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        MATHEMATICS FOR AIR CREW TRAINEES      26

From the fact that time is equal to distance divided by rate, the unknown time each ship took can be expressed thus (Rule 2):
     42/r=time in hours for bomber. 
     56/r+90=time in hours for pursuit ship.
     The problem states that the two times are 
     equal, hence 
(Rule 3) 42/r=56/r+90.
This equates hours to hours (Rule 4).
Cross-multiplying the equation gives
                 56r=42r+3,780
or               56r-42r=3.780
hence              14r=3,780
and     r=270 mph, the speed of the bomber  
                                         Answers. 
consequently  r+90=360 mph, the speed of the pursuit ship.
(2) Example: Two airplanes start out toward each other from two towns 570 miles apart. They meet after 2 hours. If the speed of one is 15 mph more than the speed of the other, find the speed of each. 
Solution: Analysis of the problem reveals that the total distance traveled by the two airplanes is 570 miles. This distance may be set up in terms of the unknown rates and known times. 
Let x=mph of slower airplane. Then x+15=mph of other airplane. Since d=rt, set up the distances in terms of the unknown and place them equal to the known distance:
                 2x+2(x+15)=570
                  2x+2x+30=570
                   4x=570-30
                     4x=540
                   x=135 mph             Answers.
                  x+15=150 mph 
(3) Example: An airplane flew from March Field to Moffett Field at the rate of 150 mph. It returned the following day at the rate of 200 mph and required 40 minutes less time to make the trip. How far is it from March Field to Moffett Field?
Solution: The fact used in making the equation in this case is that the distance is the same both ways. The distance will not be solved for directly. 
Let x=number of hours for trip from March Field to Moffett Field.

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