document.write( "Question 1192602: In going over 750 kms, it takes a passenger plane 2 hour less time to fly with a tailwind of 50 kph than against a headwind travelling at the same speed. How fast was the passenger plane flying at each way of the course? \n" ); document.write( "
Algebra.Com's Answer #824488 by greenestamps(13200)![]() ![]() You can put this solution on YOUR website! \n" ); document.write( "With the wording exactly as shown, the problem can't be answered, because we don't know the exact distance. (The problem says the plane went OVER 750km; it doesn't say it went 750km.) \n" ); document.write( "Assuming the distance is in fact exactly 750km.... \n" ); document.write( "The plane's speed one direction is its speed in still air plus the wind speed; in the other direction it is its speed in still air minus the wind speed. Since the wind speed is 50km/h, the difference in the two speeds is 100km/h. \n" ); document.write( "If a formal algebraic solution is not required, the problem can be solved quickly by playing with numbers. 150*5=750, and 250*3=750; so 5 hours at 150km/h and 3 hours at 250km/h satisfy the given conditions. \n" ); document.write( "For a solution using formal algebra, I would let the two times in hours be x and x+2, so that the two speeds are 750/x and 750/(x+2). Then I would write and solve an equation that says the difference in speeds at the two times is 100km/h -- literally, \"the speed for the shorter time is 100km/h more than the speed for the longer time\": \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "The negative value for time of course makes no sense; so x=3 and x+2=5. \n" ); document.write( "The two speeds are then 750/3=250 and 750/5=150. \n" ); document.write( "ANSWERS: 250km/h with the wind; 150km/h against the wind \n" ); document.write( " \n" ); document.write( " |