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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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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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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) Option 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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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 BSo, total distance traveled = 3dTime = distance/speed time of 1st journey = d/x time of 2nd journey = d/ytime of 3rd journey = d/zTotal time = d/x + d/y + dzTo simplify, rewrite with common denominator: dyz/xyz + dxz/xyz + dxy/xyzSo, 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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14 Feb 2016, 23:18
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 ... thegmat/
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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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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) 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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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 rearranging  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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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) \(\frac{x+y+z}{3} = \frac{6}{3} = 2\)......... wrong2) \(\frac{3xyz}{(xy+yz+zx)} = \frac{18}{11}\)............... correctAnswer 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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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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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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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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