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# Is xy < x^2*y^2? (1) xy > 0 (2) x + y = 1

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Manager
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Is xy < x^2*y^2? (1) xy > 0 (2) x + y = 1  [#permalink]

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19 Oct 2018, 09:42
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65% (hard)

Question Stats:

52% (01:38) correct 48% (01:46) wrong based on 61 sessions

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Is $$xy < x^2y^2$$?

(1) xy > 0
(2) x + y = 1

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Director
Joined: 19 Oct 2013
Posts: 516
Location: Kuwait
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Re: Is xy < x^2*y^2? (1) xy > 0 (2) x + y = 1  [#permalink]

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19 Oct 2018, 14:19
PriyankaPalit7 wrote:
Is $$xy < x^2y^2$$?

1) xy>0
2) x+y=1

X and Y are of the same sign.

If x = 2 and y = 2

Then 4 < 16 yes

If x = 1/2 and y = 1/2

1/4 < 1/16 no.

Insufficient.

Eliminate A/D

Statement 2)

If x = -1

Y= 1
Then the statement -1 < 1 yes

If x = 1/2 and y = 1/2

Then the answer is xy < x^2 y^2 is no

Combined it is sufficient.

1/4 < 1/16 no.

Posted from my mobile device
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Re: Is xy < x^2*y^2? (1) xy > 0 (2) x + y = 1  [#permalink]

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19 Oct 2018, 23:16
Hi,

Slightly tricky question.

We can solve this using either Plugging in values for x and y OR we can do the mathematical approach.

Let’s see the mathematical approach.

Question:

Is x * y < x^2 * y^ 2 ?

Let’s simplify the question a bit.

Is 0 < x^2 * y^ 2 - x * y?

Is 0 < x * y (x * y - 1)?

If M = x * y, then

Is 0 < M * (M - 1)?

Answer to the question would be YES if M < 0 or M > 1 i.e., “x” and “y” have alternate signs OR “x” and “y” both positive or both negative and their product greater than 1.

Answer to the question would be NO if 0 < M < 1 i.e., “x” and “y” both positive or both negative and their product less then 1.

Statement I is insufficient:

x*y >0

i.e., M > 0

Answer to the question, would be YES or NO.

If M = 1/4

i.e., let’s say X = ½ and Y = ½, then answer to the question would be NO.

If M = 2

i.e., let’s say X = 2 and Y = 1, then answer to the question would be YES.

So not sufficient.

Statement II is insufficient:

x+y=1

If M = 1/4

let’s say X = ½ and Y = ½, then answer to the question would be NO.

If M = -2

i.e., let’s say X = 2 and Y = -1, then answer to the question would be YES.

So not sufficient.

Together it is sufficient.

x*y > 0

and x+y = 1

Only way we could achieve this is,

x and y has to lie between 0 and 1.

i.e., M has to be between 0 and 1. So answer to the question would be NO.

So together it is sufficient.

Hope it is clear.
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Is xy < x^2*y^2? (1) xy > 0 (2) x + y = 1  [#permalink]

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03 Jun 2019, 23:02
PriyankaPalit7 wrote:
Is $$xy < x^2y^2$$?

(1) xy > 0
(2) x + y = 1

Always we should use our basic inference from a qn stem.

qn is asking whether xy and (1-xy) are of opposite sign?
or the range of xy?
so as per modified qn stem in mind
option 1 says only one sided limit of xy
option 2 says the sum of x and y,however we need the range of xy,so again dillema in deriving the actual conclusion.
lets combine the upper limiting value can be judged from the 2nd statement,how? whether it is -19+20,or 0.5+0.5 ,the multiplication will result a value that is always <1
so the qn that whether xy(1-xy) is less than 0 will pop up in your mind as yes
because xy>0 and (1-xy) is positive (from 2nd statement) and the result is confirmed NO.
so OA-C
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Re: Is xy < x^2*y^2? (1) xy > 0 (2) x + y = 1  [#permalink]

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04 Jun 2019, 10:35
PriyankaPalit7 wrote:
Is $$xy < x^2y^2$$?

(1) xy > 0
(2) x + y = 1

I enjoyed this problem! It tests a lot of different Data Sufficiency skills in ways that aren't obvious at first.

First of all, understand the question stem.

There are a few different ways to handle the question stem, here. You could try to simplify it with math, but there's a trick. You can't just divide both sides by xy, because you don't know whether xy is positive or negative. If it's positive, you wouldn't have to flip the inequality sign. If it's negative, you would have to flip it. Since you don't know either way, you aren't allowed to do that division.

Instead, in this situation, try subtracting a term from both sides:

Is $$xy < x^2y^2$$?

Is $$0 < x^2y^2 - xy$$?

Is $$xy(xy-1) > 0$$?

OR, you can use a "decoding" kind of approach, and try to figure out what the question was asking you in plain English. When is xy less than (xy)^2? Well, if xy is negative, the answer would definitely be "yes". Also, if xy is a large positive number, the answer would be "yes" as well, because large positive numbers get bigger when you square them. In fact, the only situation where the answer would be "no" is if xy is between 0 and 1, inclusive. So the question is really asking, "is xy between 0 and 1?"

Now, approach the statements.

Statement 1: First, suppose that you used the math approach. You now know that xy is positive, so try plugging in some positive values for xy.

If xy = 0.5, then xy(xy-1) = 0.5(0.5-1) = -0.25, which is NOT greater than 0.
If xy = 100, then xy(xy-1) = 100(99) = 9900, which IS greater than 0.

So, the statement is insufficient.

Or, suppose that you used the "decoding" approach to the question. This statement tells you that xy is positive, but it doesn't tell you whether it's between 0 and 1, so it's not sufficient.

Statement 2: Similarly, suppose that you used the math approach. You know that x + y = 1. Try some values.

x = 0, y = 1: xy(xy-1) = 0(0-1) = 0, which is NOT greater than 0.
x = 0.5, y = 0.5: xy(xy-1) = 0.25(-0.75), which is NOT greater than 0.
x = 100, y = -99: xy(xy-1) = -9900(-9901), which IS greater than 0.

Or, suppose that you "decoded". Could xy be between 0 and 1? Yes, because x and y could both be decimals between 0 and 1. Or, xy could be negative, for instance if x is negative and y is positive. So, the answer could be yes or no, and the statement is insufficient.

Statements 1 and 2 together:

Things get a little simpler at this point! x and y can't both be negative, because then they can't sum to 1. So, x and y have to both be positive. They also have to be between 0 and 1, because their sum needs to be 1. So the answer to the question is "yes" and the statements are sufficient together.
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Re: Is xy < x^2*y^2? (1) xy > 0 (2) x + y = 1   [#permalink] 04 Jun 2019, 10:35
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