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I tried the following

For statement 1, x = 2 and y = 1/2, x*y = 2*1/2 = 1 (an integer) No
x= 2 and y = 1, x*y = 2*1 = 2(also an integer) Yes

For statement 2, x = 2.5,y=1.5, x+y = 2.5+1.5 (an integer) No
x= 2, y = 1, x+y = 2+1 = 3( an integer) Yes

Combining 1+2, we might arrive at an answer E, how ever consider the possibility that x = 0 and y = 0 , both are integers
and this would satisfy both the conditions and so the answer is C according to me.
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I tried the following

For statement 1, x = 2 and y = 1/2, x*y = 2*1/2 = 1 (an integer) No
x= 2 and y = 1, x*y = 2*1 = 2(also an integer) Yes

For statement 2, x = 2.5,y=1.5, x+y = 2.5+1.5 (an integer) No
x= 2, y = 1, x+y = 2+1 = 3( an integer) Yes

Combining 1+2, we might arrive at an answer E, how ever consider the possibility that x = 0 and y = 0 , both are integers
and this would satisfy both the conditions and so the answer is C according to me.

Hi,
you can always find values of x and y as an integers satisfying the given conditions..
BUT we have to be sure that there are no values fittin as non integer, ecause the Q is a 'must ' Q..
hope it helped
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Lets x = -square root 2
y = square root 2

(1) xy = -2. Insufficient

(2) x+y = 0. Insufficient

(1)+(2), Also gives no information regarding whether x and y are integers or not

E is correct
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For the product xy to be an integer, x and y both necessarily may not be integers
eg: x=0.2 and y=10 gives xy=2

But for x+y to be an integer, both must necessarily be integers
eg: 1+0 = 1, 2+(-2)=0

Therefore, Statement (2) is sufficient.
Answer (B)
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1) If both are integers, we will get an integer. However, let x=0.5 and y=6 xy=3
INSUFFICIENT

2) Once again, if both are integers it will yield an integer. However, let x = 0.5 and y = 0.5 and then x+y = 1
INSUFFICIENT

1+2) Does not give us anything new.

Answer E
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imo C

1+2= on combining provides....y(y+int) = int

so Y will be integer

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Stmt 1
2/3 * 3/2 = 1 (Integer)

hence insufficient.

Stmt 2

1/3+2/3 = 1 (Integer)

hence insufficient.

Combining both if we take x & y both as root2
We get yes & no.

Hence E is the answer
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I'm going with C.

When you combine 1 and 2, there is no way you can have fractions for x and y. So it will be C.
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Bunuel
Are x and y integers?

(1) The product xy is an integer.
(2) x + y is an integer.

This problem turned out to be more challenging than I though it would be. C is not correct. Oddly enough Princeton Review also says that the OA is C and is wrong about it. The correct answer is E.

If \(x=\sqrt{2}\) and \(y=-\sqrt{2}\) then the answer is NO but if \(x = y = 1\), the answer is YES. Not sufficient.

Answer: E.
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