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If x and y are integers, is y an even number?

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Math Revolution GMAT Instructor
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If x and y are integers, is y an even number?  [#permalink]

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New post 04 Jan 2018, 00:26
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[GMAT math practice question]

If \(x\) and \(y\) are integers, is \(y\) an even number?

1) \(y=x^2+3x+2\)
2) \(xy\) is an even number

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Re: If x and y are integers, is y an even number?  [#permalink]

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New post 04 Jan 2018, 00:45
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MathRevolution wrote:
[GMAT math practice question]

If \(x\) and \(y\) are integers, is \(y\) an even number?

1) \(y=x^2+3x+2\)
2) \(xy\) is an even number


(1) y can be written as x^2 + 3x + 2 = x(x+3) + 2. Now since x and x+3 are two integers having a difference of '3', one of them will be odd and the other will be even. So their product, x(x+3) will be Even. And when we add '2', which is again an even number, the result will be Even only. So y is even. Sufficient.

(2) Product of x & y is even, this means that both of them cannot be odd (because then product will be odd). At least one of them will be even, so we can have one of x/y even & other odd, OR we could have both x/y even. Can't say whether y will be even or odd. So not sufficient.

Hence A answer
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Re: If x and y are integers, is y an even number?  [#permalink]

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New post 04 Jan 2018, 07:42
MathRevolution wrote:
[GMAT math practice question]

If \(x\) and \(y\) are integers, is \(y\) an even number?

1) \(y=x^2+3x+2\)
2) \(xy\) is an even number


---------

Stmt 1: y=x2+3x+2 => y= (x+1)(x+2)
which is product of 2 consecutive integers, thus one of them have to be even thus y is even => Sufficient

Stmt 2: xy = even no. implies either x, y or both could be even => Not sufficient to determine whether y is even or not
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Re: If x and y are integers, is y an even number?  [#permalink]

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New post 07 Jan 2018, 17:09
=>

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

Since we have 2 variables (x and y) and 0 equations, C is most likely to be the answer. So, we should consider conditions 1) and 2) together first.

Conditions 1) and 2)
Case 1: \(x\) is odd
Since \(xy\) is even, \(y\) is even.

Case 2: \(x\) is even.
Since \(x^2, 3x\) and \(2\) are even, \(y =x^2+3x+2\) is even.
Since we have a unique answer, both conditions together are sufficient.

Since this is an integer question (one of the key question areas), we should also consider choices A and B by CMT 4(A).

Condition 1):
There are two cases to consider.
Case 1: \(x\) is even
Since \(x^2, 3x\) and 2 are even, \(y =x^2+3x+2\) is even.

Case 2: \(x\) is odd
Since \(x^2+3x\) is even and \(2\) is even, \(y =x^2+3x+2\) is even.
Since we have a unique answer, condition 1) is sufficient.


Condition 2):
If \(x = 1\) and \(y = 2, y\) is even.
If \(x = 2\) and \(y = 1, y\) is odd.
Since we do not have a unique answer, condition 2) is not sufficient.

The answer is A.

Normally, in problems which require 2 equations, such as those in which the original conditions include 2 variables, or 3 variables and 1 equation, or 4 variables and 2 equations, each of conditions 1) and 2) provide an additional equation. In these problems, the two key possibilities are that C is the answer (with probability 70%), and E is the answer (with probability 25%). Thus, there is only a 5% chance that A, B or D is the answer. This occurs in common mistake types 3 and 4. Since C (both conditions together are sufficient) is the most likely answer, we save time by first checking whether conditions 1) and 2) are sufficient, when taken together. Obviously, there may be cases in which the answer is A, B, D or E, but if conditions 1) and 2) are NOT sufficient when taken together, the answer must be E.

Answer: A
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Re: If x and y are integers, is y an even number? &nbs [#permalink] 07 Jan 2018, 17:09
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