SOLUTION: At 8.00 AM a doctor leaves Sanford on a northbound train. Her husband, noticing that the doctor forgot her briefcase, boards another northbound train at 10.00 AM, and travelling

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Question 873960: At 8.00 AM a doctor leaves Sanford on a northbound train. Her husband, noticing that the doctor forgot her briefcase, boards another northbound train at 10.00 AM, and travelling 18 mph faster than the first train, the second train overtakes the first train at 4.00 PM . find the speeds of the trains, and the distance they traveled ?
Found 2 solutions by mananth, ghostrecog:
Answer by mananth(16946) (Show Source):
You can put this solution on YOUR website!
speed of doctor's train be x
leaves at 8.00 meets husbad at 4.00 time = 8 hours
husband speed = (x+18)
husband time = 6 hours
distance is same
6(x+18)=8x
6x+108=8x
2x=108
x=54 mph
doctor's train 54 mph
Husband's train 54+18 = 72 mph
distance = 54 *8 = 432 miles

Answer by ghostrecog(4) (Show Source):
You can put this solution on YOUR website!
Let x the be the speed of the first train.
Then as per above question , x+18 be the speed of second train.
http://www.algebra.com/cgi-bin/display-illustration.mpl?tutor=ghostrecog&illustration_name=Solution1.jpg&size=thumbnail
As per the above table.
First train starts at 8 am and meets second train at 4 pm. It means it has taken 8 hrs to to reach that point.
similarly the second train meets the first train at 6hrs.
As the distance traveled by both the trains are equal. Then the equation from the above table as follows.
8*x=6*(x+18)
8x=6x+108
2x=108
x=54
Therefore First train traveled with the speed of 54 mph.
Then the second train x+18=54+18=72 mph.
Now we got the speed of first train and second train. Need to find the distance traveled by the both train.
Distance=speed X Time
Distance=54X8
=432 miles