<PRE><B>Question 277914</B>
Driving his car at 50 mi/hr, jose can cover a certain distance in 20 min less than when he drives at 45 mi/hr . Find that distance.
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Faster Rate DATA:
rate = 50 mph ; distance = x miles : time = d/r = x/50 hrs
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Slower Rate DATA:
rate = 45 mph ; distance = x miles ; time = d/r = x/45 hrs
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Equation:
slower time - faster time = 1/3 hr
x/45 - x/50 = 1/3
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3*50x - 3*45x = 45*50
15x = 45*50
x = 150 miles
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Cheers,
Stan H.</PRE>