SOLUTION: After riding at a steady speed for 40 miles, a bicylist had a flat tire and walked 5 miles to a repair shop. The cycling rate was 4 times faster than the walking rate. If the tim
Algebra ->
Customizable Word Problem Solvers
-> Travel
-> SOLUTION: After riding at a steady speed for 40 miles, a bicylist had a flat tire and walked 5 miles to a repair shop. The cycling rate was 4 times faster than the walking rate. If the tim
Log On
Question 148362: After riding at a steady speed for 40 miles, a bicylist had a flat tire and walked 5 miles to a repair shop. The cycling rate was 4 times faster than the walking rate. If the time spent cycling and walking was 5 hours, at what rate was the cyclist riding? Answer by ptaylor(2198) (Show Source):
You can put this solution on YOUR website! Distance(d) equals Rate(r) times Time(t) or d=rt; r=d/t and t=d/r
Let r=walking rate
Then 4r=cycling rate
Time spent walking=5/r
Time spent cycling=40/4r=10/r
Now we are told that the sum of the above times is 5 hours, so:
5/r + 10/r=5 multiply each term by r
5+10=5r
15=5r divide each side by 5
r=3 mph------------------walking rate
4r=4*3=12 mph----------------cycling rate
ck
40/12 + 5/3=5
40/12+20/12=5
60/12=5
5=5
Hope this helps------ptaylor
