SOLUTION: Two cyclists drive a distance of 15 kilometres. One of the cyclists drives 10 km h faster than the other one and arrives at their destination 20 minutes earlier. What are the spe

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Question 290060: Two cyclists drive a distance of 15 kilometres. One of the cyclists
drives 10 km h faster than the other one and arrives at their destination
20 minutes earlier. What are the speeds of the cyclists?
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
Two cyclists drive a distance of 15 kilometres.
One of the cyclists drives 10 km h faster than the other one and arrives at
their destination 20 minutes earlier.
What are the speeds of the cyclists?
:
Let s = speed of the slow cyclist
then
(s+10) = speed of the faster one
:
Change 20 min to 1%2F3 hr
:
Write a time equation: Time = dist/speed
:
slow time - fast time = 1%2F3 hr
15%2Fs - 15%2F%28%28s%2B10%29%29 = 1%2F3
Multiply by 3s(s+10), results:
3(15)(s-10) - 3(15)s = s(s+10)
:
45(s+10) - 45s = s(s+10)
:
45s + 450 - 45s = s^2 + 10s
:
0 = s^2 + 10s - 450
:
Use the quadratic formula to find s:
x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
In this equation, x=s; a=1, b=10; c=-450
s+=+%28-10+%2B-+sqrt%2810%5E2-4%2A1%2A-450+%29%29%2F%282%2A1%29+
:
s+=+%28-10+%2B-+sqrt%28100+-%28-1800%29+%29%29%2F2+
;
s+=+%28-10+%2B-+sqrt%28100+%2B+1800%29%29%2F2+
:
s+=+%28-10+%2B-+sqrt%281900+%29%29%2F2+
Positive solution
s+=+%28-10+%2B+43.589%29%2F2+
s = 33.589%2F2
s = 16.7945 ~ 16.8km/h is the slow cyclist
and
26.8 km/h is the fast cyclist
;
:
Check solution by finding the times
Slow: 15/16.8 = .89 hrs
Fast: 15/26.8 = .56 hrs
------------------------
difference is: .33 hrs which is about 20 min