SOLUTION: Michael averaged running a certain distance 1 mph faster than David ran the same distance. If it took Michael 36 minutes to run the distance and it took David 4 minutes longer to r

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Question 1048956: Michael averaged running a certain distance 1 mph faster than David ran the same distance. If it took Michael 36 minutes to run the distance and it took David 4 minutes longer to run the same distance, how far did the boys run?
I don't even know how to begin to set up this problem into an equation, other than distance=rateXtime.
Found 2 solutions by josgarithmetic, MathTherapy:
Answer by josgarithmetic(39625) (Show Source):
You can put this solution on YOUR website!
According to RT=D for the basic constant travel rates rule, here is an the tabulation for the problem description.

Speed unit of MILES per HOUR, time unit as HOURS, and distance as MILES: SPEED TIME DISTANCE Michael r+1 36/60 d David r 4/60+36/60 d

If you understand that, then simplify the time values.

Speed unit of MILES per HOUR, time unit as HOURS, and distance as MILES: SPEED TIME DISTANCE Michael r+1 d David r d

Next is formulate two equations and solve.
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Answer by MathTherapy(10555) (Show Source):
You can put this solution on YOUR website!
Michael averaged running a certain distance 1 mph faster than David ran the same distance. If it took Michael 36 minutes to run the distance and it took David 4 minutes longer to run the same distance, how far did the boys run?
I don't even know how to begin to set up this problem into an equation, other than distance=rateXtime.

Let distance be D
Then Michael�s speed = , or , or , or
David�s speed = , or , or , or , or
We then get:
10D = 9D + 6 ------- Multiplying by LCD, 6
10D � 9D = 6
D, or distance =
Nothing complex....it's that easy! You want DISTANCE, so solve for nothing else but DISTANCE!