SOLUTION: I've been working on this problem for hours and still can't seem to find a solution, could you please help if possible?
A person drives to a destination at a rate of thirty mph
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A person drives to a destination at a rate of thirty mph
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Question 947456: I've been working on this problem for hours and still can't seem to find a solution, could you please help if possible?
A person drives to a destination at a rate of thirty mph and returns over the same route at fifty mph. How far is the destination if the time returning is one hour less than the time going? Found 2 solutions by vleith, MathTherapy:Answer by vleith(2983) (Show Source):
You can put this solution on YOUR website! Distance = Speed * Time
You know the distance is the same going both ways. You are given the speed both ways and a relationship of the time.
Let T be the number of hours outbound
DistanceOut = SpeedOut * timeOut
DistanceOut = 30*T
time coming back is 1 hour less than time going out = T-1
DistanceBack = SpeedBack * TimeBack
DistanceBack = 50 * (T-1)
Distance is the same both ways. So
So the time outbound is 5/2 hours = 2.5 Hours
Which says the time back is 1.5 hours.
To check your answer, does 30*2.5 = 50*1.5 ??
Get it now??
Now go out and have some fun this weekend!
You can put this solution on YOUR website!
I've been working on this problem for hours and still can't seem to find a solution, could you please help if possible?
A person drives to a destination at a rate of thirty mph and returns over the same route at fifty mph. How far is the destination if the time returning is one hour less than the time going?
Let the distance be D
Then outbound time =
Inbound time =
Therefore,
5D – 150 = 3D ------ Multiplying by LCD, 150
5D – 3D = 150
2D = 150
D, or distance = , or mph
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