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Re: If ab >0, is (ab)^2 < (ab)^(1/2) ? [#permalink]
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Abhi077 wrote:
If \(ab >0\), is \((ab) ^2\) < \(\sqrt{ab}\)

1.\(\frac{4}{a}\) > \(7b\)
2. \(a-16 > -16\)


With the correction, it's much easier to analyze the question algebraically. As I explained above, the only way (ab)^2 < √(ab) is true is if 0 < ab < 1, so that's what we want to know: is ab < 1?

Using only Statement 1, we have two cases:

- if a > 0, then we can multiply both sides of the inequality by a without reversing the inequality. So we then have

4/a > 7b
4 > 7ab
4/7 > ab

So in this case, it is indeed true that ab < 1.

- if a < 0, then we can again multiply both sides of the inequality by a, but because we're multiplying by a negative number, we must reverse the inequality:

4/a > 7b
4 < 7ab
4/7 < ab

So when a (and also b of course, since ab > 0) is negative, then the value of ab can be anything greater than 4/7, and can indeed be larger than 1. It is easy enough to confirm this just by choosing numbers: we might have, say, that a = -4 and b = -100, for example, in which case 4/a > 7b is clearly true, since -1 > -700, and in this case ab is much larger than 1.

Statement 2 alone is clearly not sufficient, since it only tells us a > 0. But once we know that a > 0, that rules out the second case above in our analysis of Statement 1. So we know only the first case is possible, and that ab < 4/7, so the two statements together are sufficient, and the answer is C.
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Re: If ab >0, is (ab)^2 < (ab)^(1/2) ? [#permalink]
Abhi077 wrote:
If \(ab >0\), is \((ab) ^2\) < \(\sqrt{ab}\)

1.\(\frac{4}{a}\) > \(7b\)
2. \(a-16 > -16\)


So i would say, nice :-o , Lets see, given is important here ab> 0

Lets manipulate :roll:\((ab)^4\) < ab

take common and it becomes => (ab) {\((ab)^3\) -1 } < 0,For representation purpose only (x) (y) < 0

Now for the above to be true, it can be that x > 0 & y< 0 or x < 0 & y > 0, So from given, only bold is valid

which means we have to only focus on (ab)^3 < 1 part

from 1) \(\frac{4}{a}\) > \(7b\)

Can be true when a=-1, b=-1, this makes the question as No
Can be true when a=1, b=-1, this makes the question as Yes

From 2) a-16 > -16, this can only be true when a> 0, dont know anything about B

When we combine we can directly pinpoint on the condition that

a > 0 and b < 0

Sufficient to answer the question

C
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Re: If ab >0, is (ab)^2 < (ab)^(1/2) ? [#permalink]
Abhi077 wrote:
If \(ab >0\), is \((ab) ^2<\sqrt{ab}\)

(1) \(\frac{4}{a}>7b\)
(2) \(a-16 > -16\)


\(ab>0…{{a,b}}=same.sign≠0\)
\((ab) ^2<\sqrt{ab}…ab=x…x^2<\sqrt{x}\)?
rearrange: \(x^2<\sqrt{x}…x^4<x…x(x^3-1)<0\)?
if \(x<0\) is \((x^3-1)>0…x^3>1…x>1…ab>1\); invalid, since \(x<0\);
if \(x>0\) is \((x^3-1)<0…x^3<1…x<1…ab<1\)? find this.

(1) \(\frac{4}{a}>7b\):
if \({{a,b}}=positive…\frac{4}{a}>7b…4/7>ab…ab<1\)
if \({{a,b}}=negative…\frac{4}{-a}>7•(-b)…4/7<ab…ab>1\)
two answers insufic.

(2) \(a-16 > -16\): \(a>0…b>0\) but we don't know if \(ab<1\), insufi.

(1&2): \(a>0\) then \({{a,b}}=positive…\frac{4}{a}>7b…4/7>ab…ab<1\); sufic.

Answer (C)
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Re: If ab >0, is (ab)^2 < (ab)^(1/2) ? [#permalink]
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