document.write( "Question 157512: A boat can go 33 mph in still water. It takes as long to go 300 miles upstream as it does to go downstream 360 miles. How fast is the current?\r
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document.write( "Will someone please explain this to me? \n" );
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Algebra.Com's Answer #116079 by ptaylor(2198)![]() ![]() You can put this solution on YOUR website! Sure! \n" ); document.write( "Distance(d) equals Rate(r) times Time(t) or d=rt; r=d/t and t=d/r\r \n" ); document.write( "\n" ); document.write( "Let r=rate (speed ) of the current\r \n" ); document.write( "\n" ); document.write( "Time to travel upstream=300/(33-r) {note: when we go against the current we must subtract the rate of the current) \n" ); document.write( "Time to travel downstream=360/(33+r) (note: downstream we add the rate of the current)\r \n" ); document.write( "\n" ); document.write( "Now we are told that the above two times are the same, so:\r \n" ); document.write( "\n" ); document.write( "300/(33-r)=360/(33+r) multiply each side by (33-r)(33+r) (or cross-multiply) \n" ); document.write( "300(33+r)=360(33-r) divide each side by 60 (I did this just to reduce the size of the numbers that we have to deal with) \n" ); document.write( "5(33+r)=6(33-r) get rid of parens \n" ); document.write( "165+5r=198-6r subtract 165 from and add 6r to both sides \n" ); document.write( "165-165+5r+6r=198-165-6r+6r collect like terms \n" ); document.write( "11r=33 divide each side by 11 \n" ); document.write( "r=3 mph---------------------------------------speed of the current\r \n" ); document.write( "\n" ); document.write( "CK \n" ); document.write( "300/(33-3)=360/(33+3) \n" ); document.write( "300/30=360/36 \n" ); document.write( "10=10\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Hope this helps---ptaylor \n" ); document.write( " |