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High 700 level rate problem, any1 got good advice for such..

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High 700 level rate problem, any1 got good advice for such..

by benjiboo » Fri Sep 14, 2012 6:11 pm
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?

A) (x+y+z)/3
B) (3xyz)/(xy+yz+zx)
C) (xyz)/(x+y+z)
D) (xy+yz+zx)/(x+y+z)
E) [3(x+y+z)]/(xyz)

[spoiler]Ans: B[/spoiler]

This is a tough one... GMAT experts want to give this one a try? :)
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by truplayer256 » Fri Sep 14, 2012 7:09 pm
This is not too bad of a problem if you remember the following:

Average speed = (total distance)/(total time)

Assume the distance from town A to B was d, then it follows that:

d/x = Since this is distance over speed, it is the time it took Jacob to drive from Town A to Town B.

d/y = Time it took Jacob to drive from Town B to Town A.

d/z = Time it took Jacob to drive back to town B after he returned to town A.

Average speed of entire trip = (Total distance for entire trip)/(Total time for entire trip)

= (d + d + d)/(d/x + d/y + d/z) = (3d)/(dyz + dxz + dyx)/(xyz)

= (3xyz)/(yz + xz + yx)

Choose B.
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by Javoni » Fri Sep 14, 2012 8:27 pm
Here we go, it is not that tough as long as you remember what the average speed is, to this end:

Average Speed = Total Distance / Total Time Spent

Here in our case we have:

From A to B; B to A and A to B again:

A-->>>>>> - B Average speed = D/t1 = X

A - <<<<<---B Average speed = D/t2 = Y

A---->>>>> - B Average speed = D/t3 = Z

Average rate for entire trip would be = Total Distance/(t1+t2+t3), hence we get:

Average rate = 3D /(D/X + D/Y + D/Z), a slight manipulation gets us to the expression:

3XYZ/(YZ+XZ+XY), THUS (B)

Please, correct me if I went awry.


benjiboo 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?

A) (x+y+z)/3
B) (3xyz)/(xy+yz+zx)
C) (xyz)/(x+y+z)
D) (xy+yz+zx)/(x+y+z)
E) [3(x+y+z)]/(xyz)

[spoiler]Ans: B[/spoiler]

This is a tough one... GMAT experts want to give this one a try? :)
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by GMATGuruNY » Fri Sep 14, 2012 11:27 pm
benjiboo 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?

A) (x+y+z)/3
B) (3xyz)/(xy+yz+zx)
C) (xyz)/(x+y+z)
D) (xy+yz+zx)/(x+y+z)
E) [3(x+y+z)]/(xyz)

[spoiler]Ans: B[/spoiler]

This is a tough one... GMAT experts want to give this one a try? :)
Let the distance = 12 miles.
Let x = 2 miles per hour, y = 3 miles per hour, and z = 6 miles per hour.
Total time for the 3 trips = 12/2 + 12/3 + 12/6 = 12 hours.
Average rate for the 3 trips = (12+12+12)/12 = 3 miles per hour. This is our target.

Now we plug x=2, y=3, and z=6 into the answers to which yields our target of 3.
Only B works:
(3xyz)/(xy+yz+zx) = (3*2*3*6) / (2*3 + 3*6 + 6*2) = (3*2*3) / (1+3+2) = 3.

The correct answer is B.
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by Brent@GMATPrepNow » Sat Sep 15, 2012 7:44 am
benjiboo 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?

A) (x+y+z)/3
B) (3xyz)/(xy+yz+zx)
C) (xyz)/(x+y+z)
D) (xy+yz+zx)/(x+y+z)
E) [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) = B

Cheers,
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by Zoser » Fri Feb 17, 2017 6:42 pm
= (d + d + d)/(d/x + d/y + d/z) = (3d)/(dyz + dxz + dyx)/(xyz)

= (3xyz)/(yz + xz + yx)
Can you please explain how you get rid of D in the last step?

Thanks
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by GMATGuruNY » Sat Feb 18, 2017 3:49 am
benjiboo 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?

A) (x+y+z)/3
B) (3xyz)/(xy+yz+zx)
C) (xyz)/(x+y+z)
D) (xy+yz+zx)/(x+y+z)
E) [3(x+y+z)]/(xyz)
To make the algebra easier, let the distance in each direction = 1.

