SOLUTION: An actress has to be on set, which is 21 miles away at 7am. If she wakes up at 5:30 am and does her make up for 48 minutes, how fast must she drive to get on to work on time? * ste

Algebra ->  Rate-of-work-word-problems -> SOLUTION: An actress has to be on set, which is 21 miles away at 7am. If she wakes up at 5:30 am and does her make up for 48 minutes, how fast must she drive to get on to work on time? * ste      Log On


   



Question 982009: An actress has to be on set, which is 21 miles away at 7am. If she wakes up at 5:30 am and does her make up for 48 minutes, how fast must she drive to get on to work on time? * step by step method please! THX!
Found 2 solutions by jim_thompson5910, MathLover1:
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
Add 5:30 to 0:48

5:30
0:48
-----
6:18

So she does her makeup, and when that is finished, she takes off at 6:18 am.

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The time difference from 6:18 am to 7:00 am is 42 minutes, which is 42/60 = 0.7 hrs

She must drive 21 miles for 0.7 hours

d = 21
r = unknown
t = 0.7

Use the equation d = r*t to solve for r


d = r*t

21 = r*0.7

21/0.7 = r*0.7/0.7 divide both sides by 0.7 to isolate r

30 = r

r = 30

She must drive at least 30 mph to make it on time. I put "at least" in there to mean that if she drives faster than 30 mph, then she'll have a bit of time to spare.

Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

distance is:
d=21mil
time:
from 5%3A30am+ to 7am is 1.5h
subtract the time she need for make up
1.5h-48mi=1.5h-0.8h=0.7h
so, to drive on set she has+t=0.7h
since d=s%2At, we have
s=d%2Ft.........substitute 21mil for d and 0.7h for t
s=21mil%2F0.7h
s=210mil%2F7h
s=30%28mil%2Fh%29
so, she must drive 30%28mil%2Fh%29 to get on to work on time