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GMAT Diagnostic Test Question 43 Field: algebra Difficulty 700

Rating:

If \(x\) and \(y\) are integers, is \(|x|>|y|\)?

1. \(|x| = |y+1|\) 2. \(x^y = x! + |y|\)

A. Statement (1) ALONE is sufficient, but Statement (2) ALONE is not sufficient B. Statement (2) ALONE is sufficient, but Statement (1) ALONE is not sufficient C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient D. EACH statement ALONE is sufficient E. Statements (1) and (2) TOGETHER are NOT sufficient
_________________

Statement (1): Given that \(|x| = |y+1|\) If \(x\) and \((y+1)\) both have same signs (either positive or negative), then \(x = y+1\) and \(x>y\), which makes \(x\) and \(y\) consecutive integers. If \(x\) and \((y+1)\) both do not have the same signs (one is negative and the other is positive), then \(x = -(y+1)\) or \((x + y + 1) = 0\). In that case, either one could be greater in absolute value. S1 is not sufficient by itself.

Statement (2): If \(x^y = x! + |y|\), \(x\) and \(y\) both have to be positive because \(x! + |y|\) is always a positive integer. If \(y\) is negative, \(x^y\) can never be an integer. So for \(x^y\) to be an integer, \(y\) has to be positive. Let's see what values satisfy the equation: If \(x = 1\), and \(y = 0\), \(x^y = x! + |y| = 1\). If \(x = 2\), and \(y = 2\), \(x^y = x! + |y| = 4\). So \(x\) and \(y\) have at least two possibilities. S2 is not sufficient by itself either.

Statement (1)+Statement (2): Because of Statement (2), \(x\) and \(y\) have to be positive, in which case, by Statement (1) \(x > y\). Sufficient. Therefore, the answer is C.
_________________

I think the question is still not correct, it should be given that x and y are positive integers (not consecutive). otherwise, we can not ignore the value of y = -10.

This question had some revisions and now it should be OK. I don't see any problem with the OE. B is the answer and GMAT TIGER explained it all. Is anything still unclear? Please read the OE carefully.

sher676 wrote:

I also dont get d as the answer. The question does not say that x and y are positive or integers. So A can't be sufficient.

Final OA/OE after a couple of changes and updates. Let me know if there is still an issue.

GMAT TIGER wrote:

Explanation

Rating:

Official Answer: B

Statement 1: \(xy = 6\) is sufficient to answer the question because solving the equation gives \(y = 3\) when \(x\) is 2 and \(y\) is -10 when \(x\) is \(-\frac{3}{5}\).

\(5x = y + 7\) \(x = \frac{y + 7}{5}\) ................i \(xy = 6\) ........................ ii

If \(y = - 10\), \(x = -3/5\). In this case, \(x - y = -3/5 - (-10) = 47/5\). Yes. If \(y = 3\), \(x = 2\). In this case, \(x - y = 2 - 3 = -1\). No.

In each case (x-y) > 0 and < 0. Hence statement 1 is not sufficient.

Statement 2: If \(x\) and \(y\) are consecutive integers, \(x = 2\) and \(y = 3\). When x and y are consecutive integers, either x = y+1 or y = x+1 is possible.

(i) If x = y+1 \(5x = y + 7\) \(5(y+1) = y + 7\) \(4y = 2\) \(y = 1/2\). Then \(x = 3/2\). However this is not possible becase x and y are even not integers. So this option is not valid.

(ii) If y = x+1 \(5x = y + 7\) \(5x = x + 1 + 7\) \(4x = 8\) \(x = 2\) Then, \(y = 3\). This is valid because x and y are consecutive integers.

So (x-y) = 2-3 = -1. Hence statement 2 is sufficient.

I think you're mixing the data from S1 with the data given in S2. Neither the question stem nor the S1 states that \(x\) is an integer. So the OE is right.

Do you agree?

pk wrote:

in your solution for xy=6, yo found y=-10 and x =-3/5. But this is not possible because X is not an integer. So answer should be D (I guess).

The question and OA/OE have been completely changed and now all the posts have nothing to do with the current question.

Maybe it's time to create a new thread for question 43.

Also, I think the current OE could be made clearer:

GMAT TIGER wrote:

Explanation Official Answer: С

Statement (1): Given that \(|x| = |y+1|\) If \(x\) and \((y+1)\) both have same signs (either positive or negative), then \(x = y+1\) and \(x>y\), which makes \(x\) and \(y\) consecutive integers. If \(x\) and \((y+1)\) both do not have the same signs (one is negative and the other is positive), then \(x = -(y+1)\) or \((x + y + 1) = 0\). In that case, either one could be greater in absolute value. S1 is not sufficient by itself.

Statement (2): If \(x^y = x! + |y|\), \(x\) and \(y\) both have to be positive because \(x! + |y|\) is always a positive integer. If \(y\) is negative, \(x^y\) can never be an integer. So for \(x^y\) to be an integer, \(y\) has to be positive. Let's see what values satisfy the equation: If \(x = 1\), and \(y = 0\), \(x^y = x! + |y| = 1\). If \(x = 2\), and \(y = 2\), \(x^y = x! + |y| = 4\). So \(x\) and \(y\) have different values. S2 is not sufficient by itself either.

Statement (1)+Statement (2): \(x\) and \(y\) are not equal and \(x > y\). Sufficient. Therefore its C.

It would be better to write:

Quote:

If \(x = 1\), and \(y = 0\), \(x^y = x! + |y| = 1\). If \(x = 2\), and \(y = 2\), \(x^y = x! + |y| = 4\). So \(x\) and \(y\) have at least two possibilities. S2 is not sufficient by itself either.

Statement (1)+Statement (2): Because of Statement (2), \(x\) and \(y\) have to be positive, in which case, by Statement (1) \(x > y\). Sufficient. Therefore, the answer is C.

Does anyone know how to easily derive from statement (2) the following possibilities: \(x = 1\) and \(y = 0\), and \(x = 2\) and \(y = 2\) ? I don't think it's that obvious. So it's easy to miss one possibility and answer the question differently.

Can someone tell me if x! is defined if x is negative? I thought it was.

If it is defined, then I don't see how in ii) both x and y need to be positive. I can see why y needs to positive in order to make the left side of the equation an integer, but not x.

Can someone tell me if x! is defined if x is negative? I thought it was.

If it is defined, then I don't see how in ii) both x and y need to be positive. I can see why y needs to positive in order to make the left side of the equation an integer, but not x.

Based on the example presented and the logic I think it should be 'y' is non-negative rather than y must be positive. Ie y=0 can leave x^y=1 as a positive integer as shown in the S2 example.? (doesn't change the answer). is this correct?

Does anyone know how to easily derive from statement (2) the following possibilities: \(x = 1\) and \(y = 0\), and \(x = 2\) and \(y = 2\) ? I don't think it's that obvious. So it's easy to miss one possibility and answer the question differently.

i didnt see an answer to this q. is there an easier way to get to 0,1 or 2? thanks.