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X and Y are positive numbers. If (X^2)*Y>1 and

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X and Y are positive numbers. If (X^2)*Y>1 and [#permalink] New post 02 Sep 2004, 08:20
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A
B
C
D
E

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X and Y are positive numbers. If (X^2)*Y>1 and (Y^2)*X<1, which of the following must be true?

I. X>1
II. (X^2)>1
III. X*Y>1

A) III only
B) I and II only
C) I and III only
D) II and III only
E) I, II and III

The answer could be worked out but it would be quite tedious. Would appreciate if someone could come up with a very fast and intelligent way of solving this question. The shortest will be the best. :)

Thanks in advance,

Cheers!
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 [#permalink] New post 02 Sep 2004, 08:44
I got B and it took me about 2 min. although I did it intuitively.
Since X & Y are positive, statements 1 and 2 are same
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 [#permalink] New post 02 Sep 2004, 08:46
Is the answer C??? If it is i'll make an attempt to put my reasoning into words coz kinda dubious......
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 [#permalink] New post 02 Sep 2004, 09:02
My answer would be E. Don't know what other easy way to solve the problem. Any ideas?
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 [#permalink] New post 02 Sep 2004, 09:13
Let X = 2 and Y = 1/3
From first condition: 2^2 * 1/3 = 4/3 --> ok
From second condition: (1/3)^2 * 2 = 4/9 --> ok
Yet, 1/3 * 2 = 2/3 and it is less than 1
III needs not be true
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 [#permalink] New post 02 Sep 2004, 09:23
We have X^2*Y>1 and Y^2*X<1
=>X^2*Y>1 and -Y^2*X>-1 (the > becomes < when the sign changes).
In Inequalities you can add these two equations PROVIDED the > or < signs are same for both the equations.

So we now have X^2*Y - Y^2*X > 0
=>XY*(X-Y) > 0
=> XY > 0 or X>Y

So XY > 0 not >1 So III need not be true.

let X= 4/3 and Y = 3/4. This will satisfy X^2*Y>1 and Y^2*X<1

I and II must be true. While III can be XY=1.

Paul had a great point - if I is true, II is true

So B is the answer.
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 [#permalink] New post 02 Sep 2004, 10:14
I got B as answer too.

I just used the number picking method.

Maybe it is easier to write first:

y^2 * x < 1 < x^2 * y

Then pick numbers.

III may be true but is not necessary.

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 [#permalink] New post 03 Sep 2004, 19:26
Yes, the answer is B.

Actually, what I wanted is to solve this fastest possible and most accurate possible. Plug in numbers could be quite tedious (especially if one do not know what are the possible ranges of numbers to plug in).

My working is this:

(1): X^2*Y>1
(2): Y^2*X<1


Now, before we proceed further, we must understand that from answer choices (A) to (E), there is none that permits us to choose "None of the above"! In addition, if condition (I): X>1 is true, then condition (II): X^2>1 must also be true.

So, if we look at the answer choices again, we can eliminate (C) and (D). And among (A), (B) and (E), we just have to prove that (A) is not NECESSARILY true to eliminate BOTH (A) and (E), leaving us with (B). So, let's attempt to prove that (A) or (III): XY>1 is not necessarily true (actually at this juncture, we have to keep our fingers crossed and hope that we CAN really prove that (A) is not necessarily true).


Try plug in numbers that satisfied XY<1: say X= 2 (notice that I use X=2 to satisfy condition (I)!!!! I shouldn't be using a number X such that X<1)and Y=1/3. So, plug in both number and you will notice that these 2 numbers satisfy both (1) and (2)! Effectively, we have shown that (III) is not necessary true, and therefore (A) and (E) are out too!

======================================
Remk:

Venksune wrote:
We have X^2*Y>1 and Y^2*X<1
=>X^2*Y>1 and -Y^2*X>-1 (the > becomes < when the sign changes).
In Inequalities you can add these two equations PROVIDED the > or < signs are same for both the equations.

So we now have X^2*Y - Y^2*X > 0
=>XY*(X-Y) > 0

Now, it is important, at this juncture to note that the question statement already tells us that X>0 and Y>0!!! So, this inequality only provides us with the additional info that (X-Y)>0 must be true.

X>0 and Y>0 => XY>0. However, this does not mean that you can immediately conclude that XY>1 is not NECESSARILY true to satisfy (1) and (2). You still have to prove that there are numbers X,Y that satisfy 0<XY<1, in order to prove that XY>1 is not a necessary condition.


Venksume: what do you think? do you agree with me? :)
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 [#permalink] New post 03 Sep 2004, 20:01
Agree. Few points though
1. Looking at the answer choices, III flashes the most no. of times. So, the thought is to work on it first.
2. Looking at I and II, it is evident that if I is true II is true. So the idea was to focus on III first.
3. The inequality calculation and the conclusion that X>Y and XY>0, yes is addl info but gets the red flag on III for sure.
4. Without choice, we will have to now choose numbers - we can get to choose X>Y if not anything else.
  [#permalink] 03 Sep 2004, 20:01
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