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

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!!

Solution:

We are given that r_x = the ratio of robot X’s constant rate to robot Z’s constant rate. If we let A = the rate of robot X and C = the rate of robot Z, we can say:

A/C = r_x

We are also given that r_y = the ratio of robot Y’s constant rate to robot Z’s constant rate. If we let B = the rate of robot Y, we can say:

B/C = r_y

We need to determine whether C is greater than both A and B.

Statement One Alone:

r_x < r_y

Statement one tells us that A/C < B/C. We can multiply both sides by C and obtain:

A < B

Thus rate of robot X is less than the rate of robot Y. However, we still do not know whether the rate of robot Z is greater than the rate of either robot X or robot Y. Statement one is not sufficient to answer the question. We can eliminate answer choices A and D.

Statement Two Alone:

r_y < 1

Since r_y < 1, B/C < 1 or B < C.

Thus, the rate of robot Z is greater than the rate of robot Y. However, we still do not know whether the rate of robot Z is greater than the rate of robot X. Statement two is not sufficient to answer the question. We can eliminate answer choice B.

Statements One and Two Together:

From statements one and two we know that the rate of robot X is less than the rate of robot Y and that the rate of robot Z is greater than the rate of robot Y. Thus, if the rate of robot Y is less than the rate of robot Z, then the rate of robot X must also be less than the rate of robot Z. Therefore, the rate of robot Z must be the greatest.

The answer is C.

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

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.

(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.

Hence, both statements are required.

C

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...

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

statement 1 & 2 together

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

so answer is C

Rx= X/Z
Ry= Y/Z
X,Y, and Z are positive

Stmt 1: Rx < Ry
X/Z < Y/Z
therefore, X< Y and Z>1 but no relation between X,Y, and Z
N.S.

Y< 1. N.S

X<Y<1 and Z>1 sufficient
Answer is C

Posted from GMAT ToolKit

Hi Brunel - can you explain how you reached "y<z" for second statement? Thanks

Hi

I will try to explain. Ry is the ratio of Y's rate of work (which he has taken as y) to Z's rate of work (which he has taken as z). So Ry = y/z. Now in second statement it is given that Ry < 1 which means y/z < 1.

Now since z is a positive quantity (rate of working cannot be negative here), we can multiply both sides of this inequality by z (and the sign of inequality will NOT change since z is positive). So:

y/z * z < 1*z Or y < z.

Dear havoc7860
your approach is utterly right for the DS questions because you could get the answer only when you bound two statements together.
For DS the only matter is can you obtain the definite yes or no answer. Never mind that the variables are in the reversed order

I did similar mistakes that mvictor alluded by misinterpreting final ration "R" as a Work/Time = 1/R

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