Question 408154
a new schedule for a train requires it to go 270 miles in 26 minutes less time than before.
 To do this it must average 4 mph faster.
 What must its new average speed be?
:
Let s = the new average speed
then
(s-4) = the old average speed
:
Change 26 min to {{{26/60}}} = {{{13/30}}} hrs
:
write a time equation: time = dist/speed
Old time - new time = 26 min
{{{270/(s-4)}}} - {{{270/s}}} = {{{13/30}}}
:
Multiply equation by 30s(s-4) to get rid of the denominators, results:
30s(270) - 30(s-4)(270) = 13s(s-4)
8100s - 8100(s-4) = 13s^2 - 52s
8100s - 8100s + 32400 = 13s^2 - 52s
Combine to form a quadratic equation
0 = 13s^2 - 52s - 32400 
Solve this using the quadratic formula a=13, b=-52, c=-32400
:
The positive solution: s = 51.963 ~ 52 mph is the new average speed
;
:
Check solution in decimals
270/48 - 270/52 = .4327 * 60 ~ 26.0 min