If b is an even integer is b < 0 ?

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If b is an even integer is b < 0 ?

by Vincen » Fri Sep 15, 2017 7:13 pm
If b is an even integer is b < 0 ?

(1) b^2 - 4b + 4 < 16
(2) b^2 > 9

The OA is A.

I know the option (2) alone is not sufficient, but I could not determine if the statement (1) is sufficient. Can any expert help me here.

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by GMATGuruNY » Fri Sep 15, 2017 7:36 pm
Vincen wrote:If b is an even integer is b < 0 ?

(1) b^2 - 4b + 4 < 16
(2) b^2 > 9

The OA is A.

I know the option (2) alone is not sufficient, but I could not determine if the statement (1) is sufficient. Can any expert help me here.
Statement 1: b² - 4b + 4 < 16
Rephrased:
(b-2)² < 16.
If b=-2, then (b-2)² = (-2-2)² = (-4)² = 16.
If b=-4, then (b-2)² = (-4-2)² = (-6)² = 36.
If b=-6, then (b-2)² = (-6-2)² = (-8)² = 64.
Notice the PATTERN.
When b=-2, (b-2)² = 16.
As b becomes more negative, (b-2)² becomes BIGGER.
Implication:
There is no negative even value for b such that (b-2)² < 16.
Since b cannot be negative, the answer to the question stem is NO.
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by Mo2men » Sat Sep 16, 2017 6:45 am
GMATGuruNY wrote:
Vincen wrote:If b is an even integer is b < 0 ?

(1) b^2 - 4b + 4 < 16
(2) b^2 > 9

The OA is A.

I know the option (2) alone is not sufficient, but I could not determine if the statement (1) is sufficient. Can any expert help me here.
Statement 1: b² - 4b + 4 < 16
Rephrased:
(b-2)² < 16.
If b=-2, then (b-2)² = (-2-2)² = (-4)² = 16.
If b=-4, then (b-2)² = (-4-2)² = (-6)² = 36.
If b=-6, then (b-2)² = (-6-2)² = (-8)² = 64.
Notice the PATTERN.
When b=-2, (b-2)² = 16.
As b becomes more negative, (b-2)² becomes BIGGER.
Implication:
There is no negative even value for b such that (b-2)² < 16.
Since b cannot be negative, the answer to the question stem is NO.
SUFFICIENT.
Dear Mitch,

For statement 1:

If b = 0 ..........then 4 < 16

If b = 2...........then 0 < 16

If b = 4............then 4 <16

How come A is sufficient?????

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by Mo2men » Sat Sep 16, 2017 6:49 am
Mo2men wrote:
GMATGuruNY wrote:
Vincen wrote:If b is an even integer is b < 0 ?

(1) b^2 - 4b + 4 < 16
(2) b^2 > 9

The OA is A.

I know the option (2) alone is not sufficient, but I could not determine if the statement (1) is sufficient. Can any expert help me here.
Statement 1: b² - 4b + 4 < 16
Rephrased:
(b-2)² < 16.
If b=-2, then (b-2)² = (-2-2)² = (-4)² = 16.
If b=-4, then (b-2)² = (-4-2)² = (-6)² = 36.
If b=-6, then (b-2)² = (-6-2)² = (-8)² = 64.
Notice the PATTERN.
When b=-2, (b-2)² = 16.
As b becomes more negative, (b-2)² becomes BIGGER.
Implication:
There is no negative even value for b such that (b-2)² < 16.
Since b cannot be negative, the answer to the question stem is NO.
SUFFICIENT.
Dear Mitch,

For statement 1:

If b = 0 ..........then 4 < 16

If b = 2...........then 0 < 16

If b = 4............then 4 <16

How come A is sufficient?????
Sorry Mitch, I did not read your post thoroughly. :):)