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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]
@brantgmatprepnow

please elaborate statement 1
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Re: If w 0, what is the value of w? [#permalink]
Expert Reply
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.
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If w 0, what is the value of w? [#permalink]
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