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Speedboat

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by charlie33 » Sat Jul 11, 2009 7:44 am
Sorry might seem like an easy one but I don't have the answers so I thought I would make sure:

A speedboat takes 3 hours longer to go 60 miles up a river than to return. If the boat cruises at 15 miles per hour in still water, what is the rate of the current?

A. 4 mi/h
B. 3 mi/h
C. 6 mi/h
D. 5 mi/h
E. 2 mi/h
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by truplayer256 » Sat Jul 11, 2009 7:59 am
Let x= Rate of the current.

When the boat goes downstream, it goes along with the current, so its speed is 15+x miles per hour.

When the boat goes upstream, it goes against the current, so its speed is 15-x miles per hour.

Let's form an equation from the given information above.

60/15-x=60/15+x+3

(60)/15-x-(60)/15+x=3

equation simplifies to:

x^2+40x-225=0

x=5 miles per hour.
D
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by shibal » Sat Jul 11, 2009 8:14 am
why do you add 3 to the rate?
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by gmat740 » Sat Jul 11, 2009 8:56 am
why do you add 3 to the rate?
This is not rate, it is time.
And the reason why??
A speedboat takes 3 hours longer to go 60 miles up a river than to return.
Hope this helps
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by shibal » Sat Jul 11, 2009 10:57 am
hummmm i still don't get how to set up the equations....
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by truplayer256 » Sat Jul 11, 2009 3:30 pm
The reason why I added 3 to one of the algebraic expressions is answered by gmat740.

Now, Shibal, you don't get how to set up the equations?

Let me explain the process to you.

If you understood all the stuff I said about the boat going upstream and downstream and how its speed changes, then setting up the equation is simple.

The total distance of the stream is 60 miles, correct?

If you put the speed that the boat went both upstream and downstream under 60, you'll get the number of hours it took the boat to travel that distance. Take a look at this:

60 Miles/ X+15 Miles per hour---> Units: Miles/Miles per hour= Miles*hours/Miles. The miles cancel and you're left with hours.

The question tells you that the boat took 3 hours longer to travel up the stream than down the stream. So, as I said before, in order to find out how many hours it took the boat to go up the stream, you do:

60 Miles/ X-15 Miles/hour

We know that the algebraic expression above is 3 hours longer than the time it took the boat to travel downstream or 60 Miles/ X+15 Miles per hour, so:

60 Miles/ X-15 Miles/hr= 60 Miles/ X+15 Miles/her +3

I hope that helps.
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by mehravikas » Sat Jul 11, 2009 10:40 pm
You can plug in answer choices if there is a problem in setting up equations.
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by shibal » Sun Jul 12, 2009 10:31 am
truplayer256, thanks very much for your patience and explanation....

i don't know what i'm doing wrong.... when i do the exercise and set up the equations i get the following:

3= (60/15+x)+(60/15-x).... solving for x i get -3x^2 -1125......
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by GMATGuruNY » Wed May 29, 2013 7:24 am
charlie33 wrote:Sorry might seem like an easy one but I don't have the answers so I thought I would make sure:

A speedboat takes 3 hours longer to go 60 miles up a river than to return. If the boat cruises at 15 miles per hour in still water, what is the rate of the current?

A. 2mi/h
B. 3 mi/h
C. 4 mi/h
D. 5 mi/h
E. 6 mi/h
On the GMAT, the answer choices would be in ascending or descending order.
I've reordered them from least to greatest.

The boat's rate = 15 miles per hour.
Let c = the current's rate.
We can plug in the answers, which represent the value of c.

When the boat travels WITH THE CURRENT, the current INCREASES the rate of travel, so its rate must be ADDED to the boat's rate.
The answer choices imply the following options for 15+c:
17, 18, 19, 20, 21.
Only the value in red divides evenly into the distance -- 60 miles -- implying that the correct answer almost certainly is D.

Answer choice D: c = 5 miles per hour.
Time with the current = d/(15+c) = 60/(15+5) = 3 hours.
When the boat travels AGAINST the current, the current DECREASES the rate of travel, so the rate of the current must be SUBTRACTED from the boat's rate:
Time against the current = d/(15-c) = 60/(15-5) = 6 hours.
Difference between the times = 6-3 = 3 hours.
Success!

The correct answer is D.
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by Brent@GMATPrepNow » Wed May 29, 2013 8:05 am
shibal wrote:hummmm i still don't get how to set up the equations....
I like to begin with a "word equation".

A speedboat takes 3 hours longer to go 60 miles up a river than to return.
So, (time to go upstream) = (time to go downstream) + 3

Given:
- total distance is 60 miles
- boat's speed in still water = 15 mph

Let x = speed of river's current
So, the boat's downstream speed = 15 + x
The boat's upstream speed = 15 - x

We'll use the fact that time = distance/speed to rewrite our word equation.
time to go upstream) = (time to go downstream) + 3
60/(15-x) = 60/(15+x) + 3

IMPORTANT: At this point, you can either solve the equation for x, OR you can plug in the answer choices. Plugging in for x is probably faster.

For "fun," let's solve for x
60/(15-x) = 60/(15+x) + 3
Rewrite as: 60/(15-x) = 60/(15+x) + (45+3x)/(15+x)
Simplify: 60/(15-x) = (3x+105)/(15+x)
Cross multiply: 60(15+x) = (15-x)(3x+105)
Simplify: 900 + 60x = -3x^2 - 60x + 1575
Rearrange: 3x^2 + 120x - 675 = 0
Divide both sides by 3: x^2 + 40x - 225 = 0
Factor: (x - 5)(x + 45) = 0

So, x = 5 or -45
-45 is out, so the answer must be D

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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