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# Is wz(x + y) a negative even integer?

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Is wz(x + y) a negative even integer?  [#permalink]

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17 Apr 2018, 04:41
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25% (medium)

Question Stats:

83% (01:05) correct 18% (01:13) wrong based on 58 sessions

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Is wz(x + y) a negative even integer?

(1) w, x, and y are positive odd integers.
(2) z is a negative even integer.

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Re: Is wz(x + y) a negative even integer?  [#permalink]

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17 Apr 2018, 05:28
Bunuel wrote:
Is wz(x + y) a negative even integer?

(1) w, x, and y are positive odd integers.
(2) z is a negative even integer.

we are looking for 2 things:
1. negative integer
2. this integer must be negative***

statement 1: here no information regarding sign ..............not sufficient.
statement 2: here only sign is given. Z is even. Look at the terms ........wz(x+y).......regardless of the value of x , y and w ...the result will always be even negative.

Thus B is the correct answer.
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Re: Is wz(x + y) a negative even integer?  [#permalink]

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17 Apr 2018, 09:13
IMO C

Statement (1)we only know that w,x and y are positive odd integers but no info. On z hence we can have both yes and no as answer. Thus insufficient

Case1- take w,x and y as 1,2,3 and z as any negative even integer. In this case answer will be yes
Case2-take w=2 x=1/2 and y=1/3
And z as any negative even integer. In this case the answer will not even be an integer.
Hence insuff.

Combining
If w,x and y are positive intgers and z is negative even integer, then the answer will always be yes. Thus sufficient
Hence C

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Re: Is wz(x + y) a negative even integer?  [#permalink]

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17 Apr 2018, 21:59
selim wrote:
Bunuel wrote:
Is wz(x + y) a negative even integer?

(1) w, x, and y are positive odd integers.
(2) z is a negative even integer.

we are looking for 2 things:
1. negative integer
2. this integer must be negative***

statement 1: here no information regarding sign ..............not sufficient.
statement 2: here only sign is given. Z is even. Look at the terms ........wz(x+y).......regardless of the value of x , y and w ...the result will always be even negative.

Thus B is the correct answer.

What if W is also negative? The product of W and Z would then be a positive, creating anotjer possible answer. Statement 2 is insufficient. Unless there's something I'm missing

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Re: Is wz(x + y) a negative even integer?  [#permalink]

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18 Apr 2018, 04:20
Bunuel wrote:
Is wz(x + y) a negative even integer?

(1) w, x, and y are positive odd integers.
(2) z is a negative even integer.

(1) If x and y are positive odd integers, then their sum will be a positive even integer. That multiplied by w, which is another positive odd integer, will also result in a positive even integer. So w*(x+y) is a positive even integer. But we dont know about z. z could be a positive or negative integer, even or odd integer, or it could be in decimals also. So we cannot determine whether w*z*(x+y) will be a negative even integer or not. Not sufficient.

(2) z is negative even, but nothing mentioned about others. So not sufficient.

Combining, w*(x+y) is positive even integer, and that multiplied by z, which is a negative even integer, will result in a negative even integer only. Sufficient.

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Re: Is wz(x + y) a negative even integer?  [#permalink]

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19 Apr 2018, 10:30
C it is.

Clearly statement A is not sufficient as we do not have information about the remaining integer.
Statement B has no ifnormation about other 3.

Combining , we are able to answer
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Re: Is wz(x + y) a negative even integer?  [#permalink]

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20 Apr 2018, 04:17
1
Is wz(x + y) a negative even integer?

IMO, 3 variables are used here.
odd number of negative variables will result in negative number,
even number of negative variables will result in positive number.

for number to be divisible by 2. we need at least one even variable.

With this in mind, lets dive into statements,

(1) w, x, and y are positive odd integers.
Here, we don't know about sign of Z. So statement is insufficient

(2) z is a negative even integer.
Okay, but we are not still sure of sign of 'w' or sign of 'x+y'. hence not sufficient.

(1) + (2) : we know Z is negative and even. w is odd and positive,
so product of wz is (odd)(-even)= -EVEN
and (x+y) is positive even.
So overall product is : -EVEN.

Hence C.
Re: Is wz(x + y) a negative even integer? &nbs [#permalink] 20 Apr 2018, 04:17
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