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Bunuel
Rob walks from A to B at \(3 \frac{1}{2}\) kmph, then rests there for 45 minutes and finally rides back at \(7 \frac{1}{2}\) kmph. What is the distance from A to B, if the total time spent by Rob is 6 hrs 37 min.

A. 28 km
B. 14 km
C. 8 km
D. 7 km
E. 6 km


Are You Up For the Challenge: 700 Level Questions
Given: Rob walks from A to B at \(3 \frac{1}{2}\) kmph, then rests there for 45 minutes and finally rides back at \(7 \frac{1}{2}\) kmph.

Asked: What is the distance from A to B, if the total time spent by Rob is 6 hrs 37 min.

Let the distance from A to B be x km

Total time = x/(7/2) + 3/4 + x/(15/2) = 6 37/60
2x/7 + 2x/15 = 6 37/60 - 3/4 = 6 37/60 - 45/60 = 5 52/60 = 5 13/15 = 88/15
x (2/7 + 2/15) = 88/15
2x (22/7*15) = 88/15
x = 7*2 = 14 km

IMO B­
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I used a slightly less orthodox way to solve this problem a bit faster.

Quote:
Rob walks from A to B at 3.5 kmph, then rests there for 45 minutes and finally rides back at 7.5 kmph. What is the distance from A to B, if the total time spent by Rob is 6 hrs 37 min.
So the trip took 5h52 of traveling or 352 minutes.
Now we can see that 7.5 is slightly more than double 3.5, so slightly less than 1/3 of the traveling time would have been spent on the return journey.
352/3 is a bit less than 120minutes or 2h. 2h at 7.5kph is 15km, so the answer must be 14km, answer B.

Had the answer choices contained 13km this methodology wouldn't have worked, so make sure to strategies correctly.
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Distance of the Round Trip is in Equal Parts. So you can find the Average Speed by taking the Harmonic Mean of the 2 Speeds that the Equal Distance Parts were traveled at.

Avg. Speed = b = [ 2 * (7/2) * (15/2) ] / (7/2 + 15/2)

Avg. Speed = (7 * 15) / (22)

Total Time Spent Traveling = 6 hours and 37 minutes - 45 minutes rest = 5 hours and 52 minutes

Total Time Spent Traveling = 88/15 hours (in fractional form)


Avg. Speed = Total Distance / Total Time traveling

Let d = distance from A to B ---- then Total Distance = 2 * d

Avg. Speed = (7 * 15) / (22) = (2 * d) / (88/15)

Cross Multiplying

(7 * 15 * 88) / (15) = 22 * 2 * d

Cancel the 15 on the L.H.S. and Divide Each Side by 22

7 * 4 = 2 * d

28 = 2 * d

d = 14 km

A.C. -B-

there has to be a way without so many nasty calculations
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Bunuel
Rob walks from A to B at \(3 \frac{1}{2}\) kmph, then rests there for 45 minutes and finally rides back at \(7 \frac{1}{2}\) kmph. What is the distance from A to B, if the total time spent by Rob is 6 hrs 37 min.

A. 28 km
B. 14 km
C. 8 km
D. 7 km
E. 6 km


Solution:

We can let the distance between A and B = d. Excluding the resting time, the total travel time is 6 hrs 37 min - 45 min = 5 hrs 52 min. Therefore, we can create the equation:

d/3.5 + d/7.5 = 5 + 52/60

Multiplying the equation by 1575 (which is 3.5 x 7.5 x 60), we have:

450d + 210d = 7875 + 1365

660d = 9240

d = 14

Answer: B
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Another method:

Calling the 2 parts of the trip as Part 1 and Part 2, the Distance traveled is Constant for both scenarios.

This means that Speed will be Inversely Proportional to the Time spent traveling.

In this case, if the Speed from Part 1 to Part 2 were to increase by (a)/(b) ———the Time traveling would Decrease by (a) / (b + a)

Speed during Part 1 = 7/2

Speed during Part 2 = 15/2

From Part 1 to Part 2, Speed Increases by:

(15/2 - 7/2) / (7/2) = 8/7

If Speed Increases by 8/7, then Time traveling from Part 1 to Part 2 DECREASES by - 8/15

Total time traveling = 6 hr 37 min - 45 min resting = 5 hr 52 min = 5 and 13/15 hours = 88/15 hours


Let Time traveling during Part 1 = A

Let Time traveling during Part 2 = B

A + B = 88/15

Since Time spent traveling Decreases from Part 1 to Part 2 by (8/15):

B = A - (8/15)*A

Substituting for B:

A + A - (8/15)*A = 88/15 hours

(22/15)*A = 88/15

A = 4 hours spent traveling during Part 1

Speed during Part 1 = 7/2

Distance = (7/2) * 4 = 14 miles

Answer B

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