- anonymous

A boat can go 25 mph in still water. It takes as long to go 160 miles upstream as it does to go downstream 240 miles. How fast is the current?

- chestercat

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- anonymous

Formula please!

- anonymous

\[T=
\frac{D}{R}=\frac{160}{25+x}=\frac{240}{25-x}\]

- anonymous

if x = current than rate going up stream is 25-x and rate going down stream is 25+x

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## More answers

- anonymous

since
\[T=\frac{D}{R}
\] and these times are the same set them equal. then solve as a ratio
\[160(25-x)=240(25+x)\]

- anonymous

divide both sides by 80 to make life easier
\[2(25-x)=3(25+x)\]
\[20-2x=75+3x\]
\[5x=-50\]
\[x=-10\] so i probably messed up and x = 10

- anonymous

oh yes of course. it is
\[\frac{160}{25-x}=\frac{240}{25+x}\] i had it backwards sorry

- anonymous

smaller distance over slower rate = bigger distance over bigger rate. my mistake. lets solve this one

- anonymous

\[160(25+x)=240(25-x)\]
\[2(25+x)=3(25-x)\]
\[20+2x=75-3x\]
\[5x=50\]
\[x=10\]

- anonymous

now we check. if the current is 10 mpr then the rate going up stream is 25-10=15mph . \[\frac{160}{15}=10\tfrac{2}{3}\] hours

- anonymous

isn't it 5x=25?

- anonymous

yea i screwed up twice. must be early

- anonymous

\[2(25+x)=3(25-x)\]
\[50+2x=75-3x\]
\[5x=25\]
\[x=5\]

- anonymous

sorry. good eye!

- anonymous

can you do one more for me?

- anonymous

whew.
\[\frac{160}{20}=\frac{240}{30}=8\] so trip is 8 hours in either case

- anonymous

I just need a formula

- anonymous

sure

- anonymous

yeah since clearly i am unable to solve a linear equation

- anonymous

A boat can go 150 miles downstream in the same time it can go 100 miles upstream. The speed of the current is 6 miles per hour.

- anonymous

Find the speed of the boat in still water.

- anonymous

same exact eqution as before, only this time the variable will be speed, not current

- anonymous

again
\[T=\frac{D}{R}\]

- anonymous

if you put x = speed of the boat then you know that the speed going downstream is x + 6 whereas the speed going up is x - 6

- anonymous

so time going down is
\[T=\frac{150}{x+6}\] and time going up is
\[T=\frac{100}{x-6}\]

- anonymous

set them equal and solve.
\[\frac{160}{x+6}=\frac{100}{x-6}\]

- anonymous

you can do this better than i can

- anonymous

typo
\[\frac{150}{x+6}=\frac{100}{x-6}\]

- anonymous

Thanks!!!!!

- anonymous

welcome

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