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

Question

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

1)  x^2 =1
2)  y^2 =4

Omkar.kamat · Answer

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 !!

vitaliyGMAT · Answer

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.

Omkar.kamat · Answer

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 !!

AkshdeepS · Answer

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.

ashikaverma13 · Answer

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.

AkshdeepS · Answer

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.

MathRevolution · Answer

==> 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.

Therefore, the answer is D.

LeoN88 · Answer

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

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

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GMAT Data Sufficiency (DS) Questions

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

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