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.