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boat traveled

by shahab03 » Wed Sep 10, 2008 7:59 pm
A boat traveled upstream a distance of 90 miles at an average speed of (V-3) miles per hour and then travelled the same distance downstream at an average speed of (V+3) miles per hour.If the trip upstream half an hour longer than downstream, how many hours did it take the boat to travel downstream?

(*) 2.5
(*) 2.4
(*) 2.3
(*) 2.2
(*) 2.1


Upstream: V-3 miles per hour
Downstream : V+3 miles per hour
Distance: 90 miles
Hence,
Time_Upstream = 90/(V-3)
Time_Downstream = 90/(V+3)
Now, it says that

Time_Upstream = 1/2 + Time_Downstream

i.e
90/(V-3) = 1/2 + 90/(V+3)

i am stuck at this point... would someone solve this for me.... thanks
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by parallel_chase » Thu Sep 11, 2008 1:33 am
90/(V-3) = 1/2 + 90/(V+3)


90/(V-3) - 90/(V+3) = 1/2

[90(V+3) - 90(V-3)] / (V+3) (V-3) = 1/2

[90V + 270 - 90V +270] = [V^2 - 9]/2

540 *2 = V^2 - 9


1080 + 9 = V^2

1089 = V^2

V = 3*11 = 33

Downstream = 90/(v+3) = 90/36 = 10/4 = 2.5


Hope this helps.
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by willshu » Thu Oct 15, 2009 5:46 am
How did you know that the square root of 1089 is 33? I got to that point and it took me a while to figure that out. Is there a quicker way?
parallel_chase wrote:90/(V-3) = 1/2 + 90/(V+3)


90/(V-3) - 90/(V+3) = 1/2

[90(V+3) - 90(V-3)] / (V+3) (V-3) = 1/2

[90V + 270 - 90V +270] = [V^2 - 9]/2

540 *2 = V^2 - 9


1080 + 9 = V^2

1089 = V^2

V = 3*11 = 33

Downstream = 90/(v+3) = 90/36 = 10/4 = 2.5


Hope this helps.
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by aakar » Thu Oct 15, 2009 10:10 am
Can't you do it the following way:

Upstream R=v-3 T=t+.5 D = 90
Downstream R=v+3 T=t D = 90

1.) (v-3)(t+.5)= 90
2.) (v+3)(t) = 90

Set both equations equal
(v-3)(t+.5) = (v+3)(t)
multiplying through and solving for v in terms of t will get you:
v-3 = 12t

plug this into the first equation and you will get:
12t(t+.5) = 90
12t^2 + 6t = 90 you can take out 6 to make everything come out in whole numbers:
2t^2 + t - 15 = 0
(2t -5)(t+3) = 0
t = 5/2 or -3

5/2 = 2.5

Thoughts?
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by Stuart@KaplanGMAT » Thu Oct 15, 2009 11:41 am
willshu wrote:How did you know that the square root of 1089 is 33? I got to that point and it took me a while to figure that out. Is there a quicker way?
Based on his "3*11", it looks like he broke 1089 down into primes.

However, we should be able to do it by quick estimation.

Start by working with an easy number. The simplest perfect square close to 1089 is 900, which is 30*30.

We also know that the units digit of 1089 is "9". How do we get a 9? Well,

1*1 = 1,
2*2 = 4,
3*3 = 9...

there we go, so 1089 is almost certainly 33*33 (the next number that gives us a 9 is 7*7, and 37*37 is going to be way too big).
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Re: boat traveled

by Stuart@KaplanGMAT » Thu Oct 15, 2009 11:48 am
shahab03 wrote:A boat traveled upstream a distance of 90 miles at an average speed of (V-3) miles per hour and then travelled the same distance downstream at an average speed of (V+3) miles per hour.If the trip upstream half an hour longer than downstream, how many hours did it take the boat to travel downstream?

(*) 2.5
(*) 2.4
(*) 2.3
(*) 2.2
(*) 2.1

The various algebraic solutions posted all work. We can also solve through a combination of common sense and backsolving.

First, common sense. We know that r = d/t. With a distance of 90, what time will give us a rate that's not a crazy complicated number? 90/2.5 = 180/5 is an integer, but none of the other answers produces an integer for rate. So, 2.5 certainly looks like the best guess.

If we want to, we can backsolve to make sure:

if t for the return trip is 2.5 hours, then t for trip upstream is 3 hours.

upstream rate = up distance/up time = 90/3 = 30
downstream rate = down distance/down time = 90/2.5 = 180/5 = 36

According to the original info, the rates should be 6 apart (v-3 vs v+3). Is 36-30 = 6? Yes! Therefore, 2.5 is the correct answer.
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Re: boat traveled

by crackgmat007 » Fri Oct 16, 2009 8:36 pm
Stuart Kovinsky wrote: First, common sense. We know that r = d/t. With a distance of 90, what time will give us a rate that's not a crazy complicated number? 90/2.5 = 180/5 is an integer, but none of the other answers produces an integer for rate.
Stuart, is the bolded text (above) a general rule or one of the strategies for guessing?
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Re: boat traveled

by Stuart@KaplanGMAT » Sun Oct 18, 2009 12:05 am
crackgmat007 wrote:
Stuart Kovinsky wrote: First, common sense. We know that r = d/t. With a distance of 90, what time will give us a rate that's not a crazy complicated number? 90/2.5 = 180/5 is an integer, but none of the other answers produces an integer for rate.
Stuart, is the bolded text (above) a general rule or one of the strategies for guessing?
It's not a firm rule, but the numbers on the GMAT usually work out nicely - we're not being tested on our ability to do weird calculations.
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by crackgmat007 » Sun Oct 18, 2009 12:47 pm
Good point.
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by rahul.s » Sun Jan 24, 2010 2:50 am
parallel_chase wrote:90/(V-3) = 1/2 + 90/(V+3)


90/(V-3) - 90/(V+3) = 1/2

[90(V+3) - 90(V-3)] / (V+3) (V-3) = 1/2

[90V + 270 - 90V +270] = [V^2 - 9]/2

540 *2 = V^2 - 9


1080 + 9 = V^2

1089 = V^2

V = 3*11 = 33

Downstream = 90/(v+3) = 90/36 = 10/4 = 2.5


Hope this helps.
awesome stuff! thanks
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by chaq305 » Mon Aug 20, 2012 9:06 pm
Can someone give me an estimate of how hard this question is on a 6-51 scale?
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by GMATGuruNY » Mon Aug 20, 2012 9:38 pm
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by [email protected] » Sun Nov 18, 2012 11:31 am
Hi Stuart,

I almost always solve the "same distance" problems like this:

We have two equations and one distance, so I set the two equations equal

(v-3)*(t+1/2) = (v+3)*t

If I solve this I still have two unknowns. I dont understand it cause usually I only have one unknown left. Could you please help me?

Thanks!
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by shreerajp99 » Fri Dec 21, 2012 10:31 pm
Just open the brackets,ul b left with v-3=12t.Put this back in the original equation that states 12t(t+1/2) = 90.Solve the quadratic and ul get t=2.5
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