college algebra word problems

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i am stuck on the following problem. here is a pic of the problem

http://imgur.com/TSsIqiI

any help would be appreciated

Dec 3rd, 2014

They key to solving this problem is to realize that the speed of the boat is REDUCED by 5 km/hr when it's traveling UPSTREAM against the current. And it is INCREASED by 5 km/hr when it is traveling downstream with the current.

Then, if we let S = the speed of the boat in still water (which is the same as saying ZERO current)

then S - 5 = the speed of the boat going upstream

and  S+5 = the speed of the boat going downstream

The other thing we need to remember is that the TIME = Distance / Speed

So, the time to travel upstream is 195 / (S - 5)

and the time to travel downstream is 195 / (S+5)

Next, according to the problem, the difference between those two times is 50 hours.

So, 195 / (S-5) - 195 / (S+5) = 50

Now we can solve for S.  To subtract fractions we need a COMMON DENOMINATOR

195(S+5) / (S-5)(S+5) - 195(S - 5) / (S-5)(S+5) = 50

195[(S+5) - (S-5)] = 50(S-5)(S+5)

195(10) = 50(S^2 - 25)

1950 = 50(S^2 - 25)

S^2 = 1950/50 + 25 = 39+25 = 64

S = sqrt(64) = 8 km/hr

Dec 3rd, 2014

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