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If ab >0, is (ab)^2 < (ab)^(1/2) ?

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

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If \(ab >0\), is \((ab) ^2<\sqrt{ab}\)


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

(2) \(a-16 > -16\)

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Originally posted by Abhi077 on 15 Feb 2019, 10:39.
Last edited by Bunuel on 16 Feb 2019, 03:18, edited 2 times in total.
Renamed the topic and edited the question.
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Re: If ab >0, is (ab)^2 < (ab)^(1/2) ?  [#permalink]

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New post Updated on: 15 Feb 2019, 22:06
Abhi077 wrote:
If \(ab >0\), is \((ab) ^2\) < \(\sqrt{ab}\)

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


edit: the OP fixed a typo in the original post above, so this solution is to the question I have quoted above (with the typo in it) and not to the now-edited question at the top of the page (which I solve in a separate post below)

First, if we replace "ab" with "x", the question is asking:

Is x^2 <√x ?

and x^2 < √x is only true if 0 < x < 1, so the question is asking "Is ab < 1?"

Statement 1 tells us:

4/a > 7ab

Before determining if this is sufficient, notice that we know ab > 0. So the right side of this inequality is positive, which means 4/a is greater than some positive number, and therefore a itself is positive. So Statement 1 guarantees that a is positive. But that's the only information you get from Statement 2. So you don't learn anything new by combining the two statements that you don't know from Statement 1 on its own, and the OA provided cannot be correct -- the answer can only be A or E, since Statement 2 alone is clearly not sufficient.

Using Statement 1 alone (which we can rewrite, since a is positive, as 4 > 7a^2 b or as 4/7 > a^2 b if we want to), it's easy to see that ab < 1 is possible, using the values a = 0.1 and b = 1, say. But it's also easy enough to find values where ab > 1. We can take, say, a = 0.1 and b = 11.

So we can't say if ab < 1 is true, and the answer is E unless there's a typo in the question.
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Originally posted by IanStewart on 15 Feb 2019, 14:19.
Last edited by IanStewart on 15 Feb 2019, 22:06, edited 1 time in total.
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If ab >0, is (ab)^2 < (ab)^(1/2) ?  [#permalink]

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New post Updated on: 18 Feb 2019, 03:38
IanStewart wrote:
Abhi077 wrote:
If \(ab >0\), is \((ab) ^2\) < \(\sqrt{ab}\)

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


First, if we replace "ab" with "x", the question is asking:

Is x^2 <√x ?

and x^2 < √x is only true if 0 < x < 1, so the question is asking "Is ab < 1?"

Statement 1 tells us:

4/a > 7ab

Before determining if this is sufficient, notice that we know ab > 0. So the right side of this inequality is positive, which means 4/a is greater than some positive number, and therefore a itself is positive. So Statement 1 guarantees that a is positive. But that's the only information you get from Statement 2. So you don't learn anything new by combining the two statements that you don't know from Statement 1 on its own, and the OA provided cannot be correct -- the answer can only be A or E, since Statement 2 alone is clearly not sufficient.

Using Statement 1 alone (which we can rewrite, since a is positive, as 4 > 7a^2 b or as 4/7 > a^2 b if we want to), it's easy to see that ab < 1 is possible, using the values a = 0.1 and b = 1, say. But it's also easy enough to find values where ab > 1. We can take, say, a = 0.1 and b = 11.

So we can't say if ab < 1 is true, and the answer is E unless there's a typo in the question.


Yes there was a typo. I'm so sorry. just corrected it
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Originally posted by Abhi077 on 15 Feb 2019, 21:17.
Last edited by Abhi077 on 18 Feb 2019, 03:38, edited 1 time in total.
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Re: If ab >0, is (ab)^2 < (ab)^(1/2) ?  [#permalink]

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New post 15 Feb 2019, 22:02
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]

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New post 15 Feb 2019, 22:18
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]

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New post 31 Aug 2019, 09:32
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] 31 Aug 2019, 09:32
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