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# Is xy < 6 (1) x < 3 and y < 2 (2) 1/2 < x < 2/3 and y^2 < 64

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Is xy < 6 (1) x < 3 and y < 2 (2) 1/2 < x < 2/3 and y^2 < 64 [#permalink]

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24 Oct 2009, 05:36
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Is xy < 6

(1) x < 3 and y < 2

(2) 1/2 < x < 2/3 and y^2 < 64
[Reveal] Spoiler: OA

Last edited by Bunuel on 09 Apr 2012, 12:28, edited 1 time in total.
Edited the question and added the OA
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Re: Is xy <6 [#permalink]

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24 Oct 2009, 05:46
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IMO B.

stmt1: if both x and y are negative such that the product of their absolute values is > 6 then xy>6. eg. x=-8, y=-3

stmt2: x is +ve, -8<y<8.
Let us take max value of x =2/3 and max value of y=8, the product is <6.
Now, let us take max value of x=2/3 and a negative value of y=-8, the product is -ve and hence < 6.
One more try, take x=2/3 and y=5, the product is <6.
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Re: Is xy <6 [#permalink]

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24 Oct 2009, 09:37
Economist wrote:
IMO B.

stmt1: if both x and y are negative such that the product of their absolute values is > 6 then xy>6. eg. x=-8, y=-3

stmt2: x is +ve, -8<y<8.
Let us take max value of x =2/3 and max value of y=8, the product is <6.
Now, let us take max value of x=2/3 and a negative value of y=-8, the product is -ve and hence < 6.
One more try, take x=2/3 and y=5, the product is <6.

no sure about -8<y<8
y^2<64 either becomes
y<+8 or y<-8
OR
mod y < 64, meaning -64<y<64

what am i missing here?
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Re: Is xy <6 [#permalink]

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24 Oct 2009, 21:50
tihor wrote:
Economist wrote:
IMO B.

stmt1: if both x and y are negative such that the product of their absolute values is > 6 then xy>6. eg. x=-8, y=-3

stmt2: x is +ve, -8<y<8.
Let us take max value of x =2/3 and max value of y=8, the product is <6.
Now, let us take max value of x=2/3 and a negative value of y=-8, the product is -ve and hence < 6.
One more try, take x=2/3 and y=5, the product is <6.

no sure about -8<y<8
y^2<64 either becomes
y<+8 or y<-8
OR
mod y < 64, meaning -64<y<64

what am i missing here?

y^2 < 64 mean that -8 < y < 8 because min y^2 is 0 and max is close to 64. In that case, y<8 or y>-8. If y<-8, y could be -10, resulting y^2 = 100, which is >64 and violets the given info in the question.
Whenever you have y^2< 64, y is either +ve or 0, or -ve.

i. y^2 < 64 doesnot mean that -64 < y^2 < 64.
ii. "Mod y < 64, meaning -64<y<64" is not correct.
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Re: Problem : Is xy <6 (1) x < 3 and y < 2 (2) 1/2 < [#permalink]

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09 Apr 2012, 12:18
I didn't understand this one. could someone please explain why C is not the answer? thanks.
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Re: Problem : Is xy <6 (1) x < 3 and y < 2 (2) 1/2 < [#permalink]

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09 Apr 2012, 12:31
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mymbadreamz wrote:
I didn't understand this one. could someone please explain why C is not the answer? thanks.

Is $$xy<6$$?

(1) $$x<3$$ and $$y<2$$ --> now, if both $$x$$ and $$y$$ are equal to zero then $$xy=0<6$$ and the answer will be YES but if both $$x$$ and $$y$$ are small enough negative numbers, for example -10 and -10 then $$xy=100>6$$ and the answer will be NO. Not sufficient.

(2) $$\frac{1}{2}<x<\frac{2}{3}$$ and $$y^2<64$$, which is equivalent to $$-8<y<8$$ --> even if we take the boundary values of $$x$$ and $$y$$ to maixmize their product we'll get: $$xy=\frac{2}{3}*8\approx{5.3}<6$$, so the answer to the question "is $$xy<6$$?" will always be YES. Sufficient.

Answer: B.
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Re: Is xy < 6 (1) x < 3 and y < 2 (2) 1/2 < x < 2/3 and y^2 < 64 [#permalink]

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09 Apr 2012, 12:41
Bunuel, your explanations are so easy to understand. Thanks!
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Re: Is xy < 6 (1) x < 3 and y < 2 (2) 1/2 < x < 2/3 and y^2 < 64 [#permalink]

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28 May 2013, 05:28
Bumping for review and further discussion.
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Re: Problem : Is xy <6 (1) x < 3 and y < 2 (2) 1/2 < [#permalink]

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28 May 2013, 15:14
Bunuel wrote:
mymbadreamz wrote:
I didn't understand this one. could someone please explain why C is not the answer? thanks.

