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# If w 0, what is the value of w?

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Re: If w 0, what is the value of w? [#permalink]
gmatophobia can you elaborate statement 1?
we have w>0 and w<1 that is 0<w<1
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Re: If w 0, what is the value of w? [#permalink]
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Re: If w 0, what is the value of w? [#permalink]
If $$w ≠ 0$$, what is the value of $$w$$?

(1) $$w^2 – w^3 > 0……..w^2(1-w)>0$$
Now $$w^2>0$$, so $$1-w>0$$ or $$w<1$$
w could be anything less than 1.
Insufficient

(2) $$w^4 = \frac{1}{256}……….w^4=(\frac{1}{6}^4=(\frac{1}{-6})^4$$
So w is either $$\frac{1}{6}$$ or $$\frac{1}{-6}$$
Insufficient

Combined
Even after combining, w can be either $$\frac{1}{6}$$ or $$\frac{1}{-6}$$
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If w 0, what is the value of w? [#permalink]
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AnujL wrote:
gmatophobia can you elaborate statement 1?
we have w>0 and w<1 that is 0<w<1

Hey AnujL

In statement 1 we have

$$w^2–w^3$$ > 0

Let's take $$w^2$$ common

$$w^2(1-w) > 0$$

We see the product of two numbers is greater than 0 (i.e. positive), this can happen in two scenarios

1) Both numbers are negative
2) Both numbers are positive

Extending the concept here, either both $$w^2$$ & (1-w) should be negative or both should be positive.

However, $$w^2$$ cannot be negative (because square of any real number is non - negative), so we can rule out case 1 (both numbers are negative). Hence, the only scenario under which this expression can be true is when both $$w^2$$ and (1-w) are positive.

We know $$w^2$$ is positive in this case

Therefore 1 - w should be positive as well.

1-w > 0

i.e. w < 1

Note: w can be anything (i.e. it can be positive or it can be negative). We don't have any restriction on w except that the value of w should be less than 1 and it cannot be 0.

Hope this clarifies.
If w 0, what is the value of w? [#permalink]
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