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DS - 1

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Source: — Data Sufficiency |
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by JJJooe » Wed Aug 26, 2009 3:23 pm
(1) xyz = 1

x,y,z are all positive or two of them are negative
xy=1/z
z is unknown
so xy is also unknown

insufficient


(2) xz > 1
x and z are either both negative or both positive
insufficient

TOGETHER
y is positive
y=1/xz<1
0<y<1

but z is still unknown so x is unkonwn
insufficient

so choose E
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by fruti_yum » Sun Aug 30, 2009 12:55 pm
JJJooe wrote:(1) xyz = 1

x,y,z are all positive or two of them are negative
xy=1/z
z is unknown
so xy is also unknown

insufficient


(2) xz > 1
x and z are either both negative or both positive
insufficient

TOGETHER
y is positive
y=1/xz<1
0<y<1

but z is still unknown so x is unkonwn
insufficient

so choose E

IMO C

I'm sure you will agree that both statements taken indiviually are insufficient

together
we have xyz = 1
xz > 1

let xz = 2
x = 1 or 2

for xyz to equal 1
if x = 2, y = 1 and z = 1/2
or x =2, y = 1/2 and z = 1

in both cases xy is less than 1
Sufficient
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by crackgmat007 » Sun Aug 30, 2009 1:31 pm
Together
we have xyz = 1
xz > 1

let
A. x = 2, y = 1/2, z = 1
B. x = 10, y = 1/2, z = 1/5

Solving we get, xy < 1 for A and xy > 1 for B. Hence E.

Can someone solve the above algebriacally?
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Re: DS - 1

by woo » Sun Aug 30, 2009 7:33 pm
Xbond wrote:Hi there,

Could you explain in the simplest this concept and how to resolve it


Is xy > 1 ?

(1) xyz = 1
(2) xz > 1
We all know that the answer is either C or E.

First we combine condition 1 and 2.
we know that xz is greater than 1 therefore, in order for condition 1 to hold y must be equal to 1/xz.

Then xy is equal to 1/z. Now we need to see whether 1/z is greater than 1. From condition 2 we know that z can take any value, thus 1/z can be both greater than 1 and less than 1.

Therefore, IMO E.
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