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Re: Is integer y > 0?
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27 Jul 2016, 06:44
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Bunuel wrote:
Is integer y > 0?
(1) –(2 + y) > 0
(2) (2 + y)² > 0
Target question:Is integer y GREATER THAN 0?
Statement 1: –(2 + y) > 0 Expand to get: –2 - y > 0 Add y to both sides to get: -2 > y If y is LESS than -2, we can be certain that y is NEGATIVE. This means we can conclude that y is definitely not greater than 0 Since we can answer the target question with certainty, statement 1 is SUFFICIENT
Statement 2: (2 + y)² > 0 Hmmm, the square of some value is ALWAYS greater than or equal to zero. So, this statement doesn't seem to tell us much. Since this statement doesn't FEEL sufficient, I'll TEST some values. There are several values of y that satisfy statement 2. Here are two: Case a: y = 1. Here, (2 + y)² = (2 + 1)² = 9, and 9 > 0. In this case, y is GREATER THAN 0 Case b: y = -1. Here, (2 + y)² = (2 + -1)² = 1, and 1 > 0. In this case, y is NOT greater than 0 Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT
Re: Is integer y > 0?
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17 Mar 2020, 03:28
The question asks if y > 0.
Statement-1 states that –(2 + y) > 0 We know that when we multiply with -1 in an inequality the sign changes. Hence 2+y <0 or we get y<-2. And hence y >0 is false. Or, Statement is Sufficient.
Cancel BCE. Keep AD.
Statement-2 states that (2 + y)^2 > 0 But square of anything will always be greater than zero. Hence if we plug in with y=-1 or y=1, we also get the same result. In other words y can be negative or positive. Hence Insufficient.
Re: Is integer y > 0?
[#permalink]
15 Jun 2021, 11:17
As per statement 1, you can see that if you plug any value of y<-2, the inequality will hold true. So, the first statement makes it clear that y<-2 and not y>0, as being questioned.
Therefore, SUFFICIENT.
However, as per the second statement, the value on the LHS will always be true for any real value of y. Therefore, Statement 2 is NOT SUFFICIENT.
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