document.write( "Question 109824: A boat can go 10 miles against the current in the same it can go in 30 miles with the current. The current flows at 4 mph. Find the speed of the boat in no current. \n" ); document.write( "
Algebra.Com's Answer #80097 by ptaylor(2198)![]() ![]() You can put this solution on YOUR website! Let r=rate (speed) of the boat with no current \n" ); document.write( "Now we know that: \n" ); document.write( "(r-4)=rate against the current \n" ); document.write( "(r+4)=rate with the current\r \n" ); document.write( "\n" ); document.write( "distance(d)=rate(r) times time(t) or d=rt; t=d/r and r=d/t \r \n" ); document.write( "\n" ); document.write( "Time required to go 10 miles against the current=10/(r-4) \n" ); document.write( "Time to go 30 miles with the current=30/(r+4)\r \n" ); document.write( "\n" ); document.write( "----And we are told that the above two times are equal. So our equation to solve:\r \n" ); document.write( "\n" ); document.write( "10/(r-4)=30/(r+4) multiply both sides by (r-4)(r+4) or cross-multiply and we get:\r \n" ); document.write( "\n" ); document.write( "10(r+4)=30(r-4) get rid of parens (distributive law) \n" ); document.write( "10r+40=30r-120 subtract 10r from and add 120 to both sides\r \n" ); document.write( "\n" ); document.write( "10r-10r+40+120=30r-10r-120+120 collect like terms\r \n" ); document.write( "\n" ); document.write( "160=20r divide both sides by 20 \n" ); document.write( "r=8 mph-------------------------------boat's rate of speed with no current\r \n" ); document.write( "\n" ); document.write( "CK \n" ); document.write( "10/(8-4)=30/(8+4) \n" ); document.write( "10/4=30/12 \n" ); document.write( "10/4=10/4\r \n" ); document.write( "\n" ); document.write( "Hope this helps---ptaylor \n" ); document.write( " |