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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] New post 31 Mar 2009, 16:00
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C
D
E

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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

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Re: DS - Rate problem / Ration problem??? - OG 12 [#permalink] New post 31 Mar 2009, 18:03
stmnt 1 - not suffic.
a<b => x/z < y/z => x< y
nothing about z

stmnt 2 - b<1 - not suffic.
nothing about a or z

let's combine - b<1 => a<b and a<1
still not sufficient, z could be >,= or < than a and/or b, and it would still satisfy stmnt 1 and 2.

I think it's E.
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Re: DS - Rate problem / Ration problem??? - OG 12 [#permalink] New post 01 Apr 2009, 00:18
for 1 and 2
a < b < 1
x/z < y/z < 1
denominator z is greater than numerator. Hence C is answer.
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Re: DS - Rate problem / Ration problem??? - OG 12 [#permalink] New post 01 Apr 2009, 01:27
I think E
Lets take V as Robot Z's constant rate

is Robot Z's constant rate the greatest of the three?

X:Y:Z=Va:Vb:V
now to have V highest we need to know whether V is <0 or >0
V could be 2 or say, 1/2
it changes the whole picture



Robots X, Y, and Z each assemble components in their respective constant rates. If a is the ration of Robot X's constant rate to Robot Z's constant rate and b 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) a < b
2) b < 1
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Re: DS - Rate problem / Ration problem??? - OG 12 [#permalink] New post 01 Apr 2009, 15:39
My take is C

X:Z=a; Y:Z=b
Lets start with option 2 b<1
that means Z is bigger than Y.

1 says a<b so again Z is bigger than X.


Lets solve it by statements

Y= bZ Y is b times faster than Z if b is less than 1 then Z is faster on the similar note A can deduced.

@A Nitya .... Picture will not change as long as V is greater zero and V can not be negative as rate will always be positive.
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Re: DS - Rate problem / Ration problem??? - OG 12 [#permalink] New post 01 Apr 2009, 17:03
hemantsood wrote:
My take is C

X:Z=a; Y:Z=b
Lets start with option 2 b<1
that means Z is bigger than Y.

1 says a<b so again Z is bigger than X.


Lets solve it by statements

Y= bZ Y is b times faster than Z if b is less than 1 then Z is faster on the similar note A can deduced.

@A Nitya .... Picture will not change as long as V is greater zero and V can not be negative as rate will always be positive.


I think you reasoning is correct, the answer should be C, not E.
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Re: Robots X, Y, and Z each assemble [#permalink] New post 03 Jan 2011, 20:26
We need both the information if the constant rate for Z is greater or not.
1. says that constant rate of Y is greater than X. x/z < x/y
2. says that constant rate of Y is smaller than Z. y/z <1

so we need both the information.
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Re: Robots X, Y, and Z each assemble [#permalink] New post 04 Jan 2011, 02:33
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ajit257 wrote:
see attachment

Can some explain the reasoning behind this ques.


Robots X, Y, and Z each assemble components at their respective constant rates. If rx is the ratio of robot X's constant rate to robot Z's constant rate and ry 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.
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Re: Robots X, Y, and Z each assemble   [#permalink] 04 Jan 2011, 02:33
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