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If ab >0, is (ab)^2 < (ab)^(1/2) ?
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Updated on: 16 Feb 2019, 03:18
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If \(ab >0\), is \((ab) ^2<\sqrt{ab}\) (1) \(\frac{4}{a}>7b\) (2) \(a16 > 16\)
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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) ?
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Updated on: 15 Feb 2019, 22:06
Abhi077 wrote: If \(ab >0\), is \((ab) ^2\) < \(\sqrt{ab}\)
1.\(\frac{4}{a}\) > \(7ab\) 2. \(a16 > 16\) 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 nowedited 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) ?
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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. \(a16 > 16\) 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) ?
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15 Feb 2019, 22:02
Abhi077 wrote: If \(ab >0\), is \((ab) ^2\) < \(\sqrt{ab}\)
1.\(\frac{4}{a}\) > \(7b\) 2. \(a16 > 16\) 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) ?
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15 Feb 2019, 22:18
Abhi077 wrote: If \(ab >0\), is \((ab) ^2\) < \(\sqrt{ab}\)
1.\(\frac{4}{a}\) > \(7b\) 2. \(a16 > 16\) So i would say, nice , Lets see, given is important here ab> 0 Lets manipulate \((ab)^4\) < 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) a16 > 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) ?
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31 Aug 2019, 09:32
Abhi077 wrote: If \(ab >0\), is \((ab) ^2<\sqrt{ab}\)
(1) \(\frac{4}{a}>7b\) (2) \(a16 > 16\) \(ab>0…{{a,b}}=same.sign≠0\) \((ab) ^2<\sqrt{ab}…ab=x…x^2<\sqrt{x}\)? rearrange: \(x^2<\sqrt{x}…x^4<x…x(x^31)<0\)? if \(x<0\) is \((x^31)>0…x^3>1…x>1…ab>1\); invalid, since \(x<0\); if \(x>0\) is \((x^31)<0…x^3<1…x<1…ab<1\)? 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) \(a16 > 16\): \(a>0…b>0\) 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)




Re: If ab >0, is (ab)^2 < (ab)^(1/2) ?
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31 Aug 2019, 09:32






