document.write( "Question 61093This question is from textbook Intermediate Algebra
\n" ); document.write( ": The current in ther lazy river moves at a rate of 4mph. Monica'sdinghy motors 6mi upstream in the same time it take to motor 12mi downstream. What is the speed of the dinghy? \n" ); document.write( "
Algebra.Com's Answer #42014 by venugopalramana(3286)\"\" \"About 
You can put this solution on YOUR website!
The current in ther lazy river moves at a rate of 4mph. Monica'sdinghy motors 6mi upstream in the same time it take to motor 12mi downstream. What is the speed of the dinghy?
\n" ); document.write( "SPEED OF WATER = 4 MPH
\n" ); document.write( "LET SPEED OF DINGHY = S MPH
\n" ); document.write( "ITS SPEED UP STREAM = S-4 MPH
\n" ); document.write( "DISTANCE TRAVELED UPSTREAM = 6 M
\n" ); document.write( "TIME TAKEN = 6/(S-4) HRS..................1
\n" ); document.write( "ITS SPEED DOWN STREAM = S+4 MPH
\n" ); document.write( "DISATANCE TRAVELLED DOWN STREAM = 12 M
\n" ); document.write( "THEN TIME TO TRAVEL UPSTREAM = 12/(S+4) HRS...............2
\n" ); document.write( "TIMES ARE EQUAL
\n" ); document.write( "6/(S-4) = 12/(S+4)
\n" ); document.write( "12(S-4)=6(S+4)
\n" ); document.write( "12S-48=6S+24
\n" ); document.write( "6S=72
\n" ); document.write( "S=12 MPH \n" ); document.write( "
\n" );