SOLUTION: HE TRAVELLED 300KM TO ATTEND A CONFERANCE .IF THE AVARAGE SPEED OF HIS TRIP HAS BEEN INCREASED BY 10 KM PER HOUR,HE WOULD HAVE REACHED THE CONFERANCE HALL ONE HOUR EARLIER. WHAT WA

Algebra ->  Proofs -> SOLUTION: HE TRAVELLED 300KM TO ATTEND A CONFERANCE .IF THE AVARAGE SPEED OF HIS TRIP HAS BEEN INCREASED BY 10 KM PER HOUR,HE WOULD HAVE REACHED THE CONFERANCE HALL ONE HOUR EARLIER. WHAT WA      Log On


   

Question 700425: HE TRAVELLED 300KM TO ATTEND A CONFERANCE .IF THE AVARAGE SPEED OF HIS TRIP HAS BEEN INCREASED BY 10 KM PER HOUR,HE WOULD HAVE REACHED THE CONFERANCE HALL ONE HOUR EARLIER. WHAT WAS THE AVARAGE SPEED OF HIS TRIP
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
Let s = the av speed of the trip
then
(x+10) = the faster speed
:
Write a time equation, time = dist/speed
:
Actual speed time - faster speed time = 1 hr
300%2Fs - 300%2F%28%28s%2B10%29%29 = 1
multiply by s(s+10), clear the denominators, results:
300(s+10) - 300s = s(s+10)
300s + 3000 - 300s = s^2 + 10s\
A quadratic equation
0 = s^2 + 10s - 3000
Factors to
(s+60)(s-50) = 0
The positive solution
s = 50 km/h is the av speed of the trip
:
:
Check this; find the time at each speed:
300/50 = 6 hr
300/60 = 5 hr
----------------
differs: 1 hr