If you solve the inequality, you will get:
Is xy / x + y > 1/xy
statement 1 -> x = 2y
2y^2 / 3y > 1 / 2y^2 - solving it further -> 2y/3 > 1/2y^2
y = 3 gives 2y/3 > 1/2y^2
y = -3 gives 2y/3 < 1/2y^2
Answer should be C.
Abdulla wrote:OA is A
If x and y are nonzero integers, is ...
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mehravikas
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answer should not be A
If you solve the inequality, you will get:
Is xy / x + y > 1/xy
statement 1 -> x = 2y
2y^2 / 3y > 1 / 2y^2 - solving it further -> 2y/3 > 1/2y^2
y = 3 gives 2y/3 > 1/2y^2
y = -3 gives 2y/3 < 1/2y^2
Answer should be C.
If you solve the inequality, you will get:
Is xy / x + y > 1/xy
statement 1 -> x = 2y
2y^2 / 3y > 1 / 2y^2 - solving it further -> 2y/3 > 1/2y^2
y = 3 gives 2y/3 > 1/2y^2
y = -3 gives 2y/3 < 1/2y^2
Answer should be C.
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Abdulla
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mehravikas wrote:answer should not be A
If you solve the inequality, you will get:
Is xy / x + y > 1/xy
statement 1 -> x = 2y
2y^2 / 3y > 1 / 2y^2 - solving it further -> 2y/3 > 1/2y^2
y = 3 gives 2y/3 > 1/2y^2
y = -3 gives 2y/3 < 1/2y^2
Answer should be C.
Abdulla wrote:OA is A
No dude, you made some mistakes..
When you solve the inequality you will get
Is xy / x + y > xy ?
go further...
Now in statement one..
x =2y ... plug it ..
2y^2 / 3y = 2y^2 and since y^2 will never be negative then we can cross them out without flipping the sign.
Is 1/3y >1 ? Sufficient.
The answer is A
Abdulla
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mehravikas
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My mistake I read the right part as - (x^1 y ^1)^-1
But it is [(x-1)(y-1)]-1
In that case the answer will be A
But it is [(x-1)(y-1)]-1
In that case the answer will be A
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sunnyjohn
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Hi Abdullah,
when we come to following equation:
1/3Y > 1 ??
how can u assume that this is true. This will only be true, if
0<3Y<1 ==> 0<Y<1/3
but the Question said X and Y are non zero integer.
I think answer should be E.
when we come to following equation:
1/3Y > 1 ??
how can u assume that this is true. This will only be true, if
0<3Y<1 ==> 0<Y<1/3
but the Question said X and Y are non zero integer.
I think answer should be E.
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November Rain
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Hi guys,
I think the OA should indeed be A.
Unless i am wrong, (x^-1 + y^-1)^-1 > [(x^-1)(y^-1)]^-1 is the same as X+Y>XY
So, if you pick values, and consider that the integers are not 0, you will see that statement 1 is sufficient while statement 2 is not.
I think the OA should indeed be A.
Unless i am wrong, (x^-1 + y^-1)^-1 > [(x^-1)(y^-1)]^-1 is the same as X+Y>XY
So, if you pick values, and consider that the integers are not 0, you will see that statement 1 is sufficient while statement 2 is not.
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Abdulla
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Hi Sunnyjohn,sunnyjohn wrote:Hi Abdullah,
when we come to following equation:
1/3Y > 1 ??
how can u assume that this is true. This will only be true, if
0<3Y<1 ==> 0<Y<1/3
but the Question said X and Y are non zero integer.
I think answer should be E.
We know that x and y is NON zero INTEGERS, so when you plug any negative or positive INTEGER, the expression won't be greater than 1..
1/3(2) > 1 No
1/3(-2) > 1 No
The answer should be A because the result (NO.)
Abdulla
