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

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

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New post 20 Apr 2018, 00:08
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A
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[GMAT math practice question]

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

1) \(x+3y\) is even
2) \((x-1)(y-1)\) is odd

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

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New post 20 Apr 2018, 00:42
1
Statement1:

x+3y is even.

Now as MULTIPLICATION BY AN ODD INTEGER DOESN’T CHANGE THE ODD/EVEN NATURE OF AN INTEGER.

Hence x +y is even (same as x+3y or 3x+5y etc)

SUFFICIENT



Statement 2:

(x-1)(y-1) is odd

Since the product of two integers are odd only if both of them are odd,

(x-1) is odd, hence x =odd+1 = even.

Similarly, (y-1) is odd, hence y= odd+1 = even.

Hence, x+y = even+even = even

SUFFICIENT.



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

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New post 20 Apr 2018, 06:00
Top Contributor
MathRevolution wrote:
[GMAT math practice question]

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

1) \(x+3y\) is even
2) \((x-1)(y-1)\) is odd


Target question: Is x+y an even integer?

Statement 1: x+3y is even
There are 4 possible cases to consider. Let's test them all
Case a: x is ODD and y is ODD. Notice that x +3y = ODD + (3)(ODD) = ODD + ODD = EVEN. In this case, x + y = ODD + ODD = EVEN. So, the answer to the target question is YES, x+y IS even
Case b: x is ODD and y is EVEN. Notice that x +3y = ODD + (3)(EVEN) = ODD + EVEN= ODD. This does NOT meet the requirement that x+3y is even, so IGNORE case b
Case c: x is EVEN and y is ODD. Notice that x +3y = EVEN + (3)(ODD) = EVEN + ODD= ODD. This does NOT meet the requirement that x+3y is even, so IGNORE case c
Case d: x is EVEN and y is EVEN. Notice that x +3y = EVEN + (3)(EVEN) = EVEN+ EVEN= EVEN. In this case, x + y = EVEN + EVEN = EVEN. So, the answer to the target question is YES, x+y IS even
So, cases a and d are the only to possible cases.
Since both cases yields the SAME answer to the target question (YES, x+y IS even), statement 1 is SUFFICIENT

Statement 2: (x-1)(y-1) is odd
If the product of two integers is ODD, then the two integers must each be ODD
So, (x-1) must be ODD, and (y-1) must be ODD
If (x-1) is ODD, then we can be certain that x is EVEN
If (y-1) is ODD, then we can be certain that y is EVEN
If x and y are both EVEN, then x+y = EVEN + EVEN = EVEN
So, the answer to the target question is YES, x+y IS even
Since we can answer the target question with certainty, statement 2 is SUFFICIENT

Answer: D

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

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New post 22 Apr 2018, 17:27
=>

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) & 2) together first. After comparing the number of variables and the number of equations, we can save time by considering conditions 1) & 2) together first.

Conditions 1) & 2)
Since both conditions are satisfied only when x is even and y is even, x + y must be an even number.
Thus, both conditions together are sufficient.

Since this question is an integer question (one of the key question areas), CMT (Common Mistake Type) 4(A) of the VA (Variable Approach) method tells us that we should also check answers A and B.

Condition 1)
There are two ways in which x + 3y can be even.
If x and y are both even, then x + y is even.
If x and y are both odd, then x + y is even too.
Thus, condition 1) is sufficient.


Condition 2)
Since (x-1)(y-1) is odd, both x -1 and y – 1 must be odd.
So, x and y must both be even.
It follows that x + y is even.
Thus, condition 2) is sufficient too.

Therefore, D is the answer.

Answer: D

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.
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Re: If x and y are integers, is x+y an even number? &nbs [#permalink] 22 Apr 2018, 17:27
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