Question 453381
On her way TO her friend Linda drove with:
r - speed
t - time it took her to her friend
d - 300mi, distance she passed
:
rt=d -> rt=300 -> r=300/t
:
On her way back Linda drove with:
r+20  - 20mi/h faster than on her way to friend
t-5/4 - she used 5/4 of the hour less time to travel
d     - distance was the same 300 miles
:
(r+20)(t-5/4)=300
rt-5r/4+20t-25=300, multiply by 4
4rt-5r+80t-100=1200, substitute rt with 300 and r with 300/t
4*300-5*300/t+80t-100=1200, subtract 1200 from both sides
1200-1500/t+80t-100-1200=0, combine numbers
80t-1500/t-100=0, multiply by t
80t^2-100t-1500=0, divide by 20
4t^2-5t-75=0, solve using quadratic formula
:
{{{t = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
{{{t = (-(-5) +- sqrt( (-5)^2-4*4*(-75) ))/(2*4) }}}
{{{t = (5 +- sqrt( 25+1200))/8 }}}
{{{t = (5 +- 35)/8}}}
{{{t = (5+35)/8 = 5}}} or {{{t=(5-35)/8 = 3.75}}}
:
Only acceptable solution is positive number so t=5 hours on her way down.
:
Her speed on the way down was 300/5 = 60mi/h.
Her speed on the way back was 60+20 = 80mi/h.