Problem Solving

Paul drives from his apt to his parents house and back. On the trip to this parents house he travels at an average speed of 60 mph. On the return trip, Paul drives at an average speed of 80 mph. Which of the followng is the closest approximation of Pauls average speed, in mph, for the round trip?

A) 60
B) 68.6
C) 70
D) 71.4
E) 80

I would hide the answer but dont know how to

I would hide the answer but dont know how to

use the 'spoiler' tags that are available while posting.

just apply the formula for harmonic mean

average speed = 2xy(x+y)

= (2*60*80)(60+80) = 9600/ 140 = 960/ 14 which is slightly less than 70

--> Ans choice B is correct

Thanks. How do you know to use the harmonic mean vs. regular mean?

I would explain sars' formula for average speed in detail

In this problem, distance (d) is the same for onward and return journey. But speed varies. So, do time taken.
Let speed of onward journey = x
Let speed of return journey = y

Time = Distance / Speed

Total Time taken for both onward and return = T(onward) + T(return) = d/x + d/y = (d*y + d*x) / (x*y) = d (x + y) / (x*y)

Now, Average speed for entire trip = Total Distance / Total Time taken

Total Distance = (d+d) OR (2*d)
Total Time taken = d (x + y) / (x*y)

So, Average speed for entire trip = [(2*d) * (x * y)] / [d (x + y)]

d cancels out giving you ==> 2 x*y / (x+y)

Bottomline is don't confuse with terms like harmonic mean. If you know the basic distance/speed formula, you can solve this. Next time, when you see a similar problem like this, you can straight away apply this formula without working out.

Thank you SO much for your reply! I will definitely keep that in mind the next time I see a similiar problem!!

LMK27 wrote:Thanks. How do you know to use the harmonic mean vs. regular mean?

harmonic mean is used when the values are having a weightage inversely proprtional to the magnitude..
but don't get flustered by all this mumbo-jumbo, just know that you can use the harmonic mean formula to solve distance-speed-time questions which involve uphill/downhill.

Even though it may sound unncecessary, I like to refer to it as harmonic mean so that everytime I see such a problem, an alarm bell rings, and I say 'k.. harmonic mean.. coming up'

But, if things get too complicated, just remember this one formula: Average Speed = Total Distance/ Total Time

Bear in mind that for such problems, the distance will be the same going uphill and coming downhill. The time however will differ and hence the average speed is unlikely to be the straightforward average of the two speeds

Hello, though unrelated to this specific question, I use the following formulae when solving speed calculations with similar conditions:

If a certain distance (D), say from A to B is covered at 'a' km/h and the same distance is covered again from B to A in 'b' km/h, then the average speed during the whole journey is given by:

Avg Speed = 2ab/a+b

Also if t1 and t2 are the times taken from A to B and B to A, then

D = (t1+t2)(ab/a+b)
D = (t1-t2)(ab/a-b)
D = (a-b)(t1t2/t1-t2)

I have saved tons of time using these formulae and hope it will benefit atleast one of you.

Lets assume the distance one way to be 240 (LCM of 60 & 80 for ease of calculation).
It would take him 4 hours for first trip @ 60mph and 3 hours to come back @ 80mph.
ie 7 hours for 480 miles or 68.6mph.

thanks everyone is makes sense now! I appreciate all the responses.