Bunuel wrote:
russ9 wrote:
2 questions:
1) What is +ves?
2) If the stem read that x & y could be positive OR negative, that would mean that only statement one is sufficient. Is that correct?
My reasoning being, the second inequality could become \(x<+-10<y\). Meaning, x would have to be less than -10, lets call it -25 and y would have to be 4 which would yield a "no" for the question. Additionally, if we took the positive squareroot, the question stem would yield a yes. Am I thinking about this the right way?
1. +ve = positive.
2. If we were not told that x and y are positive, then the answer would be C, not A:
Is x < 10 < y?(1) x < y and xy = 100. If x=-20 and y=-5, then the answer is NO but if x=5 and y=20, then the answer is YES. Not sufficient.
Notice that from xy = 100 we can deduce that x and y have the same sign.(2) x^2 < 100 < y^2 --> -10 < x < 10 and |y|>10. So, y can be more than 10 as well as less than -10. Not sufficient.
(1)+(2) Since x < y, then y < -10 is not possible, thus y > 10. So, we have that x < 10 < y. Sufficient.
Answer: C.
Hope it's clear.
Hi Bunuel,
I see how you can prove that 1 is NOT sufficient although I'm having a hard time with #2. The equality reads x^2 < 100 < y^2. Doesn't that yield that x < +/- 10? Wouldn't that make it is x<10 or -x>10? I can tell that my signs are off but if I follow the math, they seem fine. What am I missing here? Assuming that this part of the problem is resolved(as I can see the light at the end of the tunnel), I still don't see how you get insufficient.
Assuming that the correct inequalities are -10<x<10 and 10<y<-10, are you saying that since the final inequality can be y<-10<x or x<10<y, therefore insufficient? But wouldn't we say that only x<10<y pertains to the main equation and therefore sufficient?
Hope my question is clear.