Setting up a math d=rt equation?

Question

I am having trouble understanding how to set up a specific problem. Here it is:

"You drove 56 miles one way on a service call. On the return trip, your average speed was 8 miles per hour greater and the trip took 10 fewer minutes. What was your average speed on the return trip?"

Thanks!

Answer 1

let the time taken for one way = t hours

speed = distance/time = 56/t miles/hr

then speed on return jorney = (56/t ) + 8 miles/hour

= (56 + 8t)/t

= 8(7+t)/t

time taken for return trip = t- 10/60 = t - 1/6 h

= (6t - 1)/6 hours

average speed on return journey = 56/(6t - 1)/6

= (56)(6)/(6t-1)

= 336/(6t-1) miles /hour


336/(6t-1) = 8(7+t)/t

divide by 8

42/(6t-1) = (7+t)/t

crossmultiply

42t = (6t-1)(7+t)

42t = 42t - 7 + 6t^2 - t

6t^2 - t - 7 = 0

6t^2 + 6t - 7t - 7 = 0

6t(t+1) - 7(t+1) = 0

(6t-7)(t+1)=0

6t-7 = 0, ignoring negative value

6t = 7

t = 7/6 hours

= 70 minutes

speed on return journey = 70 - 10 = 60 min = 1 hr

average speed on return journey = 56/1 = 56 miles/hour

Answer 2

x = speed to the servicecall
x + 8 is the speed on the return trip

distance = 56 miles for both directions

distance = speed * time
let t be the time it takes you to travel 56 miles to the phone service

1min = 1/6 hr
it takes you 1/6 hr less to ruturn, so the time is t - 1/6

56 = xt
56 = (x + 8) (t - 1/6)

solve for t
t = 56/x
56 = (x + 8) (56/x - 1/6)
56 = 56 - x/6 + 448/x - 4/3
0 = -x/6 + 448/x - 4/3
0 = -x^2/6 - 4x/3 + 448

use quadratic formula and you'll get x = 48 mi/hr

so the speed on the return trip is 48 + 8 = 56 mi/hr <= answer

hope this helps

Answer 3

Let x = rate going
Then x+8 = rate returning
Time going = 56/x [since t= d/r]
time returning = 56/(x+8)
time going = time returning +10
56/x = 56/(x+1) +10
You should now be able to solve for x.