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tejal777
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If x^2 + 5y = 49. Is y an integer? 04 Oct 2009, 04:25
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If x^2 + 5y = 49. Is y an integer?

(1) 1 < x < 4
(2) x^2 is an integer

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Stmt 1 insuff. We dont know whether x is an integer or not.
Stmt 2 insuff.
Combined:
If x^2 is an integer,then x is an integer.Poss values of x=2,3 which makes y an integer.IMO C.
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Bunuel
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If x^2 + 5y = 49. Is y an integer? 04 Oct 2009, 04:54
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tejal777
x^2+5y=49. Is y an integer?
1. 1<x<4
2. X^2 is an integer

Stmt 1 insuff. We dont know whether x is an integer or not.
Stmt 2 insuff.
Combined:
If x^2 is an integer, then x is an integer. Poss values of x=2,3 which makes y an integer.IMO C.
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If x^2 + 5y = 49. Is y an integer?

Note that y to be an integer 49-x^2 should be a multiple of 5.

(1) 1 < x < 4. Well, if x is an integer, 2 or 3, then the answer will be YES but x may as well be some fraction say 3/2 and in this case the answer will be NO. Not sufficient.

(2) x^2 is an integer --> clearly insufficient.

(1)+(2) Now, \(x^2\) is an integer does NOT necessarily mean that \(x\) is an integer, it may as well be some irrational number: \(\sqrt{2}\), \(\sqrt{3}\), \(\sqrt{5}\), ... Note that as these values are in the range \(1<x<4\) then they are valid values for \(x\) and for them \(y\) is not an integer. So we still could have an YES (\(x=2\) for example) and a NO (for \(x=\sqrt{2}\)) answers. Not sufficient.

Answer E.

Side note: \(x\) can be an irrational number and \(y\) still be an integer for example \(x=\sqrt{14}<4\) --> \(14+5y=49\) --> \(y=7\).

Hope it's clear.
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gettinit
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If x^2 + 5y = 49. Is y an integer? 14 Dec 2010, 21:02
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if x^2+5y=49, is Y an integer?

1) 1<x<4
2) x^2 is an integer


I have an issue with the kaplan answer here. My answer would be c because stmt one tells you x is between integers 1 and 4, but it could be a non-integer so not sufficient, and 2 is not sufficient because x could be an integer. However, together the stmts satisfy as x could only be 2 or 3 and 2^2 and 3^2 both agree with the equation making y an integer. Please advise.
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If x^2 + 5y = 49. Is y an integer? 14 Dec 2010, 21:43
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if x^2+5y=49, is Y an integer?

1) 1<x<4
2) x^2 is an integer


I have an issue with the kaplan answer here. My answer would be c because stmt one tells you x is between integers 1 and 4, but it could be a non-integer so not sufficient, and 2 is not sufficient because x could be an integer. However, together the stmts satisfy as x could only be 2 or 3 and 2^2 and 3^2 both agree with the equation making y an integer. Please advise.
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Hi!

You've misinterpreted statement (2).

From (2), we know that x^2 is an integer. Based on this, you've erroneously concluded that x must be an integer as well.

However, that's not the case. If x = root2 or root3, x^2 will still be an integer. Moreover, root2 and root3 both satisfy statement 1 as well. In fact, all we know is that:

root1 < x < root16,

so x could be root2, root3, root4, root5, ...., root 15. root4 and root9 are both integers, but none of the others are.

So, even combined, we can't determine whether y is an integer.
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If x^2 + 5y = 49. Is y an integer? 23 Jul 2011, 21:34
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May be E i think if you mean x2 = x^2

A. not sufficient ... x can be 2 in which case y is an integer..
x can be 3.3 .. and y is not integer..

B. not sufficient.. x^2 is an integer.. if x^2 = 4 .. y is an integer
but if x^2 = 3... y is not an integer..

considering both statements, again, since x is between 1 and 4.. x can be 1.73 so x^2 = 3 then y is not an integer..
or x can be 2 and x^2=4 ..so y is an integer..
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If x^2 + 5y = 49. Is y an integer? 10 Jul 2025, 10:46
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