PS Question

[MI3]

Q: 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?

Unfortunately, I don't have the answer choices for the above question and I wasn't able to solve it either, so looking for guidance on this problem.

Please advise, thanks.
M.

[cans]

d=90
let t hours to travel downstream. then t+0.5 hours to travel upstream.
d=s*t =>90=(v+3)*t=(v-3)*(t+0.5)
vt+3t = vt+.5v-3t-1.5 => v-12t-3=0 ------eqn1
90/t=v+3 => v = 90/t - 3
replace v in eqn1
90/t - 3 - 12t - 3 = 0 => -12t^2 - 6t + 90 =0
2t^2 + t - 15=0
t=[-1+-root(1+120) ]/4
as t can't be -ve, t = (-1+11)/4 = 2.5 hours
IMO 2.5 hours

[sameerballani]

2.5 hrs.

[manpsingh87]

Q: 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?

Unfortunately, I don't have the answer choices for the above question and I wasn't able to solve it either, so looking for guidance on this problem.

Please advise, thanks.
M.

90/v-3-90/v+3=1/2;
180(v+3-v+3)=(v+3)(v-3);
180*6=v^2-9;
1080+9=v^2;
v^2=1089;
and 33^2=1089;
so v=33; and v+3=36; hence time taken while traveling downstream= 90/36=2.5h