SOLUTION: You drove 56 miles on a service call for your company and it took 10 minutes longer than the return trip when you drove an average of 8 miles per hour faster. What was your average
Algebra ->
Customizable Word Problem Solvers
-> Travel
-> SOLUTION: You drove 56 miles on a service call for your company and it took 10 minutes longer than the return trip when you drove an average of 8 miles per hour faster. What was your average
Log On
Question 241510: You drove 56 miles on a service call for your company and it took 10 minutes longer than the return trip when you drove an average of 8 miles per hour faster. What was your average speed on the return trip? Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! You drove 56 miles on a service call for your company and it took 10 minutes longer than the return trip when you drove an average of 8 miles per hour faster. What was your average speed on the return trip?
---------
Service Call DATA:
distance = 56 miles ; rate = x mph ; time = d/r = 56/x hrs
================================
Return trip DATA:
distance = 56 miles ; rate = (x+8) mph ; time = 56/(x+8) hrs
------------------------------------
Equation:
go time - return time = 1/6 hr
56/x - 56/(x+8) = 1/6
1/x - 1/(x+8) = 1/336
Multiply thru by 336x(x+8) to get:
336(x+8) - 336x = x(x+8)
336*8 = x^2+8x
x^2 + 8x - 2688 = 0
(x-48)(x+56) = 0
Positive solution:
x = 48 mph (rate on the service call)
x+8 = 56 mph (rate on the return)
==============================================
Cheers,
Stan H.
(