Problem Solving

A man goes from city A to city B situated 60 kms apart by a boat. His onward journey was with the stream while the return journey was an upstream journey. It took him four and half hours to complete the round trip. If the speed of the stream is 10 km/hr, how long did it take him to complete the onward journey?

a. 3 hours
b. 3.5 hours
c. 2.25 hours
d. 1.5 hours

OA: D

According to online answers, in order to solve this problem, we eventually need to come to the understanding that:

V_avg = 2UD/(U+D)

Where:
U = B-10
D = B+10
B = still-water speed of boat

I have spent literally an hour trying to derive this equation. How is this equation true? I can't explain it, but when I use it I get the right answer...

Any help would be GREATLY appreciated. I'm so frustrated at this point, but I don't want to quit without understanding that average speed of travel equation!

Thanks!!

For your 2nd query, Vavg = Total Distance/Total Time

V_avg = 2D/(D/U+D/V) = 2/(1/U+1/V)

=> 2/V_avg=1/U+1/V => V=2UV/(U+V)

kidcorpo wrote:A man goes from city A to city B situated 60 kms apart by a boat. His onward journey was with the stream while the return journey was an upstream journey. It took him four and half hours to complete the round trip. If the speed of the stream is 10 km/hr, how long did it take him to complete the onward journey?

a. 3 hours
b. 3.5 hours
c. 2.25 hours
d. 1.5 hours

OA: D

This is clearly not a GMAT question.

First, there's only 4 answer choices; all GMAT questions have 5 answer choices.

Second, to be solvable we must assume that the speed of the boat (not counting the current) is constant. For all we know the driver went at super speed on the way up and really slowly on the way down. On the GMAT, you'll never have to make those assumptions, the question will clearly include all relevant information.

All that said, and assuming that the speed of the boat was constant, we can solve this question in 15 seconds.

Total trip took 4.5 hours.

The river current was running with the boat on the onward journey and against it on the return journey. Therefore, the onward journey will take less time than the return journed.

Since the onward journey is quicker, it will take less than half the time of the entire trip.

4.5/2 = 2.25. Only 1 answer choice is less than 2.25: choose (D).

For your 2nd query, Vavg = Total Distance/Total Time

V_avg = 2D/(D/U+D/V) = 2/(1/U+1/V)

=> 2/V_avg=1/U+1/V => V=2UV/(U+V)

anksgupta wrote:For your 2nd query, Vavg = Total Distance/Total Time

V_avg = 2D/(D/U+D/V) = 2/(1/U+1/V)

=> 2/V_avg=1/U+1/V => V=2UV/(U+V)

THANK YOU!

Stuart Kovinsky wrote:

kidcorpo wrote:A man goes from city A to city B situated 60 kms apart by a boat. His onward journey was with the stream while the return journey was an upstream journey. It took him four and half hours to complete the round trip. If the speed of the stream is 10 km/hr, how long did it take him to complete the onward journey?

a. 3 hours
b. 3.5 hours
c. 2.25 hours
d. 1.5 hours

OA: D

This is clearly not a GMAT question.

First, there's only 4 answer choices; all GMAT questions have 5 answer choices.

Second, to be solvable we must assume that the speed of the boat (not counting the current) is constant. For all we know the driver went at super speed on the way up and really slowly on the way down. On the GMAT, you'll never have to make those assumptions, the question will clearly include all relevant information.

All that said, and assuming that the speed of the boat was constant, we can solve this question in 15 seconds.

Total trip took 4.5 hours.

The river current was running with the boat on the onward journey and against it on the return journey. Therefore, the onward journey will take less time than the return journed.

Since the onward journey is quicker, it will take less than half the time of the entire trip.

4.5/2 = 2.25. Only 1 answer choice is less than 2.25: choose (D).

And this is why I shouldn't trust free online GMAT prep tests!

I will go out and buy Kaplan 800 and OG11 before wasting anymore time with these free prep questions.

Thank you for your response!