Question 170619
Jeff leaves his house on his bicycle at 8:30 am, and averages 5 mph. At 9:00 am his wife leaves, following the same path and averages 6 mph. At what time will joan catch up with Jeff?
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Jeff DATA:
rate = 4 mph ; time = x hours; distance = rt = 4x miles 
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Wife DATA:
rate = 6 mph ; time = (x-(1/2)) hrs; distance = rt = 6(x-(1/2)) miles 
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EQUATION:
distance = distance
4x = 6(x-(1/2))
4x = 6x -3
2x = 3
x = 3/2 hours
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Answer: wife will catch up to Jeff at 8:30 + 1 1/2 hrs = 10:00
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Cheers,
stan H.