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m1033512
IMO A


x = y/|y|

when y <0 , we get x = 1

when y y>0 , we get x = 1

so clear anser No

sufficient

statement 2 says x <0 , but can be be anything


so not sufficient


IMO A

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m1033512 I believe when y<0, x = \(\frac{y}{|y|}\) would give us -1
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we will have both as -y in numerator and denomintor both

it will cancel out to 1

Posted from my mobile device
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m1033512
we will have both as -y in numerator and denomintor both

it will cancel out to 1

Posted from my mobile device

m1033512
Friend we'll only have -y in the denominator, not the numerator. Numberator will stay y(which has been defined as negative by us)

for eg:- let y be -5
\(\frac{y}{|y|} = \frac{-5}{|-5|}\)
=-5/5
=-1
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but I defined two cases for y ( removed the mod)

first

y<0 , then both are negative, and will take the same value so 1

second y > 0 , both are positive and will take the same value so 1

so for any y != 0, we always get 1

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m1033512
but I defined two cases for y ( removed the mod)

first

y<0 , then both are negative, and will take the same value so 1

second y > 0 , both are positive and will take the same value so 1

so for any y != 0, we always get 1

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m1033512

Friend, I agree that you've defined two cases for y
for y > 0 it is clearly 1

but I'm talking about the case where you have taken y < 0

\(\frac{y}{|y|}\) for any number less than 0 will give us -1

I believe the mistake you're making is while removing mod; I believe you are multiplying both the numerator as well as the denominator with a -ve in the process, which you shouldn't as there is only 1 mod that is being removed.

Try solving X*|Y| = Y for any negative Y, it's the same question in different form.
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St1:
\(x = \frac{y}{|y|}\)
it shows, either x = 1 (when y >0) or -1 (when y <0)
Hence, Not SUFFICIENT

St2:\(|x| = -x\)
It means either x = 0, or x<0
Hence, Not SUFFICIENT

Combining St 1 & 2, x =-1.

SUFFICIENT

Answer C

gmatbusters
If \(y ≠ 0\), is x = -1 ?


(1) \(x = \frac{y}{|y|}\)

(2) \(|x| = -x\)


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gmatbusters
If \(y ≠ 0\), is x = -1 ?


(1) \(x = \frac{y}{|y|}\)

(2) \(|x| = -x\)


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(1) if y is -ve, then x =-1 whereas when y is +ve, then x is positive. Insufficient

(2) x is a negative number. insufficient

(1)+(2) only possible value of x is -1. Sufficient

C is correct
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