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
:
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/((s+7))}}} - 5 = {{{350/s}}}
Multiply equation by s(x+7) to get rid of the denominators:
s(s+7)*{{{350/((s+7))}}} - 5(s(s+7)) = s(s+7)*{{{350/s}}}
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 = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
a = 1; b = 7; c = -490; solve for s;
{{{s = (-7 +- sqrt(7^2- 4 * 1* -490 ))/(2*1) }}}
{{{s = (-7 +- sqrt(49 + 1960))/(2) }}}
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:
{{{350/18.91}}} = 18.51 hrs outbound
and
{{{350/25.91}}} = 13.51 hrs return; note the 5 hr difference confirms our solutions