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If y is an integer, is y≤−1?

(1) x+z2≤0
(2) y<x

1. z^2 always a positive value. even if z= 0 or -ve or +ve, So if x + z^2 <= 0, x= -ve or 0, Insufficient
2. y/,x, insufficient
on combining y will be always negative. So if x = 0, y = -1, if x = -1, y = -2. As y takes only integer values. In any case y is always <= -1.
Ans : C
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If y is an integer, is \(y\leq{-1}\)?

(1) \(x+z^2\leq{0}\)
(2) \(y<x\)

Statement 1: No mention of y at all and there are no specific hints from the question stem itself either. No sufficient.

Statement 2: Since we do not know anything about x, y can be any integer. Not sufficient.

Combining the 2: From S1, z^2 has to be positive, so for x+z^2 to be <0, x needs to be a negative number with a greater magnitude than z^2. From S2, clearly, y<x so y has to be equal to or < -1

Final answer : C
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palaknayyar, can we assume that z is an integer?

if z=.1, then y <=-.01, which does not help/would make the answer E.

However, if we assume z is an integer, then we can figure that z^2 >= 1, which then, of course, y=<-1 and we can answer C...

thoughts?
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Rtejpar
palaknayyar, can we assume that z is an integer?

if z=.1, then y <=-.01, which does not help/would make the answer E.

However, if we assume z is an integer, then we can figure that z^2 >= 1, which then, of course, y=<-1 and we can answer C...

thoughts?


If z=01, then \(x\leq -0.01<0\)

So y<x<0.
Since y is an integer, y is at the most -1 => \(y\leq -1\)
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