Jacob drove from Town A to Town B at an average rate of x miles per ho

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Also use some simple values. x=1, y=2, z=3. Let distance d = 6.
\(Avg speed = \frac{total distance}{total time}\)
= \(\frac{6*3}{(6+3+2)} = \frac{18}{11}\)

Plugging in values for x,y,z and d

1) \(\frac{x+y+z}{3} = \frac{6}{3} = 2\)......... wrong
2) \(\frac{3xyz}{(xy+yz+zx)} = \frac{18}{11}\)...............correct

Answer is B

Distance for one trip: d
Total distance: 3d

average speed = \(\frac{total distance }{ total time}\)

total time for each trip = \(\frac{d}{x}, \frac{d}{y}, \frac{d}{z}\)

total time = \(\frac{d}{x} + \frac{d}{y} + \frac{d}{z}\)

average speed = \(\frac{3d }{ d/x + d/y + d/z}\)

Multiply the numerator and denominator by xyz:

\(\frac{3xyz }{ yz + xz + xy}\)

Answer is B.

3/[(1/x)+(1/y)+(1/z)]➡
3/[(xy+yz+zx)/xyz]➡
3xyz/(xy+yz+zx)
B

pretty time consuming...
we have the same distance.
say D.
time taken for first leg: D/x
time taken for second leg: D/y
time taken for third leg: D/z
total time: D/x + D/y + D/z => after re-arranging - we get: D(yz+xz+xy)/xyz
total distance: 3D.
3D divide by D(yz+xz+xy)/xyz
we get:
3xyz/(yz+xz+xy)

Your answer is correct, but your arithmetic is not.

Your fraction should be \(\frac{18}{11}\), not \(\frac{9}{5}\). With your numbers for d, x, y, and z, answer B does indeed yield the correct answer. Cheers!

gracias. Corrected.

To solve, we can use the formula for average rate:

average rate = total distance/total time

We can let distance from Town A to Town B (or vice versa) = d, and since he went from A to B, then from B to A, and then from A to B, his total distance traveled was 3d. Recall that time = distance/rate, so the time to get from Town A to Town B = d/x, the time it takes to get from Town B back to Town A = d/y, and the time it takes to go back to Town B = d/z. Thus:

average rate = 3d/(d/x + d/y + d/z)

To combine the three fractions in the denominator, we use the common denominator xyz:

average rate = 3d/(yzd/xyz + xzd/xyx + xyd/xyz)

average rate = 3d/[(yzd + xzd + xyd)/xyz]

average rate = 3d * xyz/(yzd + xzd + xyd)

average rate = 3d * xyz/[d(yz + xz + xy)]

The ds cancel and we are left with:

3xyz/(yz + xz + xy)

Answer: B

Let the distance from A to B = 2 miles.
Let x = 1 mile per hour, y = 1 mile per hour, and z = 2 miles per hour, with the result that the total time = first leg + second leg + third leg = 2+2+1 = 5 hours and that the average speed for the entire 6-mile trip \(= \frac{total-distance}{total-time} = \frac{6}{5}\).

Since the average speed = 6/5, the denominator for the correct answer must yield a multiple of 5 when x=1, y=1 and z=2.
A quick scan of the answer choices reveals that the denominators for A, C, D and E will not yield a multiple of 5.
Only B is viable.

Don't do the mistake of taking a simple average. Average speed is equal to total distance upon a total time. Use individual distances (same in each case) and respective speeds to get individual times and add them up to get the total time.

Then it is a straight forward question.

Given: Jacob drove from Town A to Town B at an average rate of x miles per hour, then returned along the same route at y miles per hour.
Asked: If he then drove back to Town B at z miles per hour along the same route, what was Jacob’s average rate of speed for the entire trip, in miles per hour?

Let the distance from Town A to Town B be D miles

Total distance travelled = D + D + D = 3D
Total time taken = D/x + D/y + D/z = D(1/x + 1/y + 1/z) = D(xy+yz+xz)/xyz

Average speed = 3Dxyz/D(xy+yz+zx) = 3xyz/(xy+yz+zx)

IMO B

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