Question

PLEASE HELP!!! 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 1

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.

Answer 2

Answer:

4 hours (Time to go)

4.5 hours (Time to return)

Step-by-step explanation:

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)