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

The best way to go about in these question is to find out the total distance covered and the total time taken.

Assume the distance for one trip = d
Total distance covered = 3d.
Time take for 1st trip = d/x
Time take for 2nd trip = d/y
Time take for 3rd trip = d/z

Average speed = total distance/ total time

Average speed = 3d/ (d/x + d/y + d/z) = 3 / (1/x + 1/y + 1/z) = 3xyz / (yz + zx + xy)
Option B

Average speed = (total distance traveled)/(total travel time)
= (total distance)/(time of 1st journey + time of 2nd journey + time of 3rd journey)

Let d = the distance between Town A and Town B
So, total distance traveled = 3d

Time = distance/speed
time of 1st journey = d/x
time of 2nd journey = d/y
time of 3rd journey = d/z

Total time = d/x + d/y + dz
To simplify, rewrite with common denominator: dyz/xyz + dxz/xyz + dxy/xyz
So, total time = (dyz + dxz + dxy)/xyz

Average speed = (total distance)/(total time)
= 3d/[(dyz + dxz + dxy)/xyz]
= (3dxyz)/(dyz + dxz + dxy)
Divide top and bottom by d to get: (3xyz)/(yz + xz + xy)
Answer: B

Cheers,
Brent

For this and other formulas, check: https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2015/02 ... -the-gmat/
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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)

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

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!
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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
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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.
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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.


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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.
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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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