fd13ae3bbc1d68f503b56afe.png)
We know that 
are all factors of , therefore, 
because the factors have to be in an arithmetic sequence with the common difference being and is the only solution.
![\[\dfrac{60}{5}+\dfrac{60}{10}+\dfrac{60}{15}+\dfrac{60}{20}=12+6+4+3=\boxed{\textbf{(C)}\ 25}.\]](http://latex.artofproblemsolving.com/b/f/3/bf3571ed622460f90fc14a18490de8f6d857daf2.png)
Video Solution by OmegaLearn
https://youtu.be/XixU0JZ5FLk?t=395
~ pi_is_3.14
See Also
| 2017 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 22 |
Followed by Problem 24 | |
| 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 AJHSME/AMC 8 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
