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If x and y are positive integers, \(x^y^-^2\)=?
1) \(x^2\)=1
2) \(y^2\)=4
1) \(x^2\)=1
2) \(y^2\)=4
23 Dec 2016, 02:39
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Statement 1 will return |X|=1 I.e X= 1,-1. But since we know they are Positive we conclude X=1. So equation remains y-2= ??. Insufficient.
Similarly statement 2 will return y=2. Thus 2x-2 =?? Or x-1=?? , Thus it's insufficient.
Combining both we get X=1 and Y= 2... Thus Xy-2= O. Definate Value.
Answer is C
Omkar Kamat
When The Going Gets Tough, The Tough Gets Going !!
Similarly statement 2 will return y=2. Thus 2x-2 =?? Or x-1=?? , Thus it's insufficient.
Combining both we get X=1 and Y= 2... Thus Xy-2= O. Definate Value.
Answer is C
Omkar Kamat
When The Going Gets Tough, The Tough Gets Going !!
23 Dec 2016, 03:15
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MathRevolution
If x and y are positive integers, \(x^y^-^2\)=?
1) \(x^2\)=1
2) \(y^2\)=4
1) \(x^2\)=1
2) \(y^2\)=4
1) \(x^2\)=1
Gives us x = +/- 1, but we are told that x,y >0 and we have only one option x=1 and \(1^{any - power} = 1\). Sufficient.
2) \(y^2\)=4
Gives us y = +/- 2. Bu again according to the question y>0, hence y=2. We have \(x^{2-2} = x^0 = 1\). Sufficient.
