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If x^2 + 5y = 49, is y an integer? 1) 1 < x < 4 2) x^2

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If x^2 + 5y = 49, is y an integer? 1) 1 < x < 4 2) x^2 [#permalink] New post 12 Sep 2005, 17:52
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If x^2 + 5y = 49, is y an integer?

1) 1 < x < 4

2) x^2 is an integer

Kaplan had the answer to this as E -> Not enough information to solve the problem. But I thing we can solve this problem if we take both the information together. Here is how I would do it:
1) tells me that x could be 3/2, 2, 5/2, 3 etc. This is not enough to solve the problem.

2) tells me that x^2 is an integer, if I combine this with statement (1) then it implies that x has to be either 2 or 3 (since the square of the fractions would between 1 & 4 would also result in a fraction). Substituting the value of x as either 2 or 3 I can then solve the equation for y and determine that y is an integer. I don't really need to find out the exact value of y but its enough for me know whether y is an integer. Since its a DS problem, this should work fine.


Do you see anything wrong with my reasoning? Please let me know if any of you have encountered this problem and have an alternate way of solving this.
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 [#permalink] New post 12 Sep 2005, 19:04
Taking statements 1 and 2 together, we have that x is between 1 and 4, not inclusive, and x^2 is an integer.

The trap is assuming that if x^2 is an integer, then x is an integer.

For example, x could be 2, or 3... or rt(8), which is approximately 2.83.

Clearly, if x is rt(8), then 5y=49-8=41 and y is 41/5 which is not an integer.

While if x is 3, then 5y=49-9=40 and y is 40/5 which is an integer.

Therefore both taken together are insufficient. Answer E.
  [#permalink] 12 Sep 2005, 19:04
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If x^2 + 5y = 49, is y an integer? 1) 1 < x < 4 2) x^2

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