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Mathematics 11 Online
OpenStudy (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?

OpenStudy (anonymous):

Formula please!

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (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

OpenStudy (anonymous):

isn't it 5x=25?

OpenStudy (anonymous):

yea i screwed up twice. must be early

OpenStudy (anonymous):

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

OpenStudy (anonymous):

sorry. good eye!

OpenStudy (anonymous):

can you do one more for me?

OpenStudy (anonymous):

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

OpenStudy (anonymous):

I just need a formula

OpenStudy (anonymous):

sure

OpenStudy (anonymous):

yeah since clearly i am unable to solve a linear equation

OpenStudy (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.

OpenStudy (anonymous):

Find the speed of the boat in still water.

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

you can do this better than i can

OpenStudy (anonymous):

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

OpenStudy (anonymous):

Thanks!!!!!

OpenStudy (anonymous):

welcome

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