document.write( "Question 111165: A boat can go 20 mi against a current in the same time that it can go 60 mi with the current. The current is 4 mph. Find the speed of the boat in still water. \n" ); document.write( "
Algebra.Com's Answer #81096 by dolly(163)![]() ![]() ![]() You can put this solution on YOUR website! \n" ); document.write( "Let the speed of the boat be x mph \n" ); document.write( "speed of the current = 4mph\r \n" ); document.write( "\n" ); document.write( "so speed against the current = (x-4)mph \n" ); document.write( "Distance traveled against the current = 20 miles \n" ); document.write( "So time taken = distance/speed \n" ); document.write( " = 20/(x-4)\r \n" ); document.write( "\n" ); document.write( "Now speed with the current = (x+4)mph \n" ); document.write( "Distance traveled with the current = 60 miles \n" ); document.write( "So time taken = distance/speed \n" ); document.write( " = 60/(x+4)\r \n" ); document.write( "\n" ); document.write( "As the time taken is the same,\r \n" ); document.write( "\n" ); document.write( " 20/(x-4) = 60/(x+4)\r \n" ); document.write( "\n" ); document.write( "Multiplying both the sides by (x-4)(x+4),\r \n" ); document.write( "\n" ); document.write( "20(x+4) = 60(x-4)\r \n" ); document.write( "\n" ); document.write( "==> 20x + 80 = 60x - 240\r \n" ); document.write( "\n" ); document.write( "==> 240 + 80 = 60x - 20x [ collecting the like terms to one side]\r \n" ); document.write( "\n" ); document.write( "==> 320 = 40x \n" ); document.write( "==> 320/40 = x \n" ); document.write( "==> 8 = x\r \n" ); document.write( "\n" ); document.write( "Thus the speed of the boat in still water = 8mph\r \n" ); document.write( "\n" ); document.write( "Best of luck!!! \n" ); document.write( " \n" ); document.write( " |