GMAT Diagnostic Test Question 43

i think the answer should be D. We do not need any of the statements.
both x and y are consecutive integers, So , lets x = n and y = n+1;

put that in the given equation :

5n = n+1 + 7

this gives n = 2

therefore, x=2 and y = 3

so, x-y = -1..

both statements are redundant here..hence, the answer should be D..

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.

Please explain GMATTIGER

Explanation
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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

Replacing the value of \(x\) on eq ii: \(\frac{y(y + 7)}{5} = 6\)
\(y^2 + 7y = 30\)
\(y^2 + 10y - 3y - 30= 0\)
\(y (y + 10) - 3 (y + 10) = 0\)
\((y + 10) (y - 3) = 0\)
\(y = - 10\) or \(3\)

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.

Therefore B is correct.

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).

in the online test, you put the answer A instead of B, please check it.

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.

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.

Is this a 600 level question?????

Is this a 600 level question?????

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.

otherwise, for ii):

Let x= -1, y=2

x^y = x! + |y|
(-1)^2 = (-1)! + |2|
1 = -1 + 2
1 = 1

and
|x| > |y| is not true because |-1| < |2|

If this is accurate, then it changes the parameters of the two statements together. I had chosen E, not C.

Any help?

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.