word problem: speed of boat in still water

Princezz3286

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A river flows at 3mph. If a boat can travel 16 miles downstream in the same amount of time it can go 10 miles upstream, find the speed of the boat in still water.

This is how I set it up.....

R T D
3 + x t 16
3 - x t 10

(3 + x)t = 16
(3 - x)t = 10

3t + xt = 16
3t - xt = 10
-------------------
6t = 26
---------------
6

t = 4 1/3
Is this correct? Can it go 4 1/3 mph? I guess it could be realistic, but why the 1/3? Can someone please explain?
 

stapel

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What does "t" stand for? :wink:

Eliz.
 

Princezz3286

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Re:

stapel said:
What does "t" stand for?
t = time sorry, then I have to plug that into the equation and I get more crazy numbers...
(3 + x)4 1/3 = 16
117/3 + 4 1/3x = 48/3
- 48/3 - 48/3
--------------------------------
69/3 +4 1/3x = 0 69/3 = 23
- 23 -23
13/3x = -69/3 13/3 * 3/69 = 39/207

ok now I have really weird numbers........ what did I do? haha HELP!
 

jwpaine

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You are making this too hard.

Really, we don't need to include the data presented for going upstream in out calculation, as the speed of the boat does not change. However, we can use the upstream data to check our answer.

Think about standing on a moving sidewalk, where the sidewalk is moving 3mph and you are moving in the same direction, going Smph. You end up going 16m in some time (which we don't care about).
Now that you are at the end of your sidewalk, walk back the other direction going the same Smph while the sidewalk is moving you back at 3mph, you end up going 10m in the same time.

What is S?
 

Princezz3286

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s = 13........? that's it?
 

jwpaine

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Down: (S + 3) = 16 ---> S = (16 - 3) =13
Up: (S - 3) = 10 ---> S = (10 + 3) =13

:D
 

Princezz3286

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jwpaine said:
Down: (S + 3) = 16 ---> S = (16 - 3) ---> 13
Up: (S - 3) = 10 ---> S = (10 + 3) --->13
Cool! thanks! :lol:
 

kasie-tutor

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Dear Princezz3286,

Let x = the speed of the boat in still water

Assuming that the speed of the boat in still water is greater than the speed of the river,
\(\displaystyle x+3\) is the net rate of the boat going downstream and \(\displaystyle x-3\) is the rate going upstream. (If you don't assume \(\displaystyle x\) is bigger than \(\displaystyle 3\), then your boat would make no progress--it would travel backwards.)

In your D = R x T table, generally you fill out two of the columns with info from the word problem, and in the remaining column you use the formula d=rt on the two filled-in columns, and put the result in the third column. For example, if you filled in the Rate and Time coulmns, you would multiply those columns to get the Distance column.

Your main trouble is the Time column in your table. Use the formula d=rt to get the Time column after you've filled-in the Distance and Rate.

So, re-think your expressions for the rates, and instead of putting t's in your Time column, put fractions which are your distance/rate expressions.

Does that help enough? Final hint: the rate will be a whole number (not mixed number).

8-) Kasie
 

Princezz3286

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kasie-tutor said:
Let x = the speed of the boat in still water....
But if x + 3 = 10, because it is down stream..... you get 7,
if x - 3 = 16, you get 19.... that is not the same number.... Now I am really confused.........
 

galactus

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You have them backwards.

If the boat travels with the river, wouldn't it be rather intuitive that it will travel it's rate plus the river's rate?.

If it is traveling upstream, we would subtract the rates of the boat and river.

Now, since d=rt, we have t=d/r.

The problem says it travels downstram in the same amount of time as it travels upstream.

So, we can equate it this way.

\(\displaystyle \frac{16}{r+3}=\frac{10}{r-3}\)

Now, solve for r. Cross-multiply:

\(\displaystyle 16(r-3)=10(r+3)\)

\(\displaystyle 16r-48=10r+30\)

\(\displaystyle 6r=78\)

\(\displaystyle r=13\)

Just as was given before.
 

jwpaine

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kasie-tutor said:
\(\displaystyle x+3\) is the net rate of the boat going upstream and \(\displaystyle x-3\) is the rate going downstream. (If you don't assume \(\displaystyle x\) is bigger than \(\displaystyle 3\), then your boat would make no progress--it would travel backwards.)
That is incorrect :oops:

Think about my sidewalk analogy.... if you were going with the sidewalk, than why would the sidewalk speed be subtracted from your speed? :shock:
 

Princezz3286

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galactus said:
You have them backwards.....

\(\displaystyle 6r=78\)

\(\displaystyle r=13\)

Just as was given before.
ok that is whaqt I thought but if you look at how Kasie gave it that is what it said so I was worried that I still did not under stand........ Thanks Everyone!
 

kasie-tutor

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\(\displaystyle x+3\) is the net rate of the boat going upstream and \(\displaystyle x-3\) is the rate going downstream.
CORRECTION: \(\displaystyle x+3\) is the speed going downstream and \(\displaystyle x-3\) is the speed going upstream. Speeds add going down the river (work together); subtract going up the river (work against eachother).

:shock: :oops: :shock: :!:

Woops. That's a bit of dyxlesia. I reversed the words "upstream" and "downstream". I meant the reverse. I edited my earlier post to say the right thing.

:mrgreen: Kasie
 
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