Is xy < 6 ? a.x < 3 and y < 2. b.1/2 < x <

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Is xy < 6 ? a.x < 3 and y < 2. b.1/2 < x < [#permalink]

26 Mar 2010, 12:02

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Is xy < 6 ?
a.x < 3 and y < 2.
b.1/2 < x < 2/3 and y^2 < 64.

source : OG10.

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Re: DS : Inequality . [#permalink]

26 Mar 2010, 12:22

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B.

St. 1 : X < 3, y < 2

If x = 2, y=1 --> xy = 2 --> xy < 6 - Yes

If x= -2, y= -4 , xy = 8 --> xy > 6 - No

Hence, insufficient.

St.2 : 0.5 < x < 0.67 , y^2 < 64 -> -8 < y < 8

If y < 0, xy < 0, hence, xy < 6
If y > 0, x = 0.55, y = 7 --> xy= 3.85 --> xy < 6

For any value of x & y we get xy < 6. Hence, sufficient.

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Re: DS : Inequality . [#permalink]

26 Mar 2010, 12:24

sushma0805 wrote:

Is xy < 6 ?
a.x < 3 and y < 2.
b.1/2 < x < 2/3 and y^2 < 64.

source : OG10.

I think the answer is B "Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient to answer the question asked".

Option 1:
From this statement, X & Y can be -ve or +ve.
X=2, Y=1; XY=2 < 6
X=-2, Y=-5; XY=10 > 6
so, Not Sufficient

Option 2:
Here we know that X is +ve, but Y still can be -ve or +ve

when Y is -ve; XY is always -ve & < 6

when Y is +ve;
lets consider X=\frac{2}{3}, Y=8, XY=\frac{16}{3} = 5.333
so the maximum value of XY is < 5.333 <6
So, Sufficient

let me know if you need any more explanation or if you feel any thing is wrong

-Ravi

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Re: DS : Inequality . [#permalink]

26 Mar 2010, 13:54

my pleasure

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Re: DS : Inequality . [#permalink]

27 Mar 2010, 09:35

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sushma0805 wrote:

Is xy < 6 ?
a.x < 3 and y < 2.
b.1/2 < x < 2/3 and y^2 < 64.

source : OG10.

a: insufficient as x = -4 and y = -3 xy = 12 > 6
also x=-2 and y=-2 xy = 4 < 6

b: x is +ve and less than .67 but y can have value < 8 or > -8 hence xy will be < 6 always.
Hence B

Re: DS : Inequality . [#permalink] 27 Mar 2010, 09:35

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Is xy < 6 ? a.x < 3 and y < 2. b.1/2 < x <

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Is xy < 6 ? a.x < 3 and y < 2. b.1/2 < x <

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