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Bunuel
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
x=5 and y=20....answer is yes
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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Bunuel
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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Can someone please explain how we will solve taking both statements together?
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Chetsibhatia
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.


Answer: C
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