SOLUTION: Please help me solve this problem: Mr.Tan makes monthly visits to his parents in Baguio, a distance of 240km from Manila. He found out that if he increases the average speed by 10k

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Question 342950: Please help me solve this problem: Mr.Tan makes monthly visits to his parents in Baguio, a distance of 240km from Manila. He found out that if he increases the average speed by 10km per hour, he could save a total of20 minutes for the journey. Find the speed at which Mr.Tan originally traveled.
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
Mr.Tan makes monthly visits to his parents in Baguio, a distance of 240km from Manila.
He found out that if he increases the average speed by 10km per hour, he could
save a total of 20 minutes for the journey.
Find the speed at which Mr.Tan originally traveled.
:
Let s = his original speed
then
(s+10) = his faster speed
:
Change 20 min to 1%2F3 hr
:
Write a time equation, time = dist/speed
:
slow speed time - fast speed time = 1/3 hr
:
240%2Fs - 240%2F%28%28s%2B10%29%29 = 1%2F3
:
Multiply equation by 3s(s+10) to eliminate the denominators, results:
3(s+10)*240 - 3s(240) = s(s+10)
:
240(3s+30) - 720s = s^2 + 10s
:
720s + 7200 - 720s = s^2 + 10x
:
A quadratic equation
s^2 + 10s - 7200 = 0
factors to
(s+90)(s-80) = 0
the positive solution;
s = 80 km/h is his original speed
:
:
Check our solution
Find the actual time of each (90 km/hr is the faster speed)
240/80 = 3 hrs
240/90 = 2.67 hrs
--------------------
differs: .33 hr which ~ 20 min