document.write( "Question 1059270: An airplane has enough fuel for a 5 hour flight. It goes 225 mile per hour with the wind on the way out and on the return trip goes 180 mph. How far can the plane fly? \n" ); document.write( "

Algebra.Com's Answer #674343 by ankor@dixie-net.com(22740) $\"\"$ $\"About$

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An airplane has enough fuel for a 5 hour flight.
\n" ); document.write( " It goes 225 mile per hour with the wind on the way out and on the return trip goes 180 mph.
\n" ); document.write( " How far can the plane fly?
\n" ); document.write( ":
\n" ); document.write( "let d = the one way distance to the point of no return
\n" ); document.write( ":
\n" ); document.write( "Write a time equation, time = dist/speed
\n" ); document.write( " time out + time back = 5 hrs
\n" ); document.write( " $\"d%2F225\"$ + $\"d%2F180\"$ = 5
\n" ); document.write( "Find the least common multiple of 225 & 180, by prime factoring
\n" ); document.write( "225: 3*3*5*5
\n" ); document.write( "180: 2*2*3*3*5
\n" ); document.write( "LCM: 2*2*3*3*5*5 = 900
\n" ); document.write( "Multiply equation by 900
\n" ); document.write( "900* $\"d%2F225\"$ + 900* $\"d%2F180\"$ = 900*5
\n" ); document.write( "Cancel the denominators
\n" ); document.write( "4d + 5d = 4500
\n" ); document.write( "9d = 4500
\n" ); document.write( "d = 4500/9
\n" ); document.write( "d = 500 mi to the point of no return
\n" ); document.write( ";
\n" ); document.write( ":
\n" ); document.write( "See if this checks out by finding the actual time each way
\n" ); document.write( "500/225 = 2.22 hrs
\n" ); document.write( "500/180 = 2.78 hrs
\n" ); document.write( "----------------------
\n" ); document.write( "total time: 5 hrs
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