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For integers a, x, and y, a = x^2 + 2x + y. Is a even?

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For integers a, x, and y, a = x^2 + 2x + y. Is a even?  [#permalink]

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New post 07 Jun 2017, 14:18
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A
B
C
D
E

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  75% (hard)

Question Stats:

48% (01:46) correct 52% (01:48) wrong based on 222 sessions

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Re: For integers a, x, and y, a = x^2 + 2x + y. Is a even?  [#permalink]

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New post 07 Jun 2017, 19:09
2
Bunuel wrote:
\(a = x^2 + 2x + y\). Is a even?

(1) y = x + 4

(2) x + 4 = 15



(1) Case 1:-
when x is odd then y =odd

let x= 1 then y = 5
then a = 1+2+5= 8---even

Case 2:-
when x =even then y = even
let x=2 then y =6
substituting a = 4+4+6 =even

both cases a= even

Case 3:-
but if x =1/2 then y = 9/2
substituting
a= 1/4 + 1 + 9/2 =23/4 ---fraction -----not even

(2) no information of y
but x is integer
insuff

combining both from (2) as x= integer
thus a must be even (from first 2 cases in (1) )

Ans C
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Re: For integers a, x, and y, a = x^2 + 2x + y. Is a even?  [#permalink]

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New post 07 Jun 2017, 23:18
2
a = x^2 + 2x + y
We don't know whether x and y are integers: and if they are, whether even or odd integers.

Statement 1. y= x+4. So a becomes:
a = x^2 + 2x + x + 4 = x^2 + 3x + 4
But we don't know if x is an integer or if it is, then even/odd integer? So Insufficient.

Statement 2. x+4=15, so x = 15-4 = 11. But we don't know about y. So Insufficient.

Combining the two statements: x = 11, y = x+4 = 15. Now we can easily check whether a is even or not. Sufficient.

Hence C answer
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Re: For integers a, x, and y, a = x^2 + 2x + y. Is a even?  [#permalink]

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New post 08 Jun 2017, 17:26
1
Given
1. y= x+4 => y = x+even=> x even -> Y even; x odd-> Y odd
substituting in the equation for x both even and odd, we get y as even only. Sufficient
2. It doesn't give anything about Y. So, not sufficient
Ans : A

ps: please correct me if I am wrong..
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Re: For integers a, x, and y, a = x^2 + 2x + y. Is a even?  [#permalink]

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New post 08 Jun 2017, 20:10
2
sasidharrs wrote:
Given
1. y= x+4 => y = x+even=> x even -> Y even; x odd-> Y odd
substituting in the equation for x both even and odd, we get y as even only. Sufficient
2. It doesn't give anything about Y. So, not sufficient
Ans : A

ps: please correct me if I am wrong..


Hi

You are assuming that x will be either even or odd. But its not given that x is an integer. Its possible that x is a number like 1.33. Then y becomes 5.33
In this case 'a' will turn out to be a decimal number, so neither even nor odd.
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Re: For integers a, x, and y, a = x^2 + 2x + y. Is a even?  [#permalink]

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New post 20 Jul 2018, 01:56
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1
a = x^2 + 2x + y

From statement 1 :

y= x+4.

a = x^2+2x+x+4
a = (x^2+x) + (2x+4) .
a = x(x+1) + 2(x+2) Now 2(x+2)is an even entity thus we just need to concentrate on the x(x+1) part.
Of any two consecutive integers, one of them is even. Thus the entity x(x+1) is also even.
We know that even + even = even.

Therefore a = even.

This statement is sufficient to answer the question.

Statement 2 :

x +4 = 15; x = 11;
Thus a = 121+22+y
a = 143 +y

if y is odd a is even. If y is even a is odd.
This statement is not sufficient.

Therefore Answer = A
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Re: For integers a, x, and y, a = x^2 + 2x + y. Is a even?  [#permalink]

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New post 20 Jul 2018, 09:49
1
Bunuel wrote:
\(a = x^2 + 2x + y\). Is a even?

