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Robots X, Y, and Z each assemble components at their respect [#permalink]

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24 Sep 2012, 05:16

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32% (01:26) wrong based on 601 sessions

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Robots X, Y, and Z each assemble components at their respective constant rates. If \(r_x\) is the ratio of Robot X's constant rate to Robot Z's constant rate and \(r_y\) is the ratio of Robot Y's constant rate to Robot Z's constant rate, is Robot Z's constant rate the greatest of the three?

(1) \(r_x<r_y\) (2) \(r_y<1\)

Practice Questions Question: 47 Page: 279 Difficulty: 600

Re: Robots X, Y, and Z each assemble components at their respect [#permalink]

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24 Sep 2012, 05:16

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Robots X, Y, and Z each assemble components at their respective constant rates. If \(r_x\) is the ratio of Robot X's constant rate to Robot Z's constant rate and \(r_y\) is the ratio of Robot Y's constant rate to Robot Z's constant rate, is Robot Z's constant rate the greatest of the three?

Let the rates of robots X, Y, and Z be x, y, and z respectively. Given: \(r_x=\frac{x}{z}\) and \(r_y=\frac{y}{z}\). Question is \(z>x\) and \(z>y\)?

(1) \(r_x<r_y\) --> \(\frac{x}{z}<\frac{y}{z}\) --> \(x<y\). Not sufficient.

(2) \(r_y<1\) --> \(\frac{y}{z}<1\) --> \(y<z\). Not sufficient.

(1)+(2) As \(x<y\) and \(y<z\) then \(x<y<z\). Sufficient.

Re: Robots X, Y, and Z each assemble components at their respect [#permalink]

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24 Sep 2012, 08:23

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Robots X, Y, and Z each assemble components at their respective constant rates. If is the ratio of Robot X's constant rate to Robot Z's constant rate and is the ratio of Robot Y's constant rate to Robot Z's constant rate, is Robot Z's constant rate the greatest of the three?

(1) INSUFF X/z < Y/z this does not allow us to pin point anything to do with z (2) INSUFF y/z<1 again, not enough, y could make 1 an hour and z could make 2 an hour or 2,000 per hour. However, we don't know anything about X Together, SUFF X/z < y/z < 1 We know that z is faster than Y and we know that Y is faster than X - the question is true _________________

If you find my post helpful, please GIVE ME SOME KUDOS!

Re: Robots X, Y, and Z each assemble components at their respect [#permalink]

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24 Sep 2012, 09:02

Bunuel wrote:

Robots X, Y, and Z each assemble components at their respective constant rates. If \(r_x\) is the ratio of Robot X's constant rate to Robot Z's constant rate and \(r_y\) is the ratio of Robot Y's constant rate to Robot Z's constant rate, is Robot Z's constant rate the greatest of the three?

(1) \(r_x<r_y\) (2) \(r_y<1\)

(1) Since \(r_x\) and \(r_y\) are RATIO of rates of X and Y wrt to Z respectively, if we compare these two the rate of Z will cancel out from both sides and we will be left with X < Y. Nothing is know about Z. Not sufficient

(2) Here, ration of rate of Y to Z is lesser than 1. Thus Y < Z. Nothing is known about X. So not sufficient.

combining (1) and (2), X < Y < Z, Z has greatest rate.

Re: Robots X, Y, and Z each assemble components at their respect [#permalink]

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28 Sep 2012, 05:12

Expert's post

Robots X, Y, and Z each assemble components at their respective constant rates. If \(r_x\) is the ratio of Robot X's constant rate to Robot Z's constant rate and \(r_y\) is the ratio of Robot Y's constant rate to Robot Z's constant rate, is Robot Z's constant rate the greatest of the three?

Let the rates of robots X, Y, and Z be x, y, and z respectively. Given: \(r_x=\frac{x}{z}\) and \(r_y=\frac{y}{z}\). Question is \(z>x\) and \(z>y\)?

(1) \(r_x<r_y\) --> \(\frac{x}{z}<\frac{y}{z}\) --> \(x<y\). Not sufficient.

(2) \(r_y<1\) --> \(\frac{y}{z}<1\) --> \(y<z\). Not sufficient.

(1)+(2) As \(x<y\) and \(y<z\) then \(x<y<z\). Sufficient.

Answer: C.

