Intermediate Algebra: A boat goes 450 mi downstream in the

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Helen

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I need help in putting this together so that I can solve it.

A boat goes 450 miles downstream in the same time it can go 400 miles upstream.
The speed of the current is 7 miles per hour. Find the speed of the boat in still water.

Please just start me out.
 
Re: Intermediate Algebra

A boat goes 450 miles downstream in the same time it can go 400 miles upstream. The speed of the current is 7 miles per hour. Find the speed of the boat in still water.
Click to expand...
What can you make use of from the expression V = D/T. ("V" is "velocity", "D" is "distance", and "T" is "time".)

The problem tells you that the times are equal.

You have the distances and the speeds in both directions.

Therefore, T = D/V upstream and downstream.

You have the two distances and the difference of the two speeds.

Can you think of a way of expressing these facts into an equality?

Give it a try and let us know how you make out.
 
Boat

TchrWill,

row #1 row #2 row #3
V = D / T

upstream 450 same

downstream 400 same

current 7 mi. per hour

Am I starting this in the right way? If you give up on me, thank you for trying. Helen
 
Re: Intermediate Algebra: A boat goes 450 mi downstream in t

Helen said:
I need help in putting this together so that I can solve it.

A boat goes 450 miles downstream in the same time it can go 400 miles upstream.
The speed of the current is 7 miles per hour. Find the speed of the boat in still water.

Please just start me out.
Click to expand...

Let r = rate of boat in still water

Going UPSTREAM, the current will slow the boat down. The rate of the boat going upstream will be r - 7 mph. The time it takes the boat to travel 400 miles upstream is 400 / (r - 7)......time = distance / rate.

Going DOWNSTREAM, the current will move the boat along faster than it would travel in still water. The rate of the boat going downstream will be r + 7 mph. The time it takes to go 450 miles downstream is 450 / (r + 7).

Now....the problem says that the two times are the same:

time to travel 400 miles upstream = time to travel 450 miles downstream

400 / (r - 7) = 450 / (r + 7)

Now....can you solve it?
 
Boat problem.

Mrspi,
I can't come up with a chart, so I will try and explain my answer.
400/(r-7) = 450/(r+7)
400-7=393
450+7=457
393 divided by 457=85mph.
Thank you for your help.
 
Not quite correct, Helen.

\(\displaystyle \L\\\frac{450}{r+7}=\frac{400}{r-7}\)

Try again. Either way, it's one fast boat.
 
Re: Boat problem.

Helen said:
400/(r-7) = 450/(r+7)
400-7=393
450+7=457
393 divided by 457=85mph.
Click to expand...
Helen, are you learning on your own, or are you a student in school?
What you've done makes absolutely no sense. If you're needing lessons, try online.
 
Speed of boat in still water.

galactus, You were very nice in your response to me. I thank you for that.
I would like to show you that I have been trying really hard and came up with this.
400(7+r) = 450(7-r)
2800+ 400r=3150-450r
50r=5950
=119 (Speed of boat in still water).

400*7=2800
450*7=3150
400r+450r=850
3150+2800=5950
5950 divided by 50=119

Am I correct on this one? Helen
 
You came up with right answer Helen: and I see you REALLY gave it your best;

400(7+r) = 450(7-r)
2800+ 400r=3150-450r : this does not result in 50r = 5950
50r=5950
=119 (Speed of boat in still water).

However, your (7 - r) should be (r - 7); like this:
450(r - 7) = 400(r + 7)
450r - 3150 = 400r + 2800
50r = 5950
r = 5950 / 50
r = 119
 
Boat problem.

Denis, Thank you for your reply. Appreciated! Helen
 
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