BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

Gmat Prep - distance/rate problem

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
Senior | Next Rank: 100 Posts
Posts: 31
Joined: Tue Jul 31, 2007 11:17 am
Location: Denver, CO

Gmat Prep - distance/rate problem

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]
Top
Source: — Problem Solving |
Junior | Next Rank: 30 Posts
Posts: 10
Joined: Mon Jun 11, 2007 2:12 am
Location: Mumbai, India
Followed by:1 members

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.
Top
Master | Next Rank: 500 Posts
Posts: 258
Joined: Mon Aug 27, 2007 12:43 pm
Thanked: 15 times

by ri2007 » Thu Oct 18, 2007 6:47 pm
Hi
Can some one explain this one pls.
thanks
Top
Master | Next Rank: 500 Posts
Posts: 460
Joined: Sun Mar 25, 2007 7:42 am
Thanked: 27 times

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
Top
Master | Next Rank: 500 Posts
Posts: 258
Joined: Mon Aug 27, 2007 12:43 pm
Thanked: 15 times

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..
Top
Senior | Next Rank: 100 Posts
Posts: 42
Joined: Wed Jun 20, 2007 8:45 am
Thanked: 1 times

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.
Top
Legendary Member
Posts: 789
Joined: Sun May 06, 2007 1:25 am
Location: Southern California, USA
Thanked: 15 times
Followed by:6 members

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?
Appetite for 700 and I scraped my plate!
Top
Senior | Next Rank: 100 Posts
Posts: 83
Joined: Mon Oct 22, 2007 4:44 am
Thanked: 1 times

EQ solution

by its_me07 » Tue Feb 12, 2008 11:08 am
can someone pls post the eq solution.
Top
Junior | Next Rank: 30 Posts
Posts: 16
Joined: Tue Feb 05, 2008 10:47 am
Thanked: 3 times

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
Top
Junior | Next Rank: 30 Posts
Posts: 16
Joined: Mon Feb 11, 2008 3:08 am
Location: chennai
Thanked: 6 times
Followed by:1 members

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
Top
User avatar
GMAT Instructor
Posts: 3225
Joined: Tue Jan 08, 2008 2:40 pm
Location: Toronto
Thanked: 1710 times
Followed by:614 members
GMAT Score:800

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?
Image

Stuart Kovinsky | Kaplan GMAT Faculty | Toronto

Kaplan Exclusive: The Official Test Day Experience | Ready to Take a Free Practice Test? | Kaplan/Beat the GMAT Member Discount
BTG100 for $100 off a full course
Top
Master | Next Rank: 500 Posts
Posts: 222
Joined: Tue Mar 25, 2008 3:52 pm
Thanked: 2 times

by vinviper1 » Tue May 27, 2008 10:12 am
This is from practice test 1 of GMAT prep
Top
Master | Next Rank: 500 Posts
Posts: 236
Joined: Fri Nov 28, 2008 9:16 am
Thanked: 9 times

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!!
Top
Master | Next Rank: 500 Posts
Posts: 211
Joined: Thu Apr 09, 2009 12:17 pm
Thanked: 12 times
GMAT Score:680

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!
Top
User avatar
Master | Next Rank: 500 Posts
Posts: 435
Joined: Sat Sep 27, 2008 2:02 pm
Location: San Jose, CA
Thanked: 43 times
Followed by:1 members
GMAT Score:720

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..
One love, one blood, one life. You got to do what you should.
https://dreambigdreamhigh.blocked/
https://gmattoughies.blocked/
Top
Post new topic Post Reply