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?

Question

Huynh

Mathematics

13 July, 03:35

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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Beatrice · Answer 1

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

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

Trip downstream took 90 v+3 = 90 33+3 = 2.5 90 v+3 = 90 33+3 = 2.5 hours.