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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] New post 24 Oct 2009, 04:36
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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, 11:28, edited 1 time in total.
Edited the question and added the OA
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Re: Is xy <6 [#permalink] New post 24 Oct 2009, 04:46
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] New post 24 Oct 2009, 08:37
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] New post 24 Oct 2009, 20:50
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] New post 09 Apr 2012, 11:18
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] New post 09 Apr 2012, 11:31
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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] New post 09 Apr 2012, 11:41
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] New post 28 May 2013, 04:28
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Re: Problem : Is xy <6 (1) x < 3 and y < 2 (2) 1/2 < [#permalink] New post 28 May 2013, 14:14
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] New post 28 May 2013, 14:23
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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] New post 24 Jul 2014, 22:34
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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] 24 Jul 2014, 22:34
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