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Is (y+1)^2 1? (1) y^2 4 (2) y+1 > 1y

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Bunuel
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Is (y+1)^2 1? (1) y^2 4 (2) y+1 > 1y [#permalink] New post   16 Apr 2021, 01:30
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45% (medium)

Question Stats:

63% (01:33) correct 37% (01:45) wrong based on 103 sessions
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Is \((y+1)^2 ≥ 1\)?


(1) \(y^2 ≥ 4\)

(2) \(y+1 > 1−y\)
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D
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D
QuantMadeEasy
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Re: Is (y+1)^2 1? (1) y^2 4 (2) y+1 > 1y [#permalink] New post   16 Apr 2021, 01:51
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Quote:
Is (y+1)^2≥1?

Rephrasing the question: Is |y+1| ≥1?
Rephrasing the question: Is y≥0 or -2≥y?

(1) y^2≥4
Simplifying y≥2 or -2≥y
Sufficient

(2) y+1>1−y
Simplifying 2y>0
y>0
Sufficient

D is correct
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Re: Is (y+1)^2 1? (1) y^2 4 (2) y+1 > 1y [#permalink] New post   17 Apr 2021, 00:57
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Is (y+1)^2 ≥1 ?
or, y+1 <=-1 or, y+1 >=1 and, y <=-2 or y>=0?

Stat1: y^2≥4
y<=-2 or y>=2, which satisfies above condition. Sufficient.

Stat2: y+1>1−y
or y> 0. again satisfies above condition. Sufficient.

So, I think D. :)
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Re: Is (y+1)^2 1? (1) y^2 4 (2) y+1 > 1y [#permalink] New post   19 Apr 2021, 01:15
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Is \((y+1)^2 ≥ 1\)?

Is \((y+1)^2 \geq{1}\)?

Take the square root: is \(|y+1|\geq 1\)?

Is \(y \geq 0\) or \(y\leq{-2}\)?

So, the question asks whether y is from the ranges shown below:




(1) \(y^2 ≥ 4\) --> \(y\leq{-2}\) or \(y\geq{2}\). The answer to the question is YES. Sufficient.

(2) \(y+1 > 1−y\) --> \(y > 0\). The answer to the question is YES. Sufficient.


Answer: D.

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Re: Is (y+1)^2 1? (1) y^2 4 (2) y+1 > 1y [#permalink] New post   30 Apr 2021, 15:16
Why can't we solve writing the formula as Y^2 + 2Y +1 , then (Y+1)(Y+1) >= 1

Then Y>=0

I know there is something wrong with this method but just don't know what it is, can someone explain why this won't work?
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Re: Is (y+1)^2 1? (1) y^2 4 (2) y+1 > 1y [#permalink]
30 Apr 2021, 15:16
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Bunuel
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