A boat traveled upstream a distance of 90 miles at an average speed of (v-3) miles per hour and then traveled the same distance downstream at an average of (v+3) miles per hour. if the trip upstream took half an hour longer than the trip downstream, how many hours did it take the boat to travel downstream?

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09-05-2018
Mathematics

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A boat traveled upstream a distance of 90 miles at an average speed of (v-3) miles per hour and then traveled the same distance downstream at an average of (v+3) miles per hour. if the trip upstream took half an hour longer than the trip downstream, how many hours did it take the boat to travel downstream?

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elsiedollawesom elsiedollawesom · Answer 1 · 2018-05-09T14:16:52+02:00

Trip upstream took 90v−390v−3 hours and trip downstream took 90v+390v+3 hours. Also given that the difference in times was 1212 hours --> 90v−3−90v+3=1290v−3−90v+3=12;

90v−3−90v+3=1290v−3−90v+3=12 --> 90(v+3)−90(v−3)v2−9=1290(v+3)−90(v−3)v2−9=12 --> 90∗6v2−9=1290∗6v2−9=12 --> v2=90∗6∗2+9v2=90∗6∗2+9 --> v2=9∗(10∗6∗2+1)v2=9∗(10∗6∗2+1) --> v2=9∗121v2=9∗121 --> v=3∗11=33v=3∗11=33;

Trip downstream took 90v+3=9033+3=2.590v+3=9033+3=2.5 hours.