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When I rephrase this I need to know is X positive. If it is, is it greater than Y (magnitude).

Statement 1 is not sufficient because.

If X was -4 and Y was -5 then the answer would be -20. However if X was 5 and why was -4 then X would be +20 In both cases X is greater than y, but the result is different. In the first case XY is smaller than Y^2. In the second case XY is greater than Y^2. Not sufficient.

Statement 2. If X was 10 and Y was 1 then XY is greater than Y^2, If X was 1 and Y was 10 then XY is less than Y^2. Not sufficient.

1+2) Since Y is positive and smaller than X, XY is greater than Y^2.

Answer C

Last edited by kys123 on 11 Feb 2012, 02:12, edited 1 time in total.

I rephrased the question as x|y|>|y| (since y^= |y|. On solving this I rephrased as x>1?

basis this rephrased version. the answer id D. however OA is C..

Have I solved the equation wrongly?

Yes, your rephrasing is wrong. You can not substitute \(y^2\) with \(|y|\), because generally they are not equal: \(\sqrt{y^2}=|y|\). Next, even if it were "is \(x|y|>|y|\)?" you still cannot reduce it by \(|y|\) and write "is \(x>1\)", as \(y\) can be zero and you cannot reduce/divide by zero. \(x|y|>|y|\) can be rephrased as \(|y|(x-1)>0\).

Is x|y|>y^2?

(1) x>y --> if \(x>y>0\) then obviously \(x|y|>y^2\) but if \(0>x>y\) then \(x|y|<0<y^2\). Not sufficient.

(2) y>0 --> \(|y|=y\). No info about x. Not sufficient.

(1)+(2) Since \(x>y>0\) then \(xy>y^2\). Sufficient.

Though we know that x>y, we don't know anything about the signs of x and y. For example:

x=2, y=1 x|y|>y^2 2|1|>1^2 2>1 Valid

x=2, y=-3 x|y|>y^2 2|-3|>-3^2 6>9 Invalid

INSUFFICIENT

(2) y>0 This tells us nothing about x. For example:

x=10, y=1 x|y|>y^2 10|1|>1^2 10>1 Valid

x=1, y=10 x|y|>y^2 1|10|>10^2 10>100 Invalid

INSUFFICIENT

1+2) x>y, y>0 ===> x>y>0 If y>0 then x is greater than zero (and y) x=10, y=9 x|y|>y^2 10|9|>9^2 90>81

Think of it like this: The RHS is the smaller number times the smaller number. The LHS is the smaller number times a number larger than the smaller number. SUFFICIENT

(2) y>0 This tells us nothing about x. INSUFFICIENT

1+2) x>y and y>0. Therefore, x>y>0. We saw in #1 that when y is greater than zero, the inequality holds true. When it is less than zero, the inequality does not hold true. Just to be sure: x>y>0 2>1>0 x|y|>y^2 2|1|>1^2 2>1

Also, take note that x|y| is y*a number larger than y. This will always be greater than y*y (y^2) so long as x is positive.

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
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I rephrased the question as x|y|>|y| (since y^= |y|. On solving this I rephrased as x>1?

basis this rephrased version. the answer id D. however OA is C..

Have I solved the equation wrongly?

Can we rephrase the question as IxI>IyI??

reason.. if we square both side xIyI>y^2 we get x^2Y^2>Y^4

X^2>Y^2 or IXI>IYI...????

No, we cannot do this. We can raise both parts of an inequality to an even power if we know that both parts of an inequality are non-negative (the same for taking an even root of both sides of an inequality).

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

I rephrased the question as x|y|>|y| (since y^= |y|. On solving this I rephrased as x>1?

basis this rephrased version. the answer id D. however OA is C..

Have I solved the equation wrongly?

Can we rephrase the question as IxI>IyI??

reason.. if we square both side xIyI>y^2 we get x^2Y^2>Y^4

X^2>Y^2 or IXI>IYI...????

No, we cannot do this. We can raise both parts of an inequality to an even power if we know that both parts of an inequality are non-negative (the same for taking an even root of both sides of an inequality).

Can we rephrase this question as is x>0 and is x>y? (This is what I did, and got the correct answer, but maybe I just got lucky).

The reason I rephrased this question as "is x>0 and x>y??", is because we know y^2 is positive, and lyl>0. Thus x has to be greater than 0 and x must be greater than y for it to be greater than y^2, because if x=y then x*lyl=y^2 and if x<Y, then x*lyl<y^2.

I rephrased the question as x|y|>|y| (since y^= |y|. On solving this I rephrased as x>1?

basis this rephrased version. the answer id D. however OA is C..

Have I solved the equation wrongly?

Ok so

Stmnt 1

Too many possibilities. X could be 4 and y could be 3 etc...

Stmnt 2

We know nothing about X - what if X is 4 and y is "3"? Or what if x is indeed 3 and y is 3

Insuff

Stmnt 1 & 2

If Y cannot be a negative value and X is greater than Y then XY will always be greater - if x is greater than y then there is no way the equation can be equal unless y is negative, which is ruled out by statement 2

I rephrased the question as x|y|>|y| (since y^= |y|. On solving this I rephrased as x>1?

basis this rephrased version. the answer id D. however OA is C..

Have I solved the equation wrongly?

Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

There 2 variables and 0 equation. Thus we need 2 equations to solve for the variables; the conditions provide 2 equations, so there is high chance that (C) will be the answer.

The question \(x|y| > y^2\) is equivalent to \(x|y| > |y|^2\) or \(|y| ( x - |y| ) > 0\). The equivalent question is if \(|y| ( x - |y| ) > 0\).

Condition 1) \(x = 2\), \(y = 1\) : It is true. \(x = 2\), \(y = 0\) : It is false. This condtion is not true.

Condition 2) This is not sufficient, since we don't know anything about \(x\).

Condition 1) & 2) Since \(y > 0\), we have \(|y| > 0\). Since \(x > y = |y| > 0\), \(x - |y| > 0\). Thus \(|y| ( x - |y| ) > 0\). Both conditions together are sufficient.

Normally for cases where we need 2 more equations, such as original conditions with 2 variables, or 3 variables and 1 equation, or 4 variables and 2 equations, we have 1 equation each in both 1) and 2). Therefore C has a high chance of being the answer, which is why we attempt to solve the question using 1) and 2) together. Here, there is 70% chance that C is the answer, while E has 25% chance. These two are the key questions. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer according to DS definition, we solve the question assuming C would be our answer hence using 1) and 2) together. (It saves us time). Obviously there may be cases where the answer is A, B, D or E.
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