SOLUTION: How do I find the answer to the following math word problem? "Mila took a trip and drove the first 20 miles at a speed of 50 mph. If she wants her average speed for the entire trip

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Question 601435: How do I find the answer to the following math word problem? "Mila took a trip and drove the first 20 miles at a speed of 50 mph. If she wants her average speed for the entire trip to be 40 mph, how fast should she drive for the last 70 miles of her trip?"

Found 2 solutions by htmentor, solver91311:
Answer by htmentor(1343) About Me  (Show Source):
You can put this solution on YOUR website!
How do I find the answer to the following math word problem? "Mila took a trip and drove the first 20 miles at a speed of 50 mph. If she wants her average speed for the entire trip to be 40 mph, how fast should she drive for the last 70 miles of her trip?"
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For the first part of the trip, the travel time = 20 mi/50 mi/h = 2/5 hr
The average speed for the entire trip is the total distance divided by the total time:
v = (d1+d2)/(t1+t2) = 40 mph = (20+70)/(2/5+t2)
Solve for t2:
90/(2/5 + t2) = 40
t2 = 9/4 - 2/5 = 37/20 hr
So the speed for the last 70 miles of the trip is 70 mi/(37/20 hr) = 37.838 mph
Check:
90/(2/5+37/20) = 90/2.25 = 40

Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!


The first part of the trip is 20 miles and the last part of the trip is 70 miles so the whole trip is 90 miles. Time equals distance divided by rate, so if you want to average 40 mph for 90 miles it will take you hours to complete the trip. You have already used hours for the first part of the trip. Calculate the amount of time you have used, subtract that amount from 2.25 hours to discover the amount of time remaining. Divide that amount of time into 70 to see how fast you need to go to cover 70 miles in the remaining amount of time. Since all of the given values are given to a whole number precision, you should round your answer to the nearest whole mile per hour; don't do any rounding until AFTER the final calculation.

John

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