Complete question :
Mr. Sawyer drove his car from his home to New York at the rate of 45 mph and returned over the same road at the rate of 40 mph. If his time returning exceeded his time going by 30 min, find his time going and his time returning.
Answer:
T1 = 4
T2 = 4.5
Explanation:
Difference in time = 0.5 hour
Time = distance / speed
Let distance = d
Time, T1 = d / 45
Time, T2 = d / 40
d/40 - d /45 = 1/2
(45d - 40d) / 100 = 2
45d - 40d = 900
5d = 900
d = 180
T1 = 180 /45 = 4 hours
T2 = 180 / 40 = 4.5 hours