Kudos points given to everyone with correct solution. Let me know if I missed someone. _________________

Re: Robots X, Y, and Z each assemble components at their respect [#permalink]

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24 Jun 2014, 00:35

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: Robots X, Y, and Z each assemble components at their respect [#permalink]

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06 Oct 2014, 12:34

I took ratios for robots X, Y, Z - 1/x, 1/y and 1/z respectively. therefore I made a mistake, and answered B since 1/y : 1/z = 1/y*z/1 = z/y and if Ry<1 then Z is less then Y...

Robots X, Y, and Z each assemble components at their respect [#permalink]

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10 Oct 2014, 11:27

mvictor wrote:

I took ratios for robots X, Y, Z - 1/x, 1/y and 1/z respectively. therefore I made a mistake, and answered B since 1/y : 1/z = 1/y*z/1 = z/y and if Ry<1 then Z is less then Y...

No its alright to do so Here's something that will help you

let the rates be 1/x for X, 1/y for Y and 1/z for Z so Rx= 1/x/1/z = z/x and similarly Ry= 1/y/1/z = z/y

statement 1 --> Rx<Ry --> z/x<z/y --> x>y (since z is constant for example z=1 , x=4 and y= 2 --> Rx<Ry --> x>y) Nothing is said about z here so insufficient

statement 2 --> Ry<1 --> z/y<1 --> z<y (cross multiply, its okay to do so here as rates cannot be negative ) nothing about x so insufficient

statement 1 & 2 together

z<y<x is z the greatest? clearly not

so answer is C

Hope it helps, if it does a Kudos will be great. Cheers!!

Re: Robots X, Y, and Z each assemble components at their respect [#permalink]

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10 Oct 2014, 11:30

Bunuel wrote:

Robots X, Y, and Z each assemble components at their respective constant rates. If \(r_x\) is the ratio of Robot X's constant rate to Robot Z's constant rate and \(r_y\) is the ratio of Robot Y's constant rate to Robot Z's constant rate, is Robot Z's constant rate the greatest of the three?

Let the rates of robots X, Y, and Z be x, y, and z respectively. Given: \(r_x=\frac{x}{z}\) and \(r_y=\frac{y}{z}\). Question is \(z>x\) and \(z>y\)?

(1) \(r_x<r_y\) --> \(\frac{x}{z}<\frac{y}{z}\) --> \(x<y\). Not sufficient.

(2) \(r_y<1\) --> \(\frac{y}{z}<1\) --> \(y<z\). Not sufficient.

(1)+(2) As \(x<y\) and \(y<z\) then \(x<y<z\). Sufficient.

Answer: C.

Kudos points given to everyone with correct solution. Let me know if I missed someone.

Hi Bunuel,

Could you kindly check my approach?

Thanks

let the rates be 1/x for X, 1/y for Y and 1/z for Z so Rx= 1/x/1/z = z/x and similarly Ry= 1/y/1/z = z/y

statement 1 --> Rx<Ry --> z/x<z/y --> x>y (since z is constant for example z=1 , x=4 and y= 2 --> Rx<Ry --> x>y) Nothing is said about z here so insufficient

statement 2 --> Ry<1 --> z/y<1 --> z<y (cross multiply, its okay to do so here as rates cannot be negative ) nothing about x so insufficient

Re: Robots X, Y, and Z each assemble components at their respect [#permalink]

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12 Dec 2015, 06:13

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________

Re: Robots X, Y, and Z each assemble components at their respect [#permalink]

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13 Dec 2015, 18:53

Expert's post

Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

Robots X, Y, and Z each assemble components at their respective constant rates. If r x is the ratio of Robot X's constant rate to Robot Z's constant rate and r y is the ratio of Robot Y's constant rate to Robot Z's constant rate, is Robot Z's constant rate the greatest of the three?

(1) r x <r y (2) r y <1

In the original condition, when you consider the ratio of X,Y,Z as x,y,z, there are 3 variables(x,y,z), which should match with the number of equations. So, you need 3 more equations. For 1) 1 equation, for 2 1 equation, which is likely to make E the answer. In 1) & 2), x/z<y/z<1 and therefore it becomes x<y<z. Since the ratio of z is the greatest, it is unique and sufficient. Therefore, the answer is C.

-> For cases where we need 3 more equations, such as original conditions with “3 variables”, or “4 variables and 1 equation”, or “5 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 80% chance that E is the answer (especially about 90% of 2 by 2 questions where there are more than 3 variables), while C has 15% chance. These two are the majority. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since E is most likely to be the answer using 1) and 2) separately according to DS definition (It saves us time). Obviously there may be cases where the answer is A, B, C or D. _________________

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