SOLUTION: if you want to travel an average of 50/mi per hour on a trip and half of the trip you go 25/mi per hour how fast do you have to travel in the second half

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Question 1050765: if you want to travel an average of 50/mi per hour on a trip and half of the trip you go 25/mi per hour how fast do you have to travel in the second half
Found 4 solutions by josgarithmetic, Alan3354, stanbon, ikleyn:
Answer by josgarithmetic(39618) About Me  (Show Source):
You can put this solution on YOUR website!
How do you measure the trip? Distance?

Try that.

Constant Travel Rate Rule is as D/T=R, D for distance and T for time.
T%2FD=1%2FR
T=D%2FR 
              RATE      TIME     DISTANCE

First          25       d%2F25       d 

Second          r       d%2Fr        d

Total                  d%2F25%2Bd%2Fr    2d

AVERAGE        50


Average rate is given and time sum will give %282d%29%2F%28d%2F25%2Bd%2Fr%29=50.
Two variables in one equation but simplification will change that.

Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
if you want to travel an average of 50/mi per hour on a trip and half of the trip you go 25/mi per hour how fast do you have to travel in the second half
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This is a "trick question."
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Pick a distance, say 100 miles.
If you go 25 mi/hr for 1/2 of it, 50 miles, it takes 2 hours.
To make 100 miles at an average speed of 50 mi/hr takes 2 hours.
Time is up.
It's not possible.

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
if you want to travel an average of 50/mi per hour on a trip and half of the trip you go 25/mi per hour how fast do you have to travel in the second half
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2nd half DATA:
rate = 25 mph ; distance = x miles ; time = x/25 hrs
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Final Results::
rate = 50 mph ; distance = 2x miles ; time = (2x)/50 hrs = x/25 hrs
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2nd half DATA:
distance = x miles ; time = (x/25)-(x/25) = 0 ; rate = x/0 = infinity
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After the 1st half at 25 mph there is zero time for the
2nd half trip if the average rate is to be 50 mph.
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Cheers,
Stan H.

Answer by ikleyn(52797) About Me  (Show Source):
You can put this solution on YOUR website!
.
I agree with Alan.

Also ignore josgarithmetic's post (which is between "wrong", "misleading", "useless" and "incomplete").