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Arithmetics - not so difficult DS problem from VStudy

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Arithmetics - not so difficult DS problem from VStudy [#permalink] New post 08 Jun 2004, 01:00
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A
B
C
D
E

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions
- Guys,
I encountered into not so difficult DS problem. I guess that VStudy answersheet gives wrong answer. (Book 1, Test 1)

Is x(y+z)>0
(1) xyz>0
(2) yz>0
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 [#permalink] New post 08 Jun 2004, 02:44
Hi,

Lets look at the second statement...
yz>0 => Case 1: y -> +ve and z ->+ve
x could be either +ve or -ve so (B) is not sufficient
for same reason (1) is also not sufficient
Assuming (1) and (2) both... again y n z could be
either +ve or -ve which means (y+z) could be either
so we cannot tell as x could also be either

Case 2: y -> -ve and z ->ve
Same reasoning!

Hence we cannot say anything... answer (E)

Another approach would be to draw a table like this

X Y Z X(Y + Z)

Statement 1 1 -1 -1 -2
(No inference) -1 1 1 -2
-1 -1 -1 2

Statement 2 -1 -1 1 0
(No inference) 1 -1 -1 -2
1 1 1 1

The combined condition is also covered among the above sample data
and no inference is possible
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 [#permalink] New post 08 Jun 2004, 03:29
thanx, manan,

the table is really helpful.
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 [#permalink] New post 08 Jun 2004, 06:58
my guess is C

from 2 we get y and z either + or - ve
from 1 we get either two of digits can be -ve and one can be +
so combine we get yz to be of same signs .
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Re: Arithmetics - not so difficult DS problem from VStudy [#permalink] New post 08 Jun 2004, 20:50
Solo wrote:
- Guys,
I encountered into not so difficult DS problem. I guess that VStudy answersheet gives wrong answer. (Book 1, Test 1)

Is x(y+z)>0
(1) xyz>0
(2) yz>0


A is not sufficient. For XYZ to be > 0, at least 2 variables has to be -ve or all 3 variables to be +ve.

B is also not sufficient. Both Y & Z can be -ve or +ve ( YZ must have same sign ) to get YZ> 0.

A & B together, let's look at condition with YZ> 0,
X Y Z YZ XYZ X(Y+Z)
1 2 3 6 6 6
1 -2 -3 6 6 -5

As you can see, X(Y+Z) can be +ve or -ve.

Thus, C is still not sufficient. The answer has to be E.
Joined: 31 Dec 1969
Location: United States
Concentration: Marketing, Other
GMAT 1: 710 Q49 V38
GMAT 2: 660 Q V
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WE: Accounting (Accounting)
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Re: Arithmetics - not so difficult DS problem from VStudy [#permalink] New post 08 Jun 2004, 20:53
Solo wrote:
- Guys,
I encountered into not so difficult DS problem. I guess that VStudy answersheet gives wrong answer. (Book 1, Test 1)

Is x(y+z)>0
(1) xyz>0
(2) yz>0


A is not sufficient. For XYZ to be > 0, at least 2 variables has to be -ve or all 3 variables to be +ve.

B is also not sufficient. Both Y & Z can be -ve or +ve ( YZ must have same sign ) to get YZ> 0.

A & B together, let's look at condition with YZ> 0,
X Y Z YZ XYZ X(Y+Z)
1 2 3 6 6 6
1 -2 -3 6 6 -5

As you can see, X(Y+Z) can be +ve or -ve.

Thus, C is still not sufficient. The answer has to be E.
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 [#permalink] New post 08 Jun 2004, 20:56
Ooops, forgot to sign my user name. It wasn't a guest.
  [#permalink] 08 Jun 2004, 20:56
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