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

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

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New post 14 Feb 2016, 01:50
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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. 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?

\(\frac{(x+y+z)}{3}\)

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

\(\frac{xyz}{(x+y+z)}\)

\(\frac{(xy+yz+zx)}{(x+y+z)}\)

\(\frac{3(x+y+z)}{xyz}\)
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Re: Jacob drove from Town A to Town B at an average rate of x miles per ho  [#permalink]

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New post 14 Feb 2016, 03:54
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Nez wrote:
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. 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?

\(\frac{(x+y+z)}{3}\)

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

\(\frac{xyz}{(x+y+z)}\)

\(\frac{(xy+yz+zx)}{(x+y+z)}\)

\(\frac{3(x+y+z)}{xyz}\)


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)
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Re: Jacob drove from Town A to Town B at an average rate of x miles per ho  [#permalink]

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New post 14 Feb 2016, 08:41
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Nez wrote:
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. 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?

\(\frac{(x+y+z)}{3}\)

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

\(\frac{xyz}{(x+y+z)}\)

\(\frac{(xy+yz+zx)}{(x+y+z)}\)

\(\frac{3(x+y+z)}{xyz}\)


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
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Re: Jacob drove from Town A to Town B at an average rate of x miles per ho  [#permalink]

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New post 14 Feb 2016, 23:18
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Nez wrote:
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. 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?

\(\frac{(x+y+z)}{3}\)

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

\(\frac{xyz}{(x+y+z)}\)

\(\frac{(xy+yz+zx)}{(x+y+z)}\)

\(\frac{3(x+y+z)}{xyz}\)



For this and other formulas, check: http://www.veritasprep.com/blog/2015/02 ... -the-gmat/
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Jacob drove from Town A to Town B at an average rate of x miles per ho  [#permalink]

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New post 02 Apr 2017, 23:05
Nezdem wrote:
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. 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?

\(\frac{(x+y+z)}{3}\)

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

\(\frac{xyz}{(x+y+z)}\)

\(\frac{(xy+yz+zx)}{(x+y+z)}\)

\(\frac{3(x+y+z)}{xyz}\)


3/[(1/x)+(1/y)+(1/z)]➡
3/[(xy+yz+zx)/xyz]➡
3xyz/(xy+yz+zx)
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Re: Jacob drove from Town A to Town B at an average rate of x miles per ho  [#permalink]

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New post 28 Apr 2017, 15:31
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)
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Jacob drove from Town A to Town B at an average rate of x miles per ho  [#permalink]

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New post Updated on: 03 Jul 2017, 08:43
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
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Originally posted by sasyaharry on 02 Jul 2017, 15:10.
Last edited by sasyaharry on 03 Jul 2017, 08:43, edited 1 time in total.
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Jacob drove from Town A to Town B at an average rate of x miles per ho  [#permalink]

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New post 02 Jul 2017, 16:48
sasyaharry wrote:
Also use some simple values. x=1, y=2, z=3. Let distance d = 6.
\(Avg speed = \frac{total distance}{total time}\)
= \(\frac{6*3}{(6+3+2)} = \frac{9}{5}\)

Plugging in values for x,y,z and d

1) \(\frac{x+y+z}{3} = \frac{6}{3} = 2\)......... wrong
2) \(\frac{3xyz}{(xy+yz+zx)} = \frac{9}{5}\)...............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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Re: Jacob drove from Town A to Town B at an average rate of x miles per ho  [#permalink]

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New post 03 Jul 2017, 08:43
genxer123 wrote:
sasyaharry wrote:
Also use some simple values. x=1, y=2, z=3. Let distance d = 6.
\(Avg speed = \frac{total distance}{total time}\)
= \(\frac{6*3}{(6+3+2)} = \frac{9}{5}\)

Plugging in values for x,y,z and d

1) \(\frac{x+y+z}{3} = \frac{6}{3} = 2\)......... wrong
2) \(\frac{3xyz}{(xy+yz+zx)} = \frac{9}{5}\)...............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!


gracias. Corrected.
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Re: Jacob drove from Town A to Town B at an average rate of x miles per ho  [#permalink]

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New post 12 Dec 2017, 18:15
Ekland wrote:
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. 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?

\(\frac{(x+y+z)}{3}\)

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

\(\frac{xyz}{(x+y+z)}\)

\(\frac{(xy+yz+zx)}{(x+y+z)}\)

\(\frac{3(x+y+z)}{xyz}\)



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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Re: Jacob drove from Town A to Town B at an average rate of x miles per ho  [#permalink]

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Re: Jacob drove from Town A to Town B at an average rate of x miles per ho   [#permalink] 25 Dec 2018, 15:52
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