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if x^2 + 5y = 49, is y an integer?

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by airan » Sat Jul 12, 2008 11:21 pm
if x^2 + 5y = 49, is y an integer?
(1) 1 < x < 4
(2) x^2 is an integer.
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Re: if x^2 + 5y = 49, is y an integer?

by parallel_chase » Mon Jul 14, 2008 5:28 am
airan wrote:if x^2 + 5y = 49, is y an integer?
(1) 1 < x < 4
(2) x^2 is an integer.

The answer is C.

Statement I, if X is 2 or 3, Y is an integer, but if X is 1.5 or 2.5 Y is not an integer. Therefore NOT SUFFICIENT.

Statement II, if X^2 is 16 Y is not an integer, but X^2 is 9 Y is an integer. NOT SUFFICIENT.


Combining Statement I & II, Only way X^2 can be an integer is
if either X= 2 or 3, therefore Y is an integer.

Hence C.

Whats the OA?
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by Dante » Mon Jul 14, 2008 6:24 am
I think that answer should be E here.
I remember encountering a problem like this where the author gave justificaiton what if X^2 = 7.

Please correct me if this understanding is wrong.
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by parallel_chase » Mon Jul 14, 2008 7:19 am
Dante wrote:I think that answer should be E here.
I remember encountering a problem like this where the author gave justificaiton what if X^2 = 7.

Please correct me if this understanding is wrong.
X^2 = 7 if we consider the statement 2 in isolation.

If we look at both the statements together.

1<X<4 and X^2 is an integer.

Only way this could be possible is if X is 2 or 3

If we take X is 2.5, 1.3, 2.2, 3.5, this would contradict with the second statement because X^2 has to be an integer.

I could be wrong as well but i dont think i have missed anything.

Let me know if you have any explanation.

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by gxliu » Mon Jul 14, 2008 7:28 am
Answer is E.
If X=Sq root of 5...This satisifies both 1 and 2. And Y is not an integer
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by jsl » Mon Jul 14, 2008 7:29 am
parallel_chase wrote:
Dante wrote:1<X<4 and X^2 is an integer.
I got C first but then I looked again and decided on E. Basically, the question doesn't explicitly state that 1 < x < 4 is an integer (or does it?). Thus, this is a bit of a cruel trick because you fell into the trap that the question presented to you...

or am i completely wrong?

I'd be interested in the answer - is it E?
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by parallel_chase » Mon Jul 14, 2008 8:00 am
jsl wrote:
parallel_chase wrote:
Dante wrote:1<X<4 and X^2 is an integer.
I got C first but then I looked again and decided on E. Basically, the question doesn't explicitly state that 1 < x < 4 is an integer (or does it?). Thus, this is a bit of a cruel trick because you fell into the trap that the question presented to you...

or am i completely wrong?

I'd be interested in the answer - is it E?
I think you are absolutely right, i fell into the trap. E indeed should be the right answer.

Thanks.
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by ildude02 » Thu Jul 17, 2008 7:37 pm
Answer is indeed E. Nice trap question.
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Intially I didn't accept...later I understood

by kranti » Tue Mar 17, 2009 9:33 pm
I chose C as the answer when I was thrown this question in Kaplan CAT1 and I could not accept that C was wrong.

On second (and deeper) look, I see that X^2 is an integer which doesn't mean that X is an integer, so I can not assume X to be an integer between 1 & 4 (i.e., 2 or 3).

The explanation provided by Kaplan, does not clarify the confusion very well.
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by hijazim » Wed Mar 18, 2009 5:48 am
The answer should b E.
It's clear that each statement alone is insufficient.

Both statements together we get:

X^2 is an integer such as 1<X^2<16

For all the integers between 1 and 16 as being the value of X^2, y will sometimes be integer and other times not.

Consider X^2=2 => y= 47/5 = 9.4 NOT INTEGER
Consider X^2=4 => y= 45/5 = 9 INTEGER

Clearly, the answer is E.
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