Question 519177
On a 42km go-kart course, A drives 0.4km/hr faster than S, but has engine trouble and stops for 30min.
 A arrives at the finish line 15min after S.
 How fast did each drive and how long did each take to finish the course?
:
Let s = S's speed
then
(s+.4) = A's speed
:
Change 30 min to .5 hrs
Change 15 min to .25 hrs
:
Write a time equation, time = dist/speed
:
A's time = S's time + .25 hr
{{{42/((s+.4))}}} + .5 = {{{42/s}}} + .25
{{{42/((s+.4))}}} = {{{42/s}}} + .25 -.5
{{{42/((s+.4))}}} = {{{42/s}}} - .25
:
multiply by s(s+.4), results:
42s = 42(s+.4) - .25s(s+.4)
42s = 42s + 16.8 - .25s^2 - .1s
:
Combine like terms on the left
.25s^2 + .1s + 42s - 42s - 16.8 = 0
.25s^2 + .1s - 16.8 = 0
;
Get rid of those decimals, multiply by 100
25s^2 + 10s - 1680 = 0
:
simplify, divide by 5
5s^2 + 2s - 336 = 0
:
you can use the quadratic formula to solve, but this will factor to:
(5s+42)(s-8) = 0
:
Positive solution is what we want here
s = 8 km/hr is S's speed
then, obviously, 
8.4 km/h is A's speed
:
:
:
See if that work out
{{{42/8.4}}} + .5 = {{{42/8}}} + .25
5 + .5 = 5.25 + .25
5.5 = 5.5; confirms our solutions