If xy 0, is (xy)^2

Question

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

(1) 4/x > 7y
(2) x − 16 > −16

chetan2u · Answer

Hi, I think answer requires to be relooked into.. If xy > 0, is (xy)^2<\sqrt{xy} (xy)^2<\sqrt{xy} ........ (xy)^2-\sqrt{xy}<0 ...... \sqrt{xy}((xy)^\frac{3}{2}-1)<0 .. we know \sqrt{xy} is positive.. so ((xy)^\frac{3}{2}-1)<0 ...... (xy)^\frac{3}{2}<1 .. which means is xy<1... (1) \frac{4}{x}> 7y 7y-\frac{4}{x}<0 .. \frac{7xy-4}{x}<0 .. so either of the numerator or denominator is -ive.. (a) It is possible that x is -ive, so 7xy-4>0 or 7xy>4 or xy>\frac{4}{7} .. ans for xy<1 CAN be YES or NO.. (b) If x is +ive, 7xy-4<0 or 7xy<4 or xy<\frac{4}{7} .. ans is YES different answers Insuff (2) x − 16 > −16 x>0.. clearly insuff combined.. If x>0, xy<\frac{4}{7}. . so our answer is YES Suff C

Mo2men · Answer

Hi,

Can I square both sides in highlighted parts? I think both are positive. Is there any tricks?

Thanks

chetan2u · Answer

Yes, you can square both sides her as both are +ive..

SPatel1992 · Answer

Hi  chetan2u  ! Can you explain the other method where we square the stem?

TheNightKing · Answer

Given: xy>0 So both positive or both negative. Approaching to check if answer is C or E: Let's look at Statement 2 first: x-16+16>-16+16 x>0 So we know x is positive. We know y has to be positive now. Statement 1: 4/x>7y. We know x is positive. So y has to be positive. 4/1>7y 4>7y 4/7>y Let's take y as 1/4 and x as 1 (xy)^2<\sqrt{xy} (1/4)^2<1/2 Sufficient.

varshas044 · Answer

Given that xy > 0 ---- x and y have the same sign x < 0 & y < 0 OR x>0 and y>0 Statement 1: 4/x > 7y -------> 4/x – 7y > 0 ----------- (4-7xy) / x > 0 Now we don’t know the sign of x hence we can not multiply both sides by x If x > 0 then xy < 0.57 -------- 0 < xy < 0.57 ---------- it will give YES as an answer to the question If x < 0 the xy > 0.57 ------- xy > 0.57---------- now this can give YES or NO as an answer to the question INSUFFICIENT Statement 2: X – 16 > -16 ----------- x > 0 ------ if x>0 then y>0 since we know that xy>0 But we don’t know where the values of xy together lie If 0< xy < 1 then we get YES to the question If xy > 1 then we can NO to the question INSUFFICIENT Statement 1+ 2 By combining these we know that x>0 and y>0 and xy>0 The only case that will work is xy < 0.57 --------- hence we can get a conclusive answer SUFFICIENT Answer – C

GMATGuruNY · Answer

Both sides of the inequality in the question stem are positive. Implication: We can safely square both sides, with the following result: Is (xy)^4 < xy ? The answer will be YES only if 00 No information about y. INSUFFICIENT. Statements combined: Since x is positive, multiplying both sides of Statement 1 by x yields the following: 4 > 7xy 4/7 > xy Since xy>0 and xy < 4/7, we know that 0

If xy > 0, is (xy)^2<√xy ?

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