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For integers a, x, and y, a = x^2 + 2x + y. Is a even?
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07 Jun 2017, 14:18
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For integers a, x, and y, \(a = x^2 + 2x + y\). Is a even? (1) y = x + 4 (2) x + 4 = 15
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Re: For integers a, x, and y, a = x^2 + 2x + y. Is a even?
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07 Jun 2017, 19:09
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= 8even 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?
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07 Jun 2017, 23:18
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 = 154 = 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?
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08 Jun 2017, 17:26
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?
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08 Jun 2017, 20:10
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?
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20 Jul 2018, 01:56
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?
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20 Jul 2018, 09:49
Bunuel wrote: \(a = x^2 + 2x + y\). Is a even?
(1) y = x + 4 (2) x + 4 = 15 Hi BunuelYou 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?
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20 Jul 2018, 12:21
Bunuel wrote: \(a = x^2 + 2x + y\). Is a even?
(1) y = x + 4
(2) x + 4 = 15 Hi Hero8888You were faster than in posting same examples BunuelCan 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?
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20 Jul 2018, 13:16
Mo2men wrote: Bunuel wrote: \(a = x^2 + 2x + y\). Is a even?
(1) y = x + 4
(2) x + 4 = 15 Hi Hero8888You were faster than in posting same examples BunuelCan 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?
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21 Jul 2018, 00:42
Bunuel wrote: For integers a, x, and y, \(a = x^2 + 2x + y\). Is a even?
(1) y = x + 4
(2) x + 4 = 15 You are all right. Veritas Prep modified the question and added the part saying that a, x, and y are integers.
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Re: For integers a, x, and y, a = x^2 + 2x + y. Is a even?
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21 Jul 2018, 01:13
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?
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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 = 15No 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




Re: For integers a, x, and y, a = x^2 + 2x + y. Is a even?
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21 Jul 2018, 20:06






