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Let the distance be D , speed of car be C , and speed of walking be W .

The first equation is : (D/C)+(D/W)=6 hr 45 min.
Second equation - (D/C)+(D/C)= 4 hr 45 min
We need to find 2D/W. The answer is 8 hr 45 min. ie option A.
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Ron walks to a viewpoint and returns to the starting point by his car and thus takes a total time of 6 hours 45 minutes. He would have gained 2 hours by driving both ways. How long would it have taken for him to walk both ways.

A. 8 h 45 min
B. 7 h 45 min
C. 6 h 45 min
D. 5 h 30 min
E. None of these

W + D = 6 : 45 ------- (I)

D + D = 4 : 45

D = 2 : 22.5 -------(II)

Using (I) and (II)

W = 6 : 45 - 2 : 22.5

W = 4 : 22.5

W + W = (4 : 22.5) * 2

8 : 45 (A)
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Bunuel
Ron walks to a viewpoint and returns to the starting point by his car and thus takes a total time of 6 hours 45 minutes. He would have gained 2 hours by driving both ways. How long would it have taken for him to walk both ways.

A. 8 h 45 min
B. 7 h 45 min
C. 6 h 45 min
D. 5 h 30 min
E. None of these

We let the walk rate = w and the car rate = c. We let the distance = d. We also convert 6 hours 45 minutes to 6 3/4 hours.

We can create the equations:

d/c + d/w = 6 3/4

and

d/c + d/c = 4 3/4


If we simply the second equation, we have:

2d/c = 19/4

d/c = 19/8

Substituting d/c = 19/8 in d/c + d/w = 6 3/4, we have:

19/8 + d/w = 6 3/4

d/w = 6 3/4 - 19/8 = 54/8 - 19/8 = 35/8

Notice that d/w is the one-way walking time, so the round-trip walking time would be:

2d/w = 2 x 35/8 = 35/4 = 8 3/4 hours = 8 hours 45 minutes

Answer: A
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Quick approach - if driving saves 2 hours, not driving but walking will use 2 hours more. Hence 6h 45 mins +2 hours is 8 hrs 45 mins
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