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Is xy>x/y?

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Source: — Data Sufficiency |
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by DanaJ » Fri Feb 13, 2009 10:48 pm
1 is insufficient since we do not know anything about whether x and y are smaller or greater than 1 and/or -1. From xy > 0, we can conclude that they have the same sign (they're both positive or negative at the same time), but that doesn't help solve the problem. Let's take some numeric examples:
a. say x = 0.5 and y = 0.8. Then xy = 0.40 and x/y = 0.5125, with xy < x/y
b. say x = 3 and y = 4. Then xy = 12 and x/y = 0.75, with xy > x/y

2 is insufficient as well, largely for the same reason.

Take the two stmts together and all you get is that you know for sure that they are both negatives. This doesn't help much either, since again, you don't know if x and y are greater than -1 or not.
So E is the correct answer.
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by earth@work » Sat Feb 14, 2009 9:30 am
thanks danaJ
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by AndreiGMAT » Tue Jul 19, 2016 6:53 am
Let's make it more challenging:

Is xy>x/y?

(1) 0<y<1
(2) xy>1
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by [email protected] » Tue Jul 19, 2016 12:16 pm
Hi AndreiGMAT,

We're asked if XY > X/Y. This is a YES/NO question. We can solve this question with a mix of TESTing VALUES and Number Properties.

1) 0 < Y < 1

Since this Fact tells us nothing about X, it's likely insufficient. We can prove this rather easily by TESTing VALUES.

IF....
X = 0
Y = 1/2
Then the answer to the question is NO.

IF....
X = -1
Y = 1/2
Then the answer to the question is YES.
Fact 1 is INSUFFICIENT

2) XY > 1

IF....
X = 2
Y = 1
Then the answer to the question is NO.

IF....
X = 1
Y = 2
Then the answer to the question is YES.
Fact 2 is INSUFFICIENT

Combined, we know...
0 < Y < 1
XY > 1

Since Y is a positive fraction, the only way for XY to be greater than 1 is if X is greater than 1. When multiplying a value greater than 1 by a positive fraction, the result is SMALLER than the initial value. When dividing a value greater than 1 by a positive fraction, the result is always BIGGER than the initial value. With these restrictions, we can deduce that XY will ALWAYS be SMALLER than X/Y, so the answer to the question is ALWAYS NO.
Combined, SUFFICIENT

Final Answer: C

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by AndreiGMAT » Tue Jul 19, 2016 1:35 pm
Hi Rich.C@EMPOWERgmat,

Thank you for the clear explanation. Can the problem be solved using algebraic way?
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by Mo2men » Wed Jul 20, 2016 8:24 am
AndreiGMAT wrote:Hi Rich.C@EMPOWERgmat,

Thank you for the clear explanation. Can the problem be solved using algebraic way?
Hi,

I think using Testing values or plug in numbers is much more easier in this case, especially that fact 1 limits your choice no a narrow range.
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by DavidG@VeritasPrep » Wed Jul 20, 2016 8:51 am
AndreiGMAT wrote:Hi Rich.C@EMPOWERgmat,

Thank you for the clear explanation. Can the problem be solved using algebraic way?
With a little logic, you can see that each statement alone is not sufficient.

Together: from statement 1, we know that y is a positive fraction. Statement 2 tells us that x and y are both positive or both negative, so if y is positive, x is also positive.

Initial question: Is xy > x/y?
Multiply both sides by y (sign won't flip because we know y is positive) to get: Is xy^2 > x?

Divide both sides by x (sign won't flip because x is positive) to get: Is y^2 > 1?

Well, y is a fraction between 0 and 1. If we square such a number, we know that result will not be greater than 1, and the answer is always NO. Together they are sufficient.
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by DavidG@VeritasPrep » Wed Jul 20, 2016 12:24 pm
For another good Data Sufficiency inequality question, see here: https://www.beatthegmat.com/is-y-x-1-x-y-t103013.html
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by Matt@VeritasPrep » Fri Jul 22, 2016 1:43 am
AndreiGMAT wrote:Hi Rich.C@EMPOWERgmat,

Thank you for the clear explanation. Can the problem be solved using algebraic way?
Let's start by rephrasing the prompt:

xy > x/y

xy - x/y > 0

(xy² - x)/y > 0

x * (y² - 1)/y > 0

To solve this, we'll likely need two things:

i) The sign of y. If y > 0, then our question becomes "Is x * (y² - 1) > 0?" but if y < 0, then our question becomes "Is x * (y² - 1) < 0?"

ii) Some information about x and (y² - 1), specifically whether they have the same sign or opposite signs.

S1 tells us x and y have the same sign. This is helpful, but it doesn't tell us about the sign of (y² - 1) or about whether y > 0. NOT SUFFICIENT

S2 tells us y < 0. This lets us rephrase the question as "Is x * (y² - 1) < 0?" Also helpful, but NOT SUFFICIENT

S1+S2 still doesn't get us there: we know x < 0, but we don't know about (y² - 1). If y = -1/2, then this is negative, but if y = -2, then it's positive. NOT SUFFICIENT, E
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