Tricky DS

[r2kins]

Q: If x and Y are integers, is x even?

A. xy + y is odd
B. 6x - 3y is odd

The answer is A

My question is:

xy + y can be evaluated in two ways

#1
xy +y = y (x+1)= odd => x is even which means this statement is sufficient!!!

#2

xy + y = odd => #2a: either xy is odd (and y is even) or #2b: xy is even (and y is odd)

In the first case,#2a, x is odd and in #2b, x is even

So, as per #2, this statement is not sufficient????

How do I get rid of this seeming paradox??

Thanks in advance..

[theCodeToGMAT]

X --> EVEN??

Statement 1:

XY + Y --> ODD
Y(X+1)
ONLY POSSIBLE CASES
O (E + 1)
SUFFICIENT

Statement 2:
6X - 3Y --> 0DD
3(2X - Y) --> ODD

O [ E - O ]
O [ O - O ]
INSUFFICIENT
Answer [spoiler]{A}[/spoiler]

[theCodeToGMAT]

Answer to your query:

for XY + Y

Case 1: XY is ODD & Y is EVEN
Since, "Y" is Even.... we cannot have "XY" as ODD .. as any number multiplied by Even the result Even.. So discard this statement


Case 2: XY is EVEN & Y is oDD
Here, we are putting "y" as already ODD..
So, we need "X" as even .. only that "XY" can be EVEN
So, "x" is EVEN..

So, Only one of the case is Valid...

[[email protected]]

Hi r2kins,

This DS question is packed with Number Properties. It's just a matter of which ones you choose to use to answer the question. Here's one way to do it:

We're told that X and Y are INTEGERS. We're asked if X is even. This is a Yes/No question.

Fact 1: XY + Y = Odd

Factor out a Y and you get...

Y(X + 1) = Odd

The ONLY way for this equation to end up ODD is if we have...
(Odd)(Odd) = Odd

So, Y = Odd and (X + 1) = Odd...... So X MUST be Even. The answer to the question is ALWAYS YES.
Fact 1 is SUFFICIENT

Fact 2: 6X - 3Y = ODD

6(INT) = EVEN, no matter what the Integer is...

So, we have....
Even - 3Y = Odd
3Y MUST be Odd, so Y MUST be Odd.

However, we don't know whether X is Even or Odd. This means the answer might be YES and it might be NO.
Fact 2 is INSUFFICIENT

Final Answer: A

GMAT assassins aren't born, they're made,
Rich

[r2kins]

Thanks Rahul. I realized after posting that XY cannot be odd if Y is even. Thanks for helping though!!

Answer to your query:

for XY + Y

Case 1: XY is ODD & Y is EVEN
Since, "Y" is Even.... we cannot have "XY" as ODD .. as any number multiplied by Even the result Even.. So discard this statement


Case 2: XY is EVEN & Y is oDD
Here, we are putting "y" as already ODD..
So, we need "X" as even .. only that "XY" can be EVEN
So, "x" is EVEN..

So, Only one of the case is Valid...

[GMATGuruNY]

Q: If x and Y are integers, is x even?

A. xy + y is odd
B. 6x - 3y is odd

x and y must be INTEGERS.
The issue at hand is EVEN VS. ODD.
Thus, each statement can be quickly evaluated by testing the following EVEN/ODD combinations:
Case 1: x=2, y=1
Case 2: x=2, y=2
Case 3: x=1, y=1
Case 4: x=1, y=2.

Statement 1: xy + y is odd
Case 1: x=2, y=1
2*1 + 1 = 3.
This works.
Thus, it is possible that x is EVEN.

Case 3: x=1, y=2
1*2 + 2 = 4.
Doesn't work.

Case 4: x=1, y=1
1*1 + 1 = 2.
Doesn't work.

Neither of the cases in which x is odd satisfies statement 1.
Thus, to satisfy statement 1, x must be EVEN.
SUFFICIENT.

Statement 2: 6x - 3y is odd
Case 1: x=2, y=1
6*2 - 3*1 = 9.
This works.
Thus, it is possible that x is EVEN.

Case 3: x=1, y=1
6*1 - 3*1 = 3.
This works.
Thus, it is possible that x is ODD.

Since x is EVEN in Case 1 but ODD in Case 2, INSUFFICIENT.

The correct answer is A.

[GMATinsight]

Q: If x and Y are integers, is x even?

A. xy + y is odd
B. 6x - 3y is odd

Question : Is x Even?

Statement 1: xy + y is odd
i.e. y(x+1) = odd
i.e. y is odd as well as x+1 is odd
i.e. y is odd and x is even
SUFFICIENT

Statement 2: 6x - 3y is odd
i.e. 3(2x-y) = Odd
i.e. 2x-y = odd
i.e. y is odd but nothing can be predicted about x. hence,
NOT SUFFICIENT

Answer: option A

[Brent@GMATPrepNow]

Q: If x and Y are integers, is x even?

A. xy + y is odd
B. 6x - 3y is odd

This is a practice question from our free GMAT video course.
Here's our video solution - https://www.gmatprepnow.com/module/gmat- ... /video/838

Cheers,
Brent

[Max@Math Revolution]

Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

If x and y are integers, is x even?

(1) xy + y is odd

(2) 6x - 3y is odd

There are 2 variables (x,y) and 2 equations, so there is high chance (C) will be the answer. When looking at the conditions together, 6x=even, 6x-3y=odd, so 3y=odd. Also, in (x+1)y=odd, x+1=odd, x=even, so the conditions seem to be sufficient and the answer looks like (C), but this is a commonly made mistake according to 4(A), we should look at the conditions separately. For condition 1, (x+1)y=odd, x+1=odd, y=odd, x=odd-1=even. So this condition is sufficient, but for condition 2, 6x-3y=odd, 6x is always even without regard to the value of x, so this condition is insufficient. The answer becomes (A).
This is a most common type of question in GMAT math DS section.

For cases where we need 2 more equation, such as original conditions with "2 variables", or "3 variables and 1 equation", or "4 variables and 2 equations", we have 1 equation each in both 1) and 2). Therefore, there is 70% chance that C is the answer, while E has 25% chance. These two are the majority. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer using 1) and 2) separately according to DS definition (It saves us time). Obviously there may be cases where the answer is A, B, D or E.

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