Since the rate from A to B = x, the time from A to B = d/r = 1/x.
Since the rate from B to A = y, the time from A to B = d/r = 1/y.
Since the rate back to B = z the time back to B = d/r = 1/z.
Total time = 1/x + 1/y + 1/z = (yz)/(xyz) + (xz)/(xyz) + (xy)/(xyz) = (xy + xz + yz)/(xyz).

Average rate = (total distance)/(total time) = (1+1+1)/[(xy + xz + yz)/(xyz)] = (3xyx)/(xy + xz + yz).

The correct answer is B.
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by Jay@ManhattanReview » Sat Feb 18, 2017 4:41 am
benjiboo 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?

A) (x+y+z)/3
B) (3xyz)/(xy+yz+zx)
C) (xyz)/(x+y+z)
D) (xy+yz+zx)/(x+y+z)
E) [3(x+y+z)]/(xyz)

[spoiler]Ans: B[/spoiler]

This is a tough one... GMAT experts want to give this one a try? :)
Hi benjiboo,

This is not a tough one. :) Many experts have put forth their approaches. Great ones. Here's mine.

Whenever the distance is the same and there are two or more speeds and we are asked to find out the average speed, do not take arithmetic mean of the speeds.

What you should do is the following.

1. Take the reciprocals of the speeds.
2. Take the arithmetic mean of the reciprocals.
3. Take the reciprocal of the arithmetic mean obtained in step 2.

Coming to the question...

We know that the distance is the same in each of the three occasions, but the speeds are different, so we can apply the above approach.

1. Take the reciprocals of the speeds => 1/x, 1/y, and 1/z
2. Take the arithmetic mean of the reciprocals.

Arithmetic mean = 1/3[1/x + 1/y + 1/z] = (xy + yz + xz)/3xyz

3. Take the reciprocal of the arithmetic mean obtained in step 2.

Average speed = Reciprocal of (xy + yz + xz)/3xyz = 3xyz / (xy + yz + xz)

The correct answer: B

Hope this helps!
Let us a see a simple question.

If Jack drives from city A to B at a uniform speed of 30 miles per and returns to A from B at a uniform speed of 40 miles per, what is his average speed for the entire journey?
Needless to state that the answer would not be the arithmetic mean of 30 and 40 = 35 miles per hour.

Apply the above steps.

1. Take the reciprocals of the speeds: 1/30 and 1/40
2. Take the arithmetic mean of the reciprocals: 1/2[1/30 + 1/40] = 1/2[7/120] = 7/240
3. Take the reciprocal of the arithmetic mean obtained in step 2: 240/7 = 34.3 miles per hour

Note that the correct answer 34.3 is less than the incorrect answer 35.

34.3 is called the harmonic mean, while 35 is the arithmetic mean. Note that harmonic mean is always less than arithmetic mean.

Hope this helps!

Relevant book: Manhattan Review GMAT Word Problems Guide

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by Matt@VeritasPrep » Thu Mar 02, 2017 5:46 pm
Not to dismiss your concerns, but this isn't high 700 level at all. It isn't really even high 500 level by the standards of the modern GMAT.

The easiest way to do this is to pick numbers for x, y, and z. Most test takers have been trained to do this by whatever course they take, and if you pick numbers the problem is simple (albeit time consuming!)

Doing it algebraically is harder and much less fun, but alas, the difficulty is only measured by how many people got the problem right, not how long it took them or how hard they found it along the way.
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by Scott@TargetTestPrep » Thu Mar 09, 2017 4:10 pm
benjiboo 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?

A) (x+y+z)/3
B) (3xyz)/(xy+yz+zx)
C) (xyz)/(x+y+z)
D) (xy+yz+zx)/(x+y+z)
E) [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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by Matt@VeritasPrep » Thu Mar 16, 2017 8:34 pm
Zoser wrote:
= (d + d + d)/(d/x + d/y + d/z) = (3d)/(dyz + dxz + dyx)/(xyz)

= (3xyz)/(yz + xz + yx)
Can you please explain how you get rid of D in the last step?

Thanks
The denominator goes like so:

(d/x + d/y + d/z) =>

xyz/xyz * (d/x + d/y + d/z) =>

(dyz + dzx + dxy)/xyz =>

d * (yz + zx + xy)/xyz

You can then cancel this d with the d that's found in the 3d in the numerator.
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by Matt@VeritasPrep » Thu Mar 16, 2017 8:35 pm
But definitely don't bother doing all that on the test! Plugging in numbers is by far the easiest way, and should be the fastest for most test takers.
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