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Source: — Data Sufficiency |
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by shankar.ashwin » Mon Nov 14, 2011 9:20 pm
p< 0 ?

Statement 1:

p^3(1-p^2) < 0

p^3 < 0 (or) 1 - p^2 < 0 -> p^2 > 1 (which implies p>1)

2 cases insufficient.

Statement 2:

p^2 < 1

Now here p can be between '0' and '1' (+ve) or -ve. Again 2 cases - Insufficient.

Together, with both statements taken together we have p^3 < 0 and p<0, common to both. Hence we can say p<0. C IMO

karthikpandian19 wrote:Is p a negative number?

(1) p^3(1 - p^2) < 0

(2) p^2 - 1 < 0
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by neelgandham » Tue Nov 15, 2011 1:40 am
Is p a negative number?
Assumption:
p3 = p^3, p2 = p^2, p^3 = a and 1-p^2 = b
(1) p^3(1 - p^2) < 0
a*b <0, Implies either a > 0 and b < 0 Or a < 0 and b > 0
If a > 0 and b < 0, then p^3 > 0 and p > 0
If a < 0 and b > 0, then p^3 < 0 and p < 0
The value of p can be -ve or +ve. Hence, insufficient
(2) p^2 - 1 < 0
Implies, p^2 < 1 => -1 < p < 1
The value of p can be -ve or +ve. Hence, insufficient
From 1 and 2,

a * b < 0 and -b < 0 (i.e. b > 0). So now we have two in equations a*b <0 and b > 0.
Implies a < 0 => p^3 < 0 => p < 0 sufficient!
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by GMATGuruNY » Tue Nov 15, 2011 6:30 am
karthikpandian19 wrote:Is p a negative number?

(1) p3(1 - p2) < 0

(2) p2 - 1 < 0
Statement 1: p³(1-p²) < 0.
p³(1+p)(1-p) < 0.
The critical points are p=0, p=-1, and p=1.
These are the only values of p where p³(1+p)(1-p) = 0.
When p is any other value, p³(1+p)(1-p) <> 0.
To determine the range of p, try one value to the left and right of each critical point.

p<-1:
Plugging p=-2 into p³(1+p)(1-p) < 0, we get:
(-2)³(1+(-2))*(1-(-2)) < 0.
-8 * -1 * 3 < 0.
24<0.
Doesn't work.
p<-1 is NOT part of the range.

-1<p<0:
Plugging x=-1/2 into p³(1+p)(1-p) < 0, we get:
(-1/2)³(1+(-1/2))*(1-(-1/2)) < 0.
(-1/8)*(1/2)*(3/2) < 0.
-3/32<0.
This works.
-1<p<0 IS part of the range.

0<p<1:
Plugging p=1/2 into p³(1+p)(1-p) < 0, we get:
(1/2)³(1+(1/2))*(1-(1/2)) < 0.
(1/8)*(3/2)*(1/2) < 0.
3/32<0.
Doesn't work.
0<p<1 is NOT part of the range.

p>1:
Plugging x=2 into p³(1+p)(1-p) < 0, we get:
(2)³(1+2)(1-2) < 0.
8*3*(-1) < 0.
-24<0.
This works.
p>1 IS part of the range.

Thus, -1<p<0 OR p>1.
INSUFFICIENT.

Statement 2: p² - 1 < 0.
p² < 1.
Thus, p must be 0 or a fraction between -1 and 1:
-1<p<1.
INSUFFICIENT.

Statements 1 and 2 combined:
From statement 1: -1<p<0 OR p>1.
From statement 2: -1<p<1.
The only range included in both statements is -1<p<0.
SUFFICIENT.

The correct answer is C.
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