document.write( "Question 332548: A turboprop plane flying with the wind flew 1,400 mi in 5 h. Flying against the wind, the plane required 7 h to travel the same distance. Find the rate of the wind and the rate of the plane in calm air. \n" ); document.write( "
Algebra.Com's Answer #238331 by galactus(183)![]() ![]() You can put this solution on YOUR website! With the wind, we add the two rates. Against the wind, we subtract them.\r \n" ); document.write( "\n" ); document.write( "Let rp=rate of plane and rw = the rate of thr wind.\r \n" ); document.write( "\n" ); document.write( "With the wind:\r \n" ); document.write( "\n" ); document.write( "Since d=rt, we have 1400=5(rp+rw)\r \n" ); document.write( "\n" ); document.write( "Against the wind:\r \n" ); document.write( "\n" ); document.write( "we have 1400=7(rp-rw)\r \n" ); document.write( "\n" ); document.write( "These reduce to \r \n" ); document.write( "\n" ); document.write( "rp+rw=280 \n" ); document.write( "rp-rw=200\r \n" ); document.write( "\n" ); document.write( "Now, from the first equation we have rp=280-rw\r \n" ); document.write( "\n" ); document.write( "Sub into the second equation:\r \n" ); document.write( "\n" ); document.write( "280-rw-rw=200\r \n" ); document.write( "\n" ); document.write( "280-2rw=200\r \n" ); document.write( "\n" ); document.write( "rw=40=rate of wind.\r \n" ); document.write( "\n" ); document.write( "That means the rate of the plane, rp, is 240.\r \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |