SOLUTION: A bicyclist started on his trip from city A to city B. In 1 hour and 36 minutes, a biker also left A and headed towards B, and he arrived there at the same time as the bicyclist. F

Algebra ->  Finance -> SOLUTION: A bicyclist started on his trip from city A to city B. In 1 hour and 36 minutes, a biker also left A and headed towards B, and he arrived there at the same time as the bicyclist. F      Log On


   

Question 972770: A bicyclist started on his trip from city A to city B. In 1 hour and 36 minutes, a biker also left A and headed towards B, and he arrived there at the same time as the bicyclist. Find the speed of the bicyclist if it is less than the speed of the biker by 32 km/hour, and the distance between the two cities is 52 km.
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
A bicyclist started on his trip from city A to city B.
In 1 hour and 36 minutes, a biker also left A and headed towards B, and he arrived there at the same time as the bicyclist.
Find the speed of the bicyclist if it is less than the speed of the biker by 32 km/hour, and the distance between the two cities is 52 km.
:
let s = the speed of the 1st biker
then
(s+32) = speed of the 2nd biker
:
change 1 hr 36 min to 1 + 36/60 = 1.6 hrs
:
Biker 1 travel time is 1.6 hrs longer than biker 2
52%2Fs - 1.6 = 52%2F%28%28s%2B32%29%29
Multiply by s(s+32), cancel the denominators
52(s+32) - 1.6s(s+32) = 52s
52x + 1664 - 1.6s^2 - 51.2s = 52s
combine like terms, on the right to form a quadratic equation
0 = 1.6s^2 + 51.2s - 1664
Finds using the quadratic formula; a=1.6; b=51.2; c=-1664
I got a positive solution of:
s = 20 km/hr for the 1st biker
then
20+32 = 52 km/hr for the 2nd biker
:
:
Check this by finding the actual travel time of each
1st biker: 52/20 = 2.6 hrs
2nd biker: 52/52 - 1.0 hrs
---------------------------
time difference: 1.6 hrs which is 1 hr 36 min