SOLUTION: On a round trip of 700km in each direction,Mr.anderson averaged 7km/h more returning than he did going.If the return trip took 5hours less than the trip going, find the time he spe

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Question 137639: On a round trip of 700km in each direction,Mr.anderson averaged 7km/h more returning than he did going.If the return trip took 5hours less than the trip going, find the time he spent traveling in each direction
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
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On a round trip of 700km in each direction,Mr.anderson averaged 7km/h more returning than he did going.If the return trip took 5hours less than the trip going, find the time he spent traveling in each direction
:
From the text we can assume the trip out was 350 km and return trip was 350 km
:
Let s = speed out
then
(s+7) = return speed
;
Write a time equation Time = Distance/speed
It says it took 5 hrs less time to return
:
350%2F%28%28s%2B7%29%29 - 5 = 350%2Fs
Multiply equation by s(x+7) to get rid of the denominators:
s(s+7)*350%2F%28%28s%2B7%29%29 - 5(s(s+7)) = s(s+7)*350%2Fs
Leaving us with:
350s + 5(s^2 + 7s) = 350(s+7)
:
350s + 5s^2 + 35s = 350s + 2450
:
Subtract 350s from both sides and arrange as a quadratic equation on the left
5s^2 + 35s - 2450 = 0
:
Simplify, divide by 5
s^2 + 7s - 490 = 0
:
We have to use the quadratic formula:
x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
a = 1; b = 7; c = -490; solve for s;
s+=+%28-7+%2B-+sqrt%287%5E2-+4+%2A+1%2A+-490+%29%29%2F%282%2A1%29+
s+=+%28-7+%2B-+sqrt%2849+%2B+1960%29%29%2F%282%29+
Do the math here and you should get a positive solution:
s = 18.91 km/h is outbound speed
:
18.91 + 7 = 25.91 km/hr return speed
:
Find the time required for each direction:
350%2F18.91 = 18.51 hrs outbound
and
350%2F25.91 = 13.51 hrs return; note the 5 hr difference confirms our solutions