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If x and y are positive integers, xy-2=?

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MathRevolution
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If x and y are positive integers, xy-2=? [#permalink] New post   23 Dec 2016, 02:31
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If x and y are positive integers, \(x^y^-^2\)=?

1) \(x^2\)=1
2) \(y^2\)=4
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D
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Omkar.kamat
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Re: If x and y are positive integers, xy-2=? [#permalink] New post   23 Dec 2016, 02:39
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

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Re: If x and y are positive integers, xy-2=? [#permalink] New post   23 Dec 2016, 03:15
MathRevolution wrote:
If x and y are positive integers, \(x^y^-^2\)=?

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.
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Re: If x and y are positive integers, xy-2=? [#permalink] New post   23 Dec 2016, 04:00
vitaliyGMAT wrote:
MathRevolution wrote:
If x and y are positive integers, \(x^y^-^2\)=?

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.

But xy-2 is asked... U get X or Y values in individual conditions. How do u get y value in cond 1 or x value in Cond 2 ??

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Re: If x and y are positive integers, xy-2=? [#permalink] New post   23 Dec 2016, 11:03
Omkar.kamat wrote:
vitaliyGMAT wrote:
MathRevolution wrote:
If x and y are positive integers, \(x^y^-^2\)=?

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.

But xy-2 is asked... U get X or Y values in individual conditions. How do u get y value in cond 1 or x value in Cond 2 ??

Omkar Kamat

When The Going Gets Tough, The Tough Gets Going !!



When x = 1 it does not matter what the power is as the rule for exponents says any number raised to "0" will yield "1" as result. So st 1 is sufficient.

For st 2, when y = 2 the exponent becomes 0, so it does not matter what x is, the expression will yield "1" as result. So st 2 is sufficient.

So D is the correct answer because both statements individually gives you "1" as value of \(x\)\(^{y-2}\)

Hope this helps.
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Re: If x and y are positive integers, xy-2=? [#permalink] New post   23 Dec 2016, 13:23
In the first condition, y could also be 1 which satisfies y>1. in that case it will be 1 raised to the power of -1.



When x = 1 it does not matter what the power is as the rule for exponents says any number raised to "0" will yield "1" as result. So st 1 is sufficient.

For st 2, when y = 2 the exponent becomes 0, so it does not matter what x is, the expression will yield "1" as result. So st 2 is sufficient.

So D is the correct answer because both statements individually gives you "1" as value of \(x\)\(^{y-2}\)

Hope this helps.[/quote]
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If x and y are positive integers, xy-2=? [#permalink] New post   23 Dec 2016, 21:20
ashikaverma13 wrote:
In the first condition, y could also be 1 which satisfies y>1. in that case it will be 1 raised to the power of -1.



When x = 1 it does not matter what the power is as the rule for exponents says any number raised to "0" will yield "1" as result. So st 1 is sufficient.

For st 2, when y = 2 the exponent becomes 0, so it does not matter what x is, the expression will yield "1" as result. So st 2 is sufficient.

So D is the correct answer because both statements individually gives you "1" as value of \(x\)\(^{y-2}\)

Hope this helps.



Had it been the case of y = 1, then also it would have remained the same.

"1" raised to any exponent remains as it is. For example if \(x = 2\) then, \(2^{-1}\) becomes \(\frac{1}{2}\)

Therefore if \(x =1\) and \(y = 1\) then, \(1^{-1}\) becomes \(\frac{1}{1}\) \(= 1\)

Hope it's clear.
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Re: If x and y are positive integers, xy-2=? [#permalink] New post   25 Dec 2016, 18:35
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==> In the original condition, there are 2 variables (x), and in order to match the number of variables to the number of equations, there must be 2 equations, and therefore C is most likely to be the answer. By solving con 1) & con 2), you get x=-1, 1 and y=-2, 2. Since x and y are positive integers, only x=1 and y=2 are possible, and con 1) becomes \(1^y^-^2=x^2^-^2=x^0=1\).

Therefore, the answer is D.
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Re: If x and y are positive integers, xy-2=? [#permalink] New post   09 Mar 2020, 22:00
Very Good Sum

x,y > 0 and Int.

#1. It says x can be 1 or -1, but we know x is +ve Int.

So, x has to be 1, answer-> 1 Sufficient

#2. y has to 2 (we know y is +ve Int.)
So, x^ (2-2) = x^0 = 1 Sufficient

E


MathRevolution wrote:
If x and y are positive integers, \(x^y^-^2\)=?

1) \(x^2\)=1
2) \(y^2\)=4
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Re: If x and y are positive integers, xy-2=? [#permalink]
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