Up the hill, down the hill (VIC)

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Up the hill, down the hill (VIC)

by chieftang » Tue Jan 03, 2012 9:51 pm
Jeff rides his bike up a hill, and then rides his bike the same distance back down the hill. On the way up, he averages X miles per hour. On the way down, he averages Y miles per hour. What is his average speed for the entire trip, in terms of X and Y?

(A) (X+Y)/2
(B) (X+Y)/(2*X*Y)
(C) (2*X*Y)/(X+Y)
(D) (X+Y)/(X*Y)
(E) (X*Y)/(X+Y)


Source: MATHCOUNTS problem bank.

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by rijul007 » Tue Jan 03, 2012 10:28 pm
Lets say the distance from top of the hill to bottom of the hill is D

Total distance = 2D

Total time taken = D/X + D/Y = D(X+Y)/XY

Average speed = 2D/[D(X+Y)/XY] = 2XY/(X+Y)

Option C

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by karthikpandian19 » Tue Jan 03, 2012 10:38 pm
Answer is C

Rate = Distance / Time

You have Rate GIVEN here, so find out the Distance and Time.

Since not given, we can assume....Distance is 100miles for Up and down each.

Time is then: 100/X hrs & 100/Y hrs (as per the formula from the top)


Total distance = 200 miles
Total time = 100/X + 100/Y
(which can be simplified as 100(X+Y)/XY

Now, Average Speed / Rate = Total Distance /Total time

When you solve for the calculation, 200 / (100(X+Y)/XY)
You will get C

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by chieftang » Tue Jan 03, 2012 11:25 pm
Ya, OK. Not terribly difficult. But let's look at two answer choices in particular:

(A) (X+Y)/2
(B) (X+Y)/(2*X*Y)
(C) (2*X*Y)/(X+Y)
(D) (X+Y)/(X*Y)
(E) (X*Y)/(X+Y)

Answer choice A & C.

Answer choice A should be recognizable as the arithmetic mean of X and Y. It does not apply to this question, however. Explanation below..

Now, how about answer choice C? This is known as the harmonic mean of X and Y, and it does apply to this question.

Rates: Arithmetic Mean vs Harmonic Mean

What's the application of each?

Well first off, when dealing with average speeds as in the question at the top of this thread, in either case the following concept applies:
  • Average speed = (Total distance) / (Total time)


In this particular problem, the harmonic mean came in to play for calculating the average speed because distances were equal, and the equation for harmonic mean was correctly derived in the solutions given by rijul007 and others in the thread.

The harmonic mean is also sometimes expressed in the format 2/((1/X)+(1/Y)) which is equivalent to answer choice C above, and sometimes described as "The number of observations divided by the reciprocal of each number in the series." It can apply to any number of rates, e.g. 3 different rates (X,Y,Z) as: 3/((1/X)+(1/Y)+(1/Z))

How about arithmetic mean, then? Well, consider the following question:

Jeff rides his bike up a hill for H hours at a rate of X miles per hour, and then rides his bike back down the hill for H hours at a rate of Y miles per hour. What is his average speed for the entire trip, in terms of X and Y?

Notice now that time is equal for both segments of the trip. So, in calculating average speed with the concept above, we would have:

Average speed = (X*H + Y*H)/(2*H) = (X+Y)/2 = arithmetic mean of X and Y. Note that the H drops off.

So, in general, when calculating an average rate...

Arithmetic means yield an average of multiple rates that occur for the same duration.

Harmonic means yield an average of multiple rates over the same unit of measure (e.g. distance).

And, of course, the rates can be of any units the GMAT chooses. Miles per hour, kilometers per hour, gallons per hour, questions per minute, etc. See also: https://www.beatthegmat.com/average-rate ... 00817.html

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by chufus » Wed Jan 04, 2012 2:42 am
Continuing from the discussions above........ Here is my solution...

Distance = d = same for both trips..

Magic Formula: Distance = Time x Speed

Lets find the time for each.

Uphill = d/x

Downhill = d/y

Total Distance = 2d
Total TIme = (d/x + d/y)
So average Speed = (2d)/(d/x + d/y) = (2xy)/(x+y)

Hence Answer C !

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by GMATGuruNY » Wed Jan 04, 2012 5:22 am
chieftang wrote:Jeff rides his bike up a hill, and then rides his bike the same distance back down the hill. On the way up, he averages X miles per hour. On the way down, he averages Y miles per hour. What is his average speed for the entire trip, in terms of X and Y?

(A) (X+Y)/2
(B) (X+Y)/(2*X*Y)
(C) (2*X*Y)/(X+Y)
(D) (X+Y)/(X*Y)
(E) (X*Y)/(X+Y)


Source: MATHCOUNTS problem bank.
Let the distance from the bottom to the top of the hill = 6 miles.
If x = 2 miles per hour, the time up the hill = d/r = 6/2 = 3 hours.
If y = 3 miles per hour, the time down the hill = d/r = 6/3 = 2 hours.
Average speed = (total distance)/(total time) = (6+6)/(3+2) = 12/5. This is our target.

A quick scan of the answer choices reveals that only C works:
(2xy)/(x+y) = (2*2*3)/(2+3) = 12/5.

The correct answer is C.
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