(1) y = x + 4
(2) x + 4 = 15


Hi Bunuel

You know better than anyone on this forum that in GMAT world statement 1 is insufficient until we get information regarding to property of x. What if I tell you that x=1/2?

\(a = {1/2}^2 + 2*1/2 + 1/2 +4\)
\(a = 5 + 3/4\) - fraction

In case if x = 0, then a = 4 - indeed even

Clearly, the question stem or OA should be modified. Thanks.
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Re: For integers a, x, and y, a = x^2 + 2x + y. Is a even?  [#permalink]

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New post 20 Jul 2018, 12:21
2
Bunuel wrote:
\(a = x^2 + 2x + y\). Is a even?

(1) y = x + 4

(2) x + 4 = 15


Hi Hero8888

You were faster than in posting same examples :-D

Bunuel

Can you please confirm what we all miss here? Does the questions miss that x & y are integers?
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Re: For integers a, x, and y, a = x^2 + 2x + y. Is a even?  [#permalink]

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New post 20 Jul 2018, 13:16
1
Mo2men wrote:
Bunuel wrote:
\(a = x^2 + 2x + y\). Is a even?

(1) y = x + 4

(2) x + 4 = 15


Hi Hero8888

You were faster than in posting same examples :-D

Bunuel

Can you please confirm what we all miss here? Does the questions miss that x & y are integers?


It seems I got the right answer by overlooking the fact that the numbers are not integers. :(
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Re: For integers a, x, and y, a = x^2 + 2x + y. Is a even?  [#permalink]

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Re: For integers a, x, and y, a = x^2 + 2x + y. Is a even?  [#permalink]

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New post 21 Jul 2018, 01:13
1
Bunuel wrote:
For integers a, x, and y, \(a = x^2 + 2x + y\). Is a even?

(1) y = x + 4

(2) x + 4 = 15


OA: A

(1) \(y= x + 4\)
\(a= x^2 + 2x + y\)
\(a=x^2 + 2x +(x+4)\)
\(a = x^2 +x+ (2x +4)\)
\(a = x(x+1) + (2x +4)\)
Case \(1\) : if \(x\) is odd, then \(x+1\) will be even.
\(a = odd(even) + (even)=even+even\)
\(a\) will be even
Case \(2\) : if \(x\) is even, then \(x+1\) will be odd.
\(a = even(odd) + (even)=even+even\)
\(a\) will be even
So Statement \(1\) alone is sufficient to answer whether \(a\) is even or not.

(2)\(x + 4 = 15\),
\(x=11\), putting \(x=11\) in question stem
\(a = x^2 + 2x + y\)
\(a = 121 + 2*11 + y\)
\(a = 143+y\)
Case\(1\) : when y is odd
\(a= odd+ odd =even\)
Case\(2\) : when y is even
\(a= odd + even= odd\)
So Statement \(2\) alone is not sufficient to answer whether \(a\) is even or not.
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Re: For integers a, x, and y, a = x^2 + 2x + y. Is a even?  [#permalink]

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New post 21 Jul 2018, 20:06
Bunuel wrote:
For integers a, x, and y, \(a = x^2 + 2x + y\). Is a even?

(1) y = x + 4

(2) x + 4 = 15



Solve the question conceptually without plugging numbers

\(a = x^2 + 2x + y\)

Since 2x is even.......the question will boil down to:

Is \(a = x^2 + y\) even? (Note: \(x^2\) does not change the nature of a number)

(2) x + 4 = 15

No info about y

Insufficient

(1) y = x + 4

y - x= 4

This mean that either both are odd or both are even....In whichever case, the answer is YES

Sufficient

Answer: A
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Re: For integers a, x, and y, a = x^2 + 2x + y. Is a even?   [#permalink] 21 Jul 2018, 20:06
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