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mirhaque
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Disagree with OA. if P^3>=-125, and P is an integer, p=? 29 Jul 2005, 17:48
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Disagree with OA.

if P^3>=-125, and P is an integer, p=?

1. 3p+1616



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Disagree with OA. if P^3>=-125, and P is an integer, p=? 29 Jul 2005, 20:14
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mirhaque
Disagree with OA.

if P^3>=-125, and P is an integer, p=?

1. 3p+16<=4
2. p^2>16
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E for me too

From the problem p>= -5, p could be -5, -4......to infiniti

From condition 1 add -16 to both sied 3p <= -12 or p <= -4, so it could be -4 to -infiniti, from the problem statement p could be -4 or -5

From condition 2 p > +/-4 so insufficient

Combining both does not help either..
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Disagree with OA. if P^3>=-125, and P is an integer, p=? 30 Jul 2005, 00:40
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From the stem P>=-5
From A) P<=-4 so it is insuficient since we have 2 values -4,-5
From B) p<-4, p>4 it is sufficient since when combined with stem we have a unique solution -5
So the ans should be B) IMHO..
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Disagree with OA. if P^3>=-125, and P is an integer, p=? 30 Jul 2005, 08:16
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BG wrote
From B) p<-4, p>4 it is sufficient since when combined with stem we have a unique solution -5

I never considered the sign change in the inequality......Sorry for the wrong answer, Thanks BG, hopefully this will help me in the future..
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Disagree with OA. if P^3>=-125, and P is an integer, p=? 30 Jul 2005, 12:04
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C

The only solution after combining 1 and 2 will be -5

B is incorrect. Consider 6,7,8
they are all >= -5 (stem condition)
and >4 ( clause 2 condition)
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Disagree with OA. if P^3>=-125, and P is an integer, p=? 30 Jul 2005, 14:33
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stem says that P>= -5 (-5,-4,-3,-2...)

1: From (1), p 4 (5,6,7,8...) or p<-4 (-5,-6,-7...) (Also out when you compare with stem)

Combining both, there are alot of intersections (-5, -6...), so (E) for me.



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