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# Robots X, Y, and Z each assemble components at their

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

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03 Jan 2011, 18:30
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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) $$r_y<1$$

Can some explain the reasoning behind this ques.
[Reveal] Spoiler: OA

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Last edited by Bunuel on 08 Feb 2012, 05:09, edited 1 time in total.
Edited the question

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Re: Robots X, Y, and Z each assemble [#permalink]

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08 Feb 2012, 05:06
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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.

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

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07 May 2013, 07:09
Does this equation work when plugging in numbers, opposed to looking at pure variables?

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

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07 May 2013, 14:10
laythesmack23 wrote:
Does this equation work when plugging in numbers, opposed to looking at pure variables?

Sure. Lets say the rates are X = 5, Y = 6 and Z = 7.

Y/Z = 6/7 -> as stated in II

X/Y = 5/6 and Y/Z = 6/7 -> 5/6< 6/7 as stated in I

Cheers
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Re: Robots X, Y, and Z each assemble [#permalink]

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22 Oct 2013, 08:00
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.

Bunuel,

Please help me clarify. The question says "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", so if it is rates, why is X's constant rate not 1/X (which is the rate of completing one unit of work, and Z's rate would therefore be 1/Z. Thus r(x) would be 1/X : 1/Z? What am I misunderstanding here?

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Re: Robots X, Y, and Z each assemble [#permalink]

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22 Oct 2013, 09:22
bulletpoint wrote:
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.

Bunuel,

Please help me clarify. The question says "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", so if it is rates, why is X's constant rate not 1/X (which is the rate of completing one unit of work, and Z's rate would therefore be 1/Z. Thus r(x) would be 1/X : 1/Z? What am I misunderstanding here?

Because we denoted rates by x , y, and z: let the rates of robots X, Y, and Z be x, y, and z respectively.
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Re: Robots X, Y, and Z each assemble components at their [#permalink]

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03 Nov 2013, 08:25
Hello Everyone,
In OG 13, the question uses "rx" and "ry" in the question stem but "r_x (x in suffix)" and "r_y(y in suffix)".
Are these typos?
Or am I supposed to guess that rx and ry of question stem has been converted to r_x and r_y in the two given options?
TIA,

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

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03 Nov 2013, 11:21
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drebellion wrote:
Hello Everyone,
In OG 13, the question uses "rx" and "ry" in the question stem but "r_x (x in suffix)" and "r_y(y in suffix)".
Are these typos?
Or am I supposed to guess that rx and ry of question stem has been converted to r_x and r_y in the two given options?
TIA,

It's a typo. x and y must be indexes in both stem and the statements.
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Re: Robots X, Y, and Z each assemble components at their [#permalink]

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03 Nov 2013, 12:06
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Also thank you for pointing out the error/typo in Diagnostic Test Q 5 (Cylindrical tank contains 36PI f3 of water...)

At lest for the second question (Cylindrical Tank...) we will never know if its a typo or the guys who had this question in their real GMAT were unfortunate!

Thanks Anyway!

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

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

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19 Jun 2016, 04:40
Assume the individual rates be X, Y and Z respectively for the robots.
Given: X/Z = $$r_x$$ and Y/Z = $$r_y$$
Required: Is Z the greatest?

Statement 1: $$r_x < r_y$$
Hence X/Z < Y/Z
Or, X < Y - (i)
We do not have any information about Z.
INSUFFICIENT

Statement 2: $$r_y < 1$$
Or Y/Z < 1
Hence Y < Z - (ii)
We do not know anything about X.
INSUFFICIENT

Combining both statements:
From (i) and (ii), we know that
X < Y and Y < Z
Hence X < Y < Z

SUFFICIENT

Correct Option: C

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

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01 Aug 2016, 02:23
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ajit257 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) $$r_y<1$$

Can some explain the reasoning behind this ques.

Fist of all we need to remember that rate can never be negative.
this makes the question super easy to deal

(1) $$r_x<r_y$$

meaning

$$\frac{x}{z}<\frac{y}{z}$$ {since neither x,y,z are rates and rate cannot be negative; therefore we can remove z from both side it easily without worrying about the sign}

so

$$x<y$$

INSUFFICINET

(2) $$r_y<1$$

meaning$$\frac{y}{z}$$ is less than 1

No info about rate of x or rate of z

INSUFFICIENT

merge both statements

$$\frac{x}{z}<\frac{y}{z}<1$$

multiply each term with z

$$z*\frac{x}{z}<z*\frac{y}{z}<z*1$$

x<y<z

Therefore Z is greatest

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

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24 Oct 2016, 09:47
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.

Hi Bunuel, I prefectly understad your solution, but why I can't reach the same conclusion if I use the notation 1/x, 1/y, 1/z to denote the 3 rates? After all the rate is output/time, therfore it should work also in this way...

Thanks

EDIT: sorry I wrote something stuopid, it works fine also using the notation 1/x etc

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

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10 Nov 2016, 23:28
Bunuel, why are we not taking the 1/x, 1/y format? shouldnt r(x) be (1/x)/1/z = z/x? and r(y) in the same manner? Thank you.

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

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10 Nov 2016, 23:42
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TheLordCommander wrote:
Bunuel, why are we not taking the 1/x, 1/y format? shouldnt r(x) be (1/x)/1/z = z/x? and r(y) in the same manner? Thank you.

In the solution I denoted rates by x , y, and z because we need to find the ratio of rates and it makes more sense to do this way. We could denoted the times of robots X, Y, and Z by x , y, and z and in this case, since the rate is reciprocal of time, the rates would be 1/x. 1/y, and 1/z but the way I did is better for this problem.
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Re: Robots X, Y, and Z each assemble components at their [#permalink]

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12 Nov 2016, 01:59
Bunuel wrote:
TheLordCommander wrote:
Bunuel, why are we not taking the 1/x, 1/y format? shouldnt r(x) be (1/x)/1/z = z/x? and r(y) in the same manner? Thank you.

In the solution I denoted rates by x , y, and z because we need to find the ratio of rates and it makes more sense to do this way. We could denoted the times of robots X, Y, and Z by x , y, and z and in this case, since the rate is reciprocal of time, the rates would be 1/x. 1/y, and 1/z but the way I did is better for this problem.

I understand Bunuel, thank you. Please review if the following solution is correct too -

If we use 1/x, 1/y and 1/z, we get rx = z/x and ry as z/y. To have the greatest rate, the denominator should be lowest; so z will have to be lower than x and y.

Statement 1 - rx<ry that means z/x<z/y => y<x. Not sufficieent.

Statement 2 - ry<1=> z/y<1 =>z<y => z<y. Not sufficient.

Statement 1+2 - z<y<x - sufficient. C.

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Re: Robots X, Y, and Z each assemble components at their   [#permalink] 12 Nov 2016, 01:59
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