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Is y^2 + 7y + xy even?

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Is y^2 + 7y + xy even?  [#permalink]

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New post 20 Aug 2014, 23:13
1
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Question Stats:

16% (02:32) correct 84% (02:36) wrong based on 266 sessions

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Is y^2 + 7y + xy even?

(1) (x + y)(x - y) is a multiple of 4
(2) (x + 2)(x - 2) is a multiple of 4


Source : 4gmat
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Re: Is y^2 + 7y + xy even?  [#permalink]

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New post 21 Aug 2014, 02:08
1
5
alphonsa wrote:
Is y^2 + 7y + xy even?

(1) (x + y)(x - y) is a multiple of 4
(2) (x + 2)(x - 2) is a multiple of 4


Source : 4gmat


Not a GMAT-like question. Every GMAT divisibility question will tell you in advance that any unknowns represent positive integers (ALL GMAT divisibility questions are limited to positive integers only).

Is y^2 + 7y + xy even?

Is y^2 + 7y + xy = y(y + 7 + x) even?

(1) (x + y)(x - y) is a multiple of 4 --> x^2 - y^2 is a multiple of 4. If x = y = 0, then the answer would be YES but if x = y = 1, then the answer would be NO. Not sufficient.

(2) (x + 2)(x - 2) is a multiple of 4 --> x^2 - 4 is a multiple of 4 --> x^2 is a multiple of 4. If x = y = 0, then the answer would be YES but if \(x =2\sqrt{2}\) and y = 1, then the answer would be NO. Not sufficient.

(1)+(2) Since from (2) x^2 is a multiple of 4 and from (1) x^2 - y^2 is a multiple of 4, then y^2 must be a multiple of 4. Now, notice that we are not told that x and y are integers. So, if x = y = 0, then the answer would be YES but if \(x =2\sqrt{2}\) and y = 2, then the answer would be NO. Not sufficient.

Answer: E.
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Re: Is y^2 + 7y + xy even?  [#permalink]

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New post 21 Aug 2014, 06:45
1
Hello bunnel

Other than using irrational numbers,
Can i take x and y as imaginary i.e 3i or 4i
as nothing is mentioned about them
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Re: Is y^2 + 7y + xy even?  [#permalink]

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New post 21 Aug 2014, 07:01
utkgogia2003 wrote:
Hello bunnel

Other than using irrational numbers,
Can i take x and y as imaginary i.e 3i or 4i
as nothing is mentioned about them


No, all numbers on the test are real numbers (GMAT only deals with real numbers). So, no you cannot use complex numbers for the GMAT problems.
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Re: Is y^2 + 7y + xy even?  [#permalink]

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New post 14 Sep 2015, 06:25
Hey guys

I have a solution to this problem which results in answer choice C, but not sure where i am going wrong in my approach, i would really appreciate if you could help me on this:

Is y^2 + 7y + xy even?

Is y^2 + 7y + xy = y(y + 7 + x) even?

So for y(y + 7 + x) to be even, either both the terms('y' and 'x+y+7') need to be even or either one term even and one term odd.('y' odd and 'x+y+7' even or vice versa.)

(x-y)(x+y) is a multiple of 4. This tells us that this term "(x-y)(x+y)" is a even number. Hence two possibilities:

Both terms '(x-y)','(x+y)' are even or either one even and one odd.

So possibilities:

1. (x+y) => even (x-y) => even
2. (x+y) => even (x-y) => odd
3. (x+y) => odd (x-y) => even

Now we see, then for x+y to be even either both x and y have to be even or both have to be odd. But now looking closely we can also determine that options 2 and 3 listed above are not possible. (x+Y) and (x-Y) would either be both odd or both even. So we rule out those possibilities. And we are left with:

(x+y) => even (x-y) => even

So X, Y could either both be even or both be odd. and hence y(x+y+7) could be both even or odd. 1st statement is not sufficient.

Looking at statement b now:

(x+2)(x-2) is a multiple of 4. no information about Y and hence not sufficient.

But if we solve statement 2, it is also even(multiple of 4), following methodology we just saw above. We can establish that (x+2) and (X-2) are both even. IF x+2 is even then x is also even.

Now combining both the statements :

statement 2: X is even

statement 1: X,y are both odd or both are even.

So we get x and y as both even.

and hence the expression y(x+y+7) as even.
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Re: Is y^2 + 7y + xy even?  [#permalink]

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New post 14 Sep 2015, 07:02
ankuragarwal1301 wrote:
Hey guys

I have a solution to this problem which results in answer choice C, but not sure where i am going wrong in my approach, i would really appreciate if you could help me on this:

Is y^2 + 7y + xy even?

