SOLUTION: Mrs Dang drove her daughter to school at the average speed of 45 miles per hour. she returend home by the same route at the average speed of 30 miles per hour. if the trip took one

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Question 167096: Mrs Dang drove her daughter to school at the average speed of 45 miles per hour. she returend home by the same route at the average speed of 30 miles per hour. if the trip took one half hour, how long did it take to get to school?, how far is the school from their home?
Found 2 solutions by josmiceli, ankor@dixie-net.com:
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
For the trip to school,
(1) d+=+r%5B1%5D%2At%5B1%5D
For the return trip,
(2) d+=+r%5B2%5D%2At%5B2%5D
where d is the distance from home to school
Given:
(3)t%5B1%5D+%2B+t%5B2%5D+=+.5hrs
(3) t%5B2%5D+=+.5+-+t%5B1%5D
r%5B1%5D+=+45 mi/hr
r%5B2%5D+=+30 mi/hr
------------------------
(1) d+=+r%5B1%5D%2At%5B1%5D
(1) d+=+45%2At%5B1%5D
(2) d+=+r%5B2%5D%2At%5B2%5D
(2) d+=+30%2At%5B2%5D
Since d is the same in both equations,
45t%5B1%5D+=+30t%5B2%5D
Substitute (3) for t%5B2%5D
45t%5B1%5D+=+30%2A%28.5+-+t%5B1%5D%29
45t%5B1%5D+=+15+-+30t%5B1%5D
75t%5B1%5D+=+15
t%5B1%5D+=+.2hrs
(3) t%5B2%5D+=+.5+-+t%5B1%5D
t%5B2%5D+=+.5+-+.2
t%5B2%5D+=+.3hrs
It took .2 hrs to get to school, or .2%2A60+=+12 minutes
Now find d
(1) d+=+45%2At%5B1%5D
(1) d+=+45%2A.2
(1) d+=+9mi
The school is 9 mi from home
check:
45t%5B1%5D+=+30t%5B2%5D
45%2A.2+=+30%2A.3
9+=+9
OK

Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
Mrs Dang drove her daughter to school at the average speed of 45 miles per hour.
she returned home by the same route at the average speed of 30 miles per hour.
if the trip took one half hour, how long did it take to get to school?,
how far is the school from their home?
:
Let d = distance to the school
:
Write a time equation: Time = dist%2Fspeed
To school time + return time = half hour
d%2F45 + d%2F30 = .5
Multiply equation by 90 to get rid of the denominators:
90*d%2F45 + 90*d%2F30 = 90(.5)
cancel out the denominators;
2d + 3d = 45
:
5d = 45
d = 45%2F5
d = 9 miles
:
Find the time to get to school (Time = dist/speed)
9/45 = .2 hr or .2(60) = 12 min
:
:
Check solution find the time to return
9/30 = .3 hr or .3(60) = 18 min; (they add up to a half hr)