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.

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.

europa2433

europa2433

step-by-step explanation: