Answer:
time in going is 4 hours, and time in returning is 4.5 hours.
Step-by-step explanation:
Let the distance be d.
Let the time taken in going to New York be 't'.
Since returning time exceeded by 30 min (i.e. 05 hr), so time returning will be t+0.5.
From Home to New York,
d=s*t, so d=45*t
From New York to Home,
d=40 * (t+0.5)
Since distance (d) from home to new york & vice versa is same, we have-
45*t=40 (t+0.5)
9t=8t+4
so, t=4 hours
Hence time in going is 4 hours, and time in returning is 4.5 hours.
4 hours (Time to go)
4.5 hours (Time to return)
Let the distance between his home and New York is d
Let the time going is t1 and the time returning is t2
As per the question, return time exceeds the going time by 30 min
t2 - t1 = 30/60 ( minutes to hours )
t2 - t1 = 1/2 Eqn⇒(1)
Rate of going is 45 mph
t1 = d/45 Eqn⇒(2)
Rate of returning is 40 mph
t2 = d/40 Eqn⇒(3)
Substituting (2) and (3) into (1)
d/40 - d/45 = 1/2
Solving the equation we get d= 180 m
Now putting the value of d in above equations
t1 = 180/45 = 4 hours (Time to go)
t2 = 180/40 = 4.5 hours (Time to return)
