DS: Inequalities

[GMAT Kolaveri]

Is p a negative number?

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

(2) p^2 - 1 < 0

Source: grockit. OA: C

[aneesh.kg]

This question can be solved very easily using 'Critical Points Method'. I have explained the method in this post:
https://www.beatthegmat.com/critical-poi ... 10450.html

[shantanu86]

Is p a negative number?

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

(2) p^2 - 1 < 0

Source: grockit. OA: C

IMO its [spoiler][C][/spoiler]

Lets analyze-

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

=> either p^3 or 1-p^2 is negative and the other one is positive.

case 1. p^3 <0 & (1-p^2)>0
=> -1< p < 0

case 2. p^3 >0 & (1-p^2)<0
=> p > 1

Thus combining both cases its not guaranteed that p< 0. Hence Insufficient

(2) p^2 - 1 < 0

=> p^2 < 1
=> -1<p < 1

Thus insufficient

(1) & (2) together
p^3(1-p^2)<0 .. i
p^2 - 1 < 0 .. ii

from .. ii
1-p^2 >0
substituting in i
p^3 < 0
=> p<0

Hence, sufficient

Thus answer is [spoiler][C][/spoiler]

Hope it helps!!

[hey_thr67]

I tried with p = 0.5 and -0.5 So, for -1 < p < 1 , the answers can be + ve or -ve.. Please explain.
In statement 1:

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

readjusting the equations,

p^3(p-1)(1+p)>0

[aneesh.kg]

Great.

Now
Identify the critical points.
Mark them on the number line.
Draw the curve.
Select the required region.

If there's a doubt, read the post on Critical Points again: https://www.beatthegmat.com/critical-poi ... 10450.html
I would like to see you attempt it properly before giving away the solution.

[GMATGuruNY]

Is p a negative number?

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

(2) p^2 - 1 < 0

Source: grockit. OA: C

The critical points method is great.

For some test-takers, plugging in numbers might be easier and more efficient.
Typically, we need to consider the following types of numbers:
Zero
One (positive and negative)
Integers (greater than positive 1 and less than negative 1)
Fractions (positive and negative)

Statement 1: p^3(1 - p^2) < 0
p=0 and p=±1 make the left-hand side equal to 0, so none of them works here.

p=-2:
(-2)³(1-(-2)²) < 0
(neg)(neg) < 0.
Doesn't work, implying that p cannot be less than -1.

p=-1/2:
(-1/2)³(1-(-1/2)²) < 0
(neg)(pos) < 0.
This works, implying that p can be a negative fraction.

p=2:
(2)³(1-2²) < 0
(pos)(neg) < 0.
This works, implying that p can be greater than 1.

p=1/2:
(1/2)³(1-(1/2)²) < 0
(pos)(pos) < 0.
Doesn't work, implying that p cannot be a positive fraction.

Since p=-1/2 and p=2 both work, INSUFFICIENT.

Statement 2, rephrased: p²<1.
Thus, p can be a negative fraction, 0, or a positive fraction.
INSUFFICIENT.

Statements 1 and 2 combined:
The only type of value that satisfies both statements is a negative fraction such as p=-1/2.
SUFFICIENT.

The correct answer is C.

An alternate way to combine the statements:
Since p²-1<0 (statement 2), 1-p²>0.
Thus, statement 1 becomes:
p³(positive) < 0.
p³<0, implying that p is a negative number.
SUFFICIENT.

[krishna239455]

Dear Aneesh
I tried to solve this problem with the trick mentioned by you.
Statement 1: Critical points considered are 0,1,-1.
Two inequalities satisfying the function are P>1 and -1<P<0
Hence P can be both +ve and -ve , Not suffeceint.

Statement 2: Critical points considered are 1 and -1.
Only One inequality satisfying the function is -1<P<1
Again P can be both +ve and -ve, Not suffeceint

After combining both the statements -1<P<0 and -1<P<1, -1<P<0 is more restrictive than other and hence can help us to conclude that P is -ve.

But what about P>1 that we got in statement 1

[aneesh.kg]

Good job Krishna!

When we COMBINE two statements, we look for those values for which both Statement(1) and Statement(2) hold true. In other words, we look for the COMMON solution.

P > 1 is a solution from Statement(1) but not from Statement(2), so we do not include it in the common solution. When you plot these values on the number line, you will see that -1 < P < 0, as you mentioned, is the only common solution to the two statements.

[krishna239455]

Thanks Aneesh !!!