Mr. Sawyer drove his car from his home to New York at the rate of 45mph and returned over the same road at the rate of 40mph. If his time returning exceeded his time going and his time returning?

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.

T1 = 4

T2 = 4.5

Step-by-step 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