Is $$xy<6$$?

(1) $$x<3$$ and $$y<2$$ --> now, if both $$x$$ and $$y$$ are equal to zero then $$xy=0<6$$ and the answer will be YES but if both $$x$$ and $$y$$ are small enough negative numbers, for example -10 and -10 then $$xy=100>6$$ and the answer will be NO. Not sufficient.

(2) $$\frac{1}{2}<x<\frac{2}{3}$$ and $$y^2<64$$, which is equivalent to $$-8<y<8$$ --> even if we take the boundary values of $$x$$ and $$y$$ to maixmize their product we'll get: $$xy=\frac{2}{3}*8\approx{5.3}<6$$, so the answer to the question "is $$xy<6$$?" will always be YES. Sufficient.

Answer: B.

i dont understand i put A....i could answer the question with that information. N why are we letting x times y = to 0 why cant we let it be equal to 1 or 2?

if a number times a number is less than 6......cant we just say use 1?
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Re: Problem : Is xy <6 (1) x < 3 and y < 2 (2) 1/2 < [#permalink]

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28 May 2013, 15:23
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madzstar wrote:
Bunuel wrote:
mymbadreamz wrote:
I didn't understand this one. could someone please explain why C is not the answer? thanks.

Is $$xy<6$$?

(1) $$x<3$$ and $$y<2$$ --> now, if both $$x$$ and $$y$$ are equal to zero then $$xy=0<6$$ and the answer will be YES but if both $$x$$ and $$y$$ are small enough negative numbers, for example -10 and -10 then $$xy=100>6$$ and the answer will be NO. Not sufficient.

(2) $$\frac{1}{2}<x<\frac{2}{3}$$ and $$y^2<64$$, which is equivalent to $$-8<y<8$$ --> even if we take the boundary values of $$x$$ and $$y$$ to maixmize their product we'll get: $$xy=\frac{2}{3}*8\approx{5.3}<6$$, so the answer to the question "is $$xy<6$$?" will always be YES. Sufficient.

Answer: B.

i dont understand i put A....i could answer the question with that information. N why are we letting x times y = to 0 why cant we let it be equal to 1 or 2?

if a number times a number is less than 6......cant we just say use 1?

On DS questions when plugging numbers, goal is to prove that the statement is not sufficient. So we should try to get a YES answer with one chosen number(s) and a NO with another.

Now, for x=y=0 we got an YES answer and for x=y=-10 we got a NO answer, thus the statement is NOT sufficient.

Of course we could use some other numbers to get an YES and a NO answers to prove that the statement is not sufficient: x=y=0 and x=y=-10 are just examples of many possible sets.

Hope it's clear.
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Re: Is xy < 6 (1) x < 3 and y < 2 (2) 1/2 < x < 2/3 and y^2 < 64 [#permalink]

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24 Jul 2014, 23:34
Hello from the GMAT Club BumpBot!

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).

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Re: Is xy < 6 (1) x < 3 and y < 2 (2) 1/2 < x < 2/3 and y^2 < 64 [#permalink]

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15 Aug 2015, 03:45
Bunuel wrote:

i dont understand i put A....i could answer the question with that information. N why are we letting x times y = to 0 why cant we let it be equal to 1 or 2?

if a number times a number is less than 6......cant we just say use 1?

On DS questions when plugging numbers, goal is to prove that the statement is not sufficient. So we should try to get a YES answer with one chosen number(s) and a NO with another.

Now, for x=y=0 we got an YES answer and for x=y=-10 we got a NO answer, thus the statement is NOT sufficient.

Of course we could use some other numbers to get an YES and a NO answers to prove that the statement is not sufficient: x=y=0 and x=y=-10 are just examples of many possible sets.

Hope it's clear.[/quote]

Bunuel

Please confirm

We can calculate the range in II

it will come from -5.28 < xy < 5.28

This is obviously less than 6

Is this right?
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Re: Is xy < 6 (1) x < 3 and y < 2 (2) 1/2 < x < 2/3 and y^2 < 64 [#permalink]

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14 Sep 2016, 12:50
Hello from the GMAT Club BumpBot!

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).

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Re: Is xy < 6 (1) x < 3 and y < 2 (2) 1/2 < x < 2/3 and y^2 < 64   [#permalink] 14 Sep 2016, 12:50
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