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MI3: Senior | Next Rank: 100 Posts; Posts: 83; Joined: Sat Feb 26, 2011 11:22 am; Thanked: 1 times; Followed by:1 members

PS Question

by MI3 » Wed Jun 01, 2011 2:30 am

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.

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Source: Beat The GMAT — Problem Solving | Wed Jun 01, 2011 2:30 am

cans: Legendary Member; Posts: 1309; Joined: Mon Apr 04, 2011 5:34 am; Location: India; Thanked: 310 times; Followed by:123 members; GMAT Score:750

by cans » Wed Jun 01, 2011 2:42 am

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

Last edited by cans on Wed Jun 01, 2011 2:45 am, edited 1 time in total.

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sameerballani: Legendary Member; Posts: 544; Joined: Mon Dec 27, 2010 8:10 am; Thanked: 45 times; Followed by:2 members

by sameerballani » Wed Jun 01, 2011 2:44 am

2.5 hrs.

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manpsingh87: Master | Next Rank: 500 Posts; Posts: 436; Joined: Tue Feb 08, 2011 3:07 am; Thanked: 72 times; Followed by:6 members

by manpsingh87 » Wed Jun 01, 2011 2:57 am

MI3 wrote: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

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