SOLUTION: Solve the problem. Tom Whig traveled 220 miles east of St Louis. For most of the trip he averaged 60 MPH, but for one period of time he was slowed to 10mph due to major accident.

Question 644193: Solve the problem.
Tom Whig traveled 220 miles east of St Louis. For most of the trip he averaged 60 MPH, but for one period of time he was slowed to 10mph due to major accident. If the total time of the travel was 7 hours, how many miles did he drive at the reduced speed?
Ans: I know I'm probably wrong but I subtracted 60-10=50mph he drove at reduced speed?
Found 2 solutions by Alan3354, stanbon:
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
Tom Whig traveled 220 miles east of St Louis. For most of the trip he averaged 60 MPH, but for one period of time he was slowed to 10mph due to major accident. If the total time of the travel was 7 hours, how many miles did he drive at the reduced speed?
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Says he drove at 10 mi/hr, not 50.
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d = r*t
t = time going 60, 7-t = time going 10 mi/hr
220 = 60t + 10(7-t)
220 = 50t + 70
t = 3 hrs @ 60 mi/hr
--> 4 hours at 10 mi/hr
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
Tom Whig traveled 220 miles east of St Louis. For most of the trip he averaged 60 MPH, but for one period of time he was slowed to 10mph due to major accident. If the total time of the travel was 7 hours, how many miles did he drive at the reduced speed?
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Faster segment DATA:
rate = 60 mph; time = t hrs ; distance = 60t miles
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Slower segment DATA:
rate = 10 mph ; time = 7-t hrs; distance = 10(7-t) miles
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Equation:
distance + distance = 220 miles
60t + 10(7-t) = 220
60t + 70 - 10t = 220
50t = 150
t = 3 hrs (time at faster speed)
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10(7-t) = 10(4) = 40 miles (distance at reduced speed)
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Cheers,
Stan H.
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