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# I got this one off a kaplan test- I disagree with the

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Manager
Joined: 19 Sep 2005
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I got this one off a kaplan test- I disagree with the [#permalink]

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23 Oct 2005, 18:37
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I got this one off a kaplan test- I disagree with the answer. here goes:

IF X^2 + 5Y = 49 is y an integer?

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

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Manager
Joined: 14 Apr 2003
Posts: 84

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Re: kaplan DS [#permalink]

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23 Oct 2005, 19:05
Jennif102 wrote:
I got this one off a kaplan test- I disagree with the answer. here goes:

IF X^2 + 5Y = 49 is y an integer?

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

Is it C?

1. alone is no suff. x =1.1 and x = 2
2. alone is not suffice X can be anything

using both if x^2 is integer then x is integer too and if between 1 and 4 then Y is an integer too.

what is OE?

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SVP
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Re: kaplan DS [#permalink]

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23 Oct 2005, 19:51
Jennif102 wrote:
IF X^2 + 5Y = 49 is y an integer?
1. 1 < x < 4
2. X^2 is an integer

E.
from i, x = 1.00001 to 3.999999999999. so not suff.
from ii, x = any integer.
from i and ii, x= 2,3,4,5,6,7,8,9,10,11,12,13,14, and 15. y could be integer or fraction.. still not enough...

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Senior Manager
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23 Oct 2005, 20:01
I agree with Himalaya. The answer I get is E.

From prior posts, it is obvious that Stmt 1 & 2 is insufficient. Combining, we could take X^2 is 7, hence x would be sqrt(7), which is between 1 & 4. However this would make 'y' a fraction (42/5).

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Manager
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23 Oct 2005, 20:13
your correct-OA is E. The sqrt7 is a good example to highlight how C is incorrect.

thx

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Manager
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23 Oct 2005, 20:28
sudhagar wrote:
I agree with Himalaya. The answer I get is E.

From prior posts, it is obvious that Stmt 1 & 2 is insufficient. Combining, we could take X^2 is 7, hence x would be sqrt(7), which is between 1 & 4. However this would make 'y' a fraction (42/5).

Please explain this to me.

If 1 < x < 4 then for what values of X will we have x^2 as an integer? I think I am not understanding this point here.

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Senior Manager
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23 Oct 2005, 22:44
jainvineet wrote:
sudhagar wrote:
I agree with Himalaya. The answer I get is E.

From prior posts, it is obvious that Stmt 1 & 2 is insufficient. Combining, we could take X^2 is 7, hence x would be sqrt(7), which is between 1 & 4. However this would make 'y' a fraction (42/5).

Please explain this to me.

If 1 < x < 4 then for what values of X will we have x^2 as an integer? I think I am not understanding this point here.

There are 2 constraints here.
1) 1 < x < 4 and
2) x^2 is an integer

Lets take x = 1, then x^2 is also 1. If x = 4, then x^2 is 16. However the question did not mention that x is an integer. It mentions only x^2 is an integer, which means x^2 = 2,3,4 ... 15. In this case, x = sqrt2, sqrt3, sqrt4 = 2, and so on. We cannot take negative values here because of (1).

Hope this helps.

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Senior Manager
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Re: kaplan DS [#permalink]

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24 Oct 2005, 00:52
I was about to answer C, till I read Sudhagar's post. X= SQRT(7). Excellent sudhagar. Now I agree it should be E.

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24 Oct 2005, 16:28
always test DS questions with real positive numbers, negative numbers, sqrts, and zero.....

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Manager
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24 Oct 2005, 16:44
Quote:
I got this one off a kaplan test- I disagree with the answer. here goes:

IF X^2 + 5Y = 49 is y an integer?

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

I think some of you guys are making this too complicated. Let's look at this in two parts by picking numbers:

1. If x=2, then 5Y = 45, and Y=9 (integer)
If x=3, then 5Y = 43, and y=43/9 (not an integer)
So statement 1 is insufficient

2. If x^2 is an integer, then you can once again use the numbers I picked for statement 1, and see that this is insufficient.

The numbers I picked for statement 1 can be used when looking at both statements at once, showing that both statements together are still insufficient, thus the answer is E.

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Manager
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24 Oct 2005, 19:55
actually bumbleeman, if X = 3 than x^2 = 9 and 5y = 49-9 = 40 and y would be 8

you have to think about radicals b/n 1 and 4 to find the correct answer

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SVP
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24 Oct 2005, 21:02
Every time when you are told that x^2 is an integer, you need to aware that there may be a trap and need to check if x itself is an integer.
_________________

Keep on asking, and it will be given you;
keep on seeking, and you will find;
keep on knocking, and it will be opened to you.

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Manager
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24 Oct 2005, 21:07
Jennif102 wrote:
actually bumbleeman, if X = 3 than x^2 = 9 and 5y = 49-9 = 40 and y would be 8

you have to think about radicals b/n 1 and 4 to find the correct answer

D'oh! This is the sort of stupid mistake that costs me points on the practice tests. In this case I will blame the way that x^2 appears on the screen here, as opposed to how it does on the exam.

Of course, despite my best efforts, I still got the correct answer of E. As HongHu pointed out, the fact that x^2 is an integer does not mean that x is an integer.

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24 Oct 2005, 21:07
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# I got this one off a kaplan test- I disagree with the

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