document.write( "Question 903958: A man can row his boat 10 kph in still water. In the same length of time that it took him to row 20 km downstream, he was able to row only half the distance upstream. What was the rate of the current? \n" ); document.write( "
Algebra.Com's Answer #548456 by richwmiller(17219)![]() ![]() You can put this solution on YOUR website! r*t=d \n" ); document.write( "(10-c)*t=10 \n" ); document.write( "10t-c*t=10 \n" ); document.write( "(10+c)*t=20 \n" ); document.write( "10+c*t=20 \n" ); document.write( "10t-ct=10 \n" ); document.write( "add \n" ); document.write( "20t=30 \n" ); document.write( "t=30/20=1.5 \n" ); document.write( "t=1.5 hr \n" ); document.write( "(10+c)*t=20 \n" ); document.write( "10t+c*t=20 \n" ); document.write( "10*1.5+c*1.5=20 \n" ); document.write( "15+c*1.5=20 \n" ); document.write( "c*1.5=5 \n" ); document.write( "c=5/1.5 \n" ); document.write( "c=10/3 \n" ); document.write( "c=3 1/3 kph current \n" ); document.write( "check \n" ); document.write( "10t+c*t=20 \n" ); document.write( "10*1.5+10/3*3/2=20 \n" ); document.write( "15+5=20 \n" ); document.write( "10t-ct=10 \n" ); document.write( "10(3/2)-10/3*3/2=10 \n" ); document.write( "15-5=10 \n" ); document.write( "ok \n" ); document.write( " \n" ); document.write( " |