SOLUTION: A bakery delivery truck leaves the Bakery at 5:00am each morning on its 140-mike route. One day the drive gets a late start and does not leave the bakery until 5:30am. To finish he

Algebra ->  Rate-of-work-word-problems -> SOLUTION: A bakery delivery truck leaves the Bakery at 5:00am each morning on its 140-mike route. One day the drive gets a late start and does not leave the bakery until 5:30am. To finish he      Log On


   

Question 1129852: A bakery delivery truck leaves the Bakery at 5:00am each morning on its 140-mike route. One day the drive gets a late start and does not leave the bakery until 5:30am. To finish her route on time the driver drives 5 miles per hour faster than Usual. At what speed does she usually drive?
Found 2 solutions by josgarithmetic, ikleyn:
Answer by josgarithmetic(39616) About Me  (Show Source):
You can put this solution on YOUR website!
r, normal speed

Usually, r*t=140, t=140%2Fr.

Late Morning, want to arrive AT the same time, so missing half hour.
%28r%2B5%29%2Ax=140
x=140%2F%28r%2B5%29


140%2Fr-140%2F%28r%2B5%29=1%2F2------solve this equation for r.

Answer by ikleyn(52776) About Me  (Show Source):
You can put this solution on YOUR website!
.
Let r be the usual speed.

Then you have this "time equation"


140%2Fr - 140%2F%28r%2B5%29 = 0.5    of an hour.


280*(r+5) - 280*r = r*(r+5)


r^2 + 5r - 1400 = 0


(r+40)*(r-35) = 0


The only meaningful is the positive root  r = 35 miles per hour.    ANSWER


Check.   140%2F35 - 140%2F%2835%2B5%29 = 4 - 3.5 = 0.5  of an hour.   ! Correct !

Solved.