document.write( "Question 198981: The distance between two cities is ninety miles, and a woman drives from one city to the other at a rate of forty-five mph. At what rate must she return if the total travel time is three hours and forty minutes? \n" ); document.write( "
Algebra.Com's Answer #149523 by ankor@dixie-net.com(22740)\"\" \"About 
You can put this solution on YOUR website!
The distance between two cities is ninety miles, and a woman drives from one
\n" ); document.write( " city to the other at a rate of forty-five mph.
\n" ); document.write( " At what rate must she return if the total travel time is three hours and forty minutes?
\n" ); document.write( ":
\n" ); document.write( "Let s = speed require to return in 3 hr 40 min
\n" ); document.write( ":
\n" ); document.write( "Write a time equation: Time = Dist/speed
\n" ); document.write( ":
\n" ); document.write( "Change time to hrs: 3 & 2/3 hrs = 11/3 hrs
\n" ); document.write( ":
\n" ); document.write( "To time + return time = 3 hr 40 min
\n" ); document.write( "\"90%2F45\" + \"90%2Fs\" = \"11%2F3\"
\n" ); document.write( "Multiply equation by 45s
\n" ); document.write( "45s*\"90%2F45\" + 45s*\"90%2Fs\" = 45s*\"11%2F3\"
\n" ); document.write( "Cancel denominators, results
\n" ); document.write( "90s + 45(90) = 15s(11)
\n" ); document.write( "90s + 4050 = 165s
\n" ); document.write( "4050 = 165s - 90s
\n" ); document.write( "4050 = 75s
\n" ); document.write( "s = \"4050%2F75\"
\n" ); document.write( "s = 54 mph required on the return trip
\n" ); document.write( ";
\n" ); document.write( ":
\n" ); document.write( "Check solution
\n" ); document.write( "90/45 + 90/54 =
\n" ); document.write( "2 + 1.67 = 3.67 hrs ~ 3 hr 40 min
\n" ); document.write( "
\n" ); document.write( "
\n" );