Difference between revisions of "2017 AMC 8 Problems/Problem 23"

m (Solution)
Line 12: Line 12:
 
<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
 +
 +
==See Also==
 +
{{AMC8 box|year=2017|num-b=22|num-a=23}}
 +
 +
{{MAA Notice}}

Revision as of 14:10, 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?

$\textbf{(A) }10\qquad\textbf{(B) }15\qquad\textbf{(C) }25\qquad\textbf{(D) }50\qquad\textbf{(E) }82$

Solution

From the question \[\dfrac{60}{x},\dfrac{60}{x+5},\dfrac{60}{x+10},\dfrac{60}{x+15}\in \mathbb{N}\] and by simple guess and check, the answer is \[\dfrac{60}{5}+\dfrac{60}{10}+\dfrac{60}{15}+\dfrac{60}{20}=\boxed{\textbf{(C)}\ 25}\] -kvedula2004

See Also

2017 AMC 8 (ProblemsAnswer KeyResources)
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. AMC logo.png

Invalid username
Login to AoPS