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Is |xy| > x^2*y^2 ? (1) 0 < x^2 < 1/4 (2) 0 < y^2 < 1/9

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Is |xy| > x^2*y^2 ? (1) 0 < x^2 < 1/4 (2) 0 < y^2 < 1/9  [#permalink]

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New post 05 Mar 2018, 00:36
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C
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E

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Question Stats:

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Re: Is |xy| > x^2*y^2 ? (1) 0 < x^2 < 1/4 (2) 0 < y^2 < 1/9  [#permalink]

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New post 05 Mar 2018, 00:39
Bunuel wrote:
Is |xy| > x^2*y^2 ?

(1) 0 < x^2 < 1/4
(2) 0 < y^2 < 1/9


9. Inequalities



For more check Ultimate GMAT Quantitative Megathread


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Re: Is |xy| > x^2*y^2 ? (1) 0 < x^2 < 1/4 (2) 0 < y^2 < 1/9  [#permalink]

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New post 05 Mar 2018, 02:05
Bunuel wrote:
Is |xy| > x^2*y^2 ?

(1) 0 < x^2 < 1/4
(2) 0 < y^2 < 1/9


both x and y should be between -1 and 1 and not equal to -1, 0 and 1.

Stmt 1: insufficient since no info about y is given
Stmt 2: insufficient since no info about x is given

Stmt 1 and 2: sufficient

IMO answer should be "C"
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Re: Is |xy| > x^2*y^2 ? (1) 0 < x^2 < 1/4 (2) 0 < y^2 < 1/9  [#permalink]

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New post 05 Mar 2018, 02:11
Bunuel wrote:
Is |xy| > x^2*y^2 ?

(1) 0 < x^2 < 1/4
(2) 0 < y^2 < 1/9


We'll show two approaches:
The first relies on number properties and is a Logical approach.

x^2*y^2 = (xy)^2.
So we're asked when the square of a number (xy)^2 is smaller than its absolute value of a number (xy).
This happens when the square is a fraction.
Written as an equation, |xy| > (xy)^2 when 0 < (xy)^2 < 1.
Looking at our statements, (1) tells us that x^2 is a fraction. Without knowing the value of y^2, this is insufficient.
Similarly, (2) gives us only the range of y^2 and is insufficient.
Combined, we know that 0 < (xy)^2 < 1, exactly what we need.

(C) is our answer.

If this confuses you, then picking numbers is a very easy way to solve the question.
This is an Alternative approach.

(1) Say x^2 = 1/9 and y = 0. Then |xy| = 0 and x^2*y^2 = 0 so the answer is NO.
We'll try to challenge this
Say x^2 = 1/9 and y = 1. Then |xy| = 1/3 and x^2*y^2 = 1/9. Then the answer is YES.
Insufficient.

(2) is entirely symmetrical - we can choose y = 1/16 and x = 0 and then y = 1/16 and x = 1
Insufficient.

Combined:
Let's say x^2 = y^2 = 1/16. Then x and y are 1/4 or -1/4 meaning that |xy| = 1/16 which is definitely larger than (1/16)*(1/16)
So our first numbers give us a YES.
We now need to challenge this by looking for numbers that give a NO.
Trying this out for a few more pairs, we can see that x^2*y^2 is always a fraction and |xy| is always a larger fraction.

Once again, (C) is our answer.
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Re: Is |xy| > x^2*y^2 ? (1) 0 < x^2 < 1/4 (2) 0 < y^2 < 1/9  [#permalink]

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New post 12 Mar 2019, 07:41
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Bunuel wrote:
Is |xy| > x²y² ?

(1) 0 < x² < 1/4
(2) 0 < y² < 1/9


Target question: Is |xy| > x²y² ?

Statement 1: 0 < x² < 1/4
No information about y
Statement 1 is NOT SUFFICIENT

Statement 2: 0 < y² < 1/9
No information about x
Statement 2 is NOT SUFFICIENT

Statements 1 and 2 combined
Statement 1 tells us that -1/2 < x < 1/2, and x ≠ 0
Statement 2 tells us that -1/3 < y < 1/3, and y ≠ 0
In other words, x and y both have magnitudes that are less than 1.

Notice that, when 0 < x < 1, then x > x²
For example, if x = 1/4, then x² = 1/16
Notice that x > x²

Likewise, if y = 2/7, then y² = 4/49
Notice that y > y²

So, if -1/2 < x < 1/2, and x ≠ 0, then we know that |x| > x²
Likewise, if -1/3 < x < 1/3, and y ≠ 0, then we know that |y| > y²

When we combine both properties, we can conclude that |xy| > x²y²
So, the answer to the target question is YES, it is true that |xy| > x²y²
Since we can answer the target question with certainty, the combined statements are SUFFICIENT

Answer: C

Cheers,
Brent
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Re: Is |xy| > x^2*y^2 ? (1) 0 < x^2 < 1/4 (2) 0 < y^2 < 1/9   [#permalink] 12 Mar 2019, 07:41
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