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Gmat Prep - distance/rate problem

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by dpatwa » Mon Oct 15, 2007 9:33 pm
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 speed of (v+3) miles per hour. If the trip upstream took half an hour longer than the trip down stream, how many hours did it take the boat to travel downstream?

A. 2.5
B. 2.4
C. 2.3
D. 2.2
E. 2.1

[spoiler]OA: A[/spoiler]
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by Gunjan99 » Mon Oct 15, 2007 10:00 pm
I think Answer is A 2.5

Here is the solution:

UP DOWN
Sp: v-3 v+3
Di: 90 90
Ti: t1 t2
now, t1 = t2 +0.5
and T = D/S
t1 = 90/(v-3)
t2 = 90/(v+3)
solving above give the value of v and than t.
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by ri2007 » Thu Oct 18, 2007 6:47 pm
Hi
Can some one explain this one pls.
thanks
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by samirpandeyit62 » Thu Oct 18, 2007 10:08 pm
Hi RI,

let t be the time take downstream

so t+.5 is the time taken upstream

hence t =90/(v+3) & t+0.5 = 90/(v-3)

here if we isolate v to the LHS of both the expressions, then we can construct a simple quadratic eqn, solving which we can get the only postive root as 2.5 which is required.
Regards
Samir
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by ri2007 » Fri Oct 19, 2007 6:18 am
thanks Samir, I was making some simple mistakes while solving the equations, got the basic ones correct then made stupid mistakes later..
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equation solution?

by tutonaranjo » Sat Feb 09, 2008 8:43 am
hi,
can someone post the equation solution on this post?
cannot solve the equation.
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im pretty confused

by resilient » Tue Feb 12, 2008 12:58 am
I dont understand what the lhs is? ALso I am not following in th emath for some reason. CAn we simplify this more or even backsolve it?
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EQ solution

by its_me07 » Tue Feb 12, 2008 11:08 am
can someone pls post the eq solution.
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by davidforsberg » Tue Feb 12, 2008 3:11 pm
can someone please post the step by step, I'm doing equations and everything and it still does not work out.
T minus 17 hours
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by siddarthd2919 » Tue Feb 12, 2008 7:17 pm
the answer is 2.5 :D
steps:
distance d=90
upstream speed=v-3
downstream speed = v+3
speed = distance/timetaken

90/(x+0.5)=v-3------1

90/x= v+3--------2

from 1 we get

v = 3+ 90/(x+0.5)

substitute v in 2 we get the quadratic equation

6x^2+3x-45=0----------3

solving 3 we get the roots as 2.5 and -3.........since time cant be negative the answer is 2.5 :idea:

correct me if i am wrong
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by Stuart@KaplanGMAT » Wed Feb 13, 2008 10:56 am
The actual algebra on this question is WAY above normal GMAT levels - where is the question from?
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by vinviper1 » Tue May 27, 2008 10:12 am
This is from practice test 1 of GMAT prep
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Re: Gmat Prep - distance/rate problem

by El Cucu » Sun Apr 19, 2009 5:53 am
dpatwa wrote: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 speed of (v+3) miles per hour. If the trip upstream took half an hour longer than the trip down stream, how many hours did it take the boat to travel downstream?

A. 2.5
B. 2.4
C. 2.3
D. 2.2
E. 2.1

[spoiler]OA: A[/spoiler]
Could someone pls. write all the steps of the equation ?
Even plugging in the correct answer (2,5) in T I can't reach a reasonable equation.

(v-3) * (3) = (v+3) * 2,5

V=33 then 36T=45T !!!!

May be some Math Guru who can think this problem without making so many calculations? Tksvm!!
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by 4seasoncentre » Sun Apr 19, 2009 6:48 am
SPEED = DIST/TIME

If Upstream trip = 2.5 hours, than downstream = 2.0 hours

For upstream
speed = distance/time
(v-3)=(90/2.5)
(v-3)=36
v=39

For downstream
speed = distance/time
(v+3)=(90/2)
(v+3)=45
V=39... and bob's your uncle!
This solution works, so stop here

but just to prove this works, let try solution B

If upstream was 2.4 hours, than downstream = 3.9 hours

For upstream
speed = distance/time
(v-3)=(90/2.4)
(v-3)=37.5
v=40.5

For downstream
speed = distance/time
(v+3)=(90/3.9)
(v+3)=23ish
V=20ish
does not work!
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by dumb.doofus » Sun Apr 19, 2009 10:12 am
Just another explanation.. I am hoping that it is simpler to understand..

We need to just equate the times taken for upstream and downstream considering that downstream it takes 0.5 hours lesser and use the speed, distance and time formula.

Time = Distance/speed

Upstream time taken = 90/(v-3) --- (1)

Downstream time taken = 90/(v+3) --- (2)

According to question, downstream is 0.5 hours lesser than upstream, this implies

90/(v-3) = 90/(v+3) + 0.5 -- (1)

6/(v^2 - 9) = 1/180

V^2 = 1089

v = 33 -- (3)

So apply (3) in (2), we get

90/(33+3) = 2.5 Hours and that's the answer..
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