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}=\boxed{\textbf{(C)}\ 25}</cmath> | <cmath>\dfrac{60}{5}+\dfrac{60}{10}+\dfrac{60}{15}+\dfrac{60}{20}=\boxed{\textbf{(C)}\ 25}</cmath> | ||
-kvedula2004 | -kvedula2004 | ||
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| + | ==See Also== | ||
| + | {{AMC8 box|year=2017|num-b=22|num-a=23}} | ||
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| + | {{MAA Notice}} | ||
Revision as of 15:10, 22 November 2017
Problem 23
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 5 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
From the question
![\[\dfrac{60}{x},\dfrac{60}{x+5},\dfrac{60}{x+10},\dfrac{60}{x+15}\in \mathbb{N}\]](http://latex.artofproblemsolving.com/e/3/e/e3e4c1027a9fb86bad9d808aac06281d4e1b7ae8.png)
and by simple guess and check, the answer is
![\[\dfrac{60}{5}+\dfrac{60}{10}+\dfrac{60}{15}+\dfrac{60}{20}=\boxed{\textbf{(C)}\ 25}\]](http://latex.artofproblemsolving.com/5/a/9/5a992894fbf432e485e126715fa390696755a230.png)
-kvedula2004
See Also
| 2017 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 22 |
Followed by Problem 23 | |
| 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. 