Is y^2 + 7y + xy = y(y + 7 + x) even?

So for y(y + 7 + x) to be even, either both the terms('y' and 'x+y+7') need to be even or either one term even and one term odd.('y' odd and 'x+y+7' even or vice versa.)

(x-y)(x+y) is a multiple of 4. This tells us that this term "(x-y)(x+y)" is a even number. Hence two possibilities:

Both terms '(x-y)','(x+y)' are even or either one even and one odd.

So possibilities:

1. (x+y) => even (x-y) => even
2. (x+y) => even (x-y) => odd
3. (x+y) => odd (x-y) => even

Now we see, then for x+y to be even either both x and y have to be even or both have to be odd. But now looking closely we can also determine that options 2 and 3 listed above are not possible. (x+Y) and (x-Y) would either be both odd or both even. So we rule out those possibilities. And we are left with:

(x+y) => even (x-y) => even

So X, Y could either both be even or both be odd. and hence y(x+y+7) could be both even or odd. 1st statement is not sufficient.

Looking at statement b now:

(x+2)(x-2) is a multiple of 4. no information about Y and hence not sufficient.

But if we solve statement 2, it is also even(multiple of 4), following methodology we just saw above. We can establish that (x+2) and (X-2) are both even. IF x+2 is even then x is also even.

Now combining both the statements :

statement 2: X is even

statement 1: X,y are both odd or both are even.

So we get x and y as both even.

and hence the expression y(x+y+7) as even.


Notice that we are not told that x and y are integers. Refer to the solution above.
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Re: Is y^2 + 7y + xy even?  [#permalink]

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New post 14 Sep 2015, 07:20
Thanks Bunuel, silly mistake, apologies for wasting the time. :(
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Re: Is y^2 + 7y + xy even?  [#permalink]

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New post 15 Sep 2015, 08:26
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.

Is y^2 + 7y + xy even?

(1) (x + y)(x - y) is a multiple of 4
(2) (x + 2)(x - 2) is a multiple of 4

Transforming the original condition and the question we have y(y+7+x)=even? and thus there are 2 variables (x,y). In order to match the number of variables and equations we need 2 equations and since there is 1 each in 1) and 2), there is high probability that C is the answer.

Using both 1) & 2) together we have (This saves us time)
x=even=4, y=2 yes, x=2sqrt2, y=2 no(since there is no guarantee that x,y are integers) therefore the conditions are not sufficient. The answer is E.

Normally for cases where we need 2 more equations, such as original conditions with 2 variable, or 3 variables and 1 equation, or 4 variables and 2 equations, we have 1 equation each in both 1) and 2). Therefore C has a high chance of being the answer, which is why we attempt to solve the question using 1) and 2) together. Here, there is 70% chance that C is the answer, while E has 25% chance. These two are the key questions. 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 according to DS definition, we solve the question assuming C would be our answer hence using ) and 2) together. (It saves us time). Obviously there may be cases where the answer is A, B, D or E.
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Re: Is y^2 + 7y + xy even?  [#permalink]

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New post 12 Oct 2015, 09:30
Hi,
What if we were told that x,y were integers.
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Re: Is y^2 + 7y + xy even?  [#permalink]

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New post 15 Mar 2016, 23:43
Hey chetan2u If in this question if the stem of the question says that x and y are both integers then B is sufficient right?
Also does GMAT really go to this extend to fool us on the even odd concept that x and y may both not be integers ?


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Is y^2 + 7y + xy even?  [#permalink]

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New post 24 Jun 2017, 02:46
Beautiful question with lot of traps .
Initially i did not know how to approach this question , but found a way to do it.
So lets get on with it

Is y^2 + 7y + xy even?

(1) (x + y)(x - y) is a multiple of 4
(2) (x + 2)(x - 2) is a multiple of 4

From 1 we can not decide whether it is sufficient , suppose x=y=0 then it is even but if you take other value x=5 y =3 then is then it is odd.
From 2 X=0 it is even as we will have only y^2 + 7y left in that case so it will be even but if take x=1 then it is even for even values of y and odd for odd value for y.

taking together these two statements y =2 then our equation becomes y^2 + 7y + xy =4+14+2x now x can take the value of fraction hence
E is the answer

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Re: Is y^2 + 7y + xy even?  [#permalink]

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