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# Is x < 10 < y? (1) x < y and xy = 100 (2) x^2 < 100 < y^2

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Re: Is x < 10 < y? (1) x < y and xy = 100 (2) x^2 < 100 < y^2 [#permalink]
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Bunuel wrote:
Is x < 10 < y?

(1) x < y and xy = 100
(2) x^2 < 100 < y^2

(1) x < y and xy = 100
Both x and y could be negative.....x=-20 and y=-5......answer is NO
Insufficient

(2) x^2 < 100 < y^2
-10<x<10 and y<-10 or y>10
x could be -5 or 5, while y can be 20 or -20
x=-5 and y=-20.....No
x=5 and y=20.....Yes
Insufficient

Combined
If -10<x<10, and y<-10 or y>10, then x<y only when y>10.
If y>10 and xy =100, x has to be less than 10..
Thus x<10<y
Sufficient
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Re: Is x < 10 < y? (1) x < y and xy = 100 (2) x^2 < 100 < y^2 [#permalink]
Bunuel wrote:
Is x < 10 < y?

(1) x < y and xy = 100
(2) x^2 < 100 < y^2

M36-81

Similar question from OG: https://gmatclub.com/forum/if-x-and-y-a ... 39873.html
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Re: Is x < 10 < y? (1) x < y and xy = 100 (2) x^2 < 100 < y^2 [#permalink]
Can someone please explain how we will solve taking both statements together?
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Re: Is x < 10 < y? (1) x < y and xy = 100 (2) x^2 < 100 < y^2 [#permalink]
Chetsibhatia wrote:
Can someone please explain how we will solve taking both statements together?

Here is slightly rephrased solution, hope it helps.

Official Solution:

Is $$x < 10 < y$$?

(1) $$x < y$$ and $$xy = 100$$:

If $$x=-20$$ and $$y=-5$$, then $$x < y$$ but $$10$$ is not between them, so the answer is NO. However, if $$x=5$$ and $$y=20$$, then the answer is YES.

Not sufficient.

(2) $$x^2 < 100 < y^2$$

The first inequality, $$x^2 < 100$$, implies $$-10 < x < 10$$. So, we are left to establish whether $$y > 10$$.

The second inequality, $$100 < y^2$$, implies $$y > 10$$ or $$y < -10$$.

So, we can have the following two cases:

a. If $$y > 10$$, then $$x < 10 < y$$ is true.

b. If $$y < -10$$, then $$x > y$$, and therefore $$x < 10 < y$$ is false.

Not sufficient.

(1)+(2) Since from (1) we have that $$x < y$$, then case b from (2), saying that $$x > y$$, cannot be true, hence we are left with case a, and therefore, $$x < 10 < y$$. Sufficient.