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If xy < 3, is x < 1?
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26 Jan 2014, 10:37
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85% (00:59) correct 15% (01:12) wrong based on 468 sessions
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Re: If xy < 3, is x < 1?
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26 Jan 2014, 10:38



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Re: If xy < 3, is x < 1?
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26 Jan 2014, 11:47
Bunuel wrote: Statement 1) y > 3 and if xy < 3, then X has to be < 1 (One of the variable has to compensate for the increase in other variable). Sufficient. Statement 2) x < 3 As there is no information of y, we cannot determine if x < 1 So insufficient. Hence Option A)
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Re: If xy < 3, is x < 1?
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26 Jan 2014, 23:03
From S1:y>3=>x=fraction/negative integer=>x<1.Sufficient From S2:x<3=>x=2,1,0...depending on y.It cannot be said whether x<1 or >1.Hence insufficient. Ans.A



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Re: If xy < 3, is x < 1?
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28 Jan 2014, 03:07
Statement (1). By number picking strategy, if y>3 and xy<3, then X is always less than 1. Sufficient
Statement (2). X can be greater or less than 1 and still xy<3 holds. Insufficient.
Answer A



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Re: If xy < 3, is x < 1?
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01 Feb 2014, 09:07



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Re: If xy < 3, is x < 1?
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19 Nov 2014, 16:18
If xy < 3, is x < 1?
(1) y > 3. If x were \(\geq{1}\), then xy would be more than or equal to 3, which would contradict the condition that xy < 3. Thus x must be less than 1. Sufficient.
(2) x < 3. Not sufficient: consider x=y=0 and x=2 and y=0. Not sufficient.
Answer: A.I'm a bit confused after analyzing another similar question with different solution In st1, why negative values of x are not being considered ? I came up with answer "E", considering negative values of x However, just when I was convinced that, values of x and y need to be assumed positive in this type of prompt, I met the following question Is xy < 6?
(1) x < 3 and y < 2. (2) 1/2 < x < 2/3 and y^2 < 64.
Data Sufficiency, Question: 68, Page: 157, OG quant review 2nd edition
SOLUTION
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 maximize 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. Here, both the positive and negative values of x and y have been considered. Bunuel can you please explain the contradiction ?



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If xy < 3, is x < 1?
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08 Jun 2017, 14:47
If xy < 3, is x < 1?
(1) y > 3
Let y = 4.......4x <3.......x <3/4< 1....Yes
Let y = 100...100x <3....x <0.03 <1...Yes
Always Yes
Sufficient (2) x < 3
Let x =1/3........1/3 y < 3 ......y < 8/3.......Answer is yes
Let x = 2...........2 y < 3........y < 2/3........Answer is NO
Insufficient
Answer: A



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Re: If xy < 3, is x < 1?
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14 Sep 2018, 17:37
Hi,
here are my two cents for this questions
We should know two rules for this for one of the method to solve algebrically
1. if a>b & c<d then From one inequality it is possible to subtract term wise another inequality of the opposite sense retaining the sense of inequality from which the other was subtracted ..
Here is what i mean Say a>b & c<d then ac > bd.
2. 1. if a>b & c>d then Two inequality having same sense may be added term wise Here is what it will be a+c>b+d
So From Stamt 1: xy < 3  (i) and y >3 (ii)
(i) (ii) Using rule 1 we have xyy<0 y(x1)<0 so wither y<0 and or x1<0 since y>3 it must be that x1<0 or x<1 So sufficient.
From Stmt 2: X<3(iii)
adding (i) and (iii) xy+x<6 x(y+1)<6 so we have x<6 and or y+1<6 or x<6 and or y<5 or we have x<3 and or y<5 we can't say if x<1 Hence insufficient.
Probus




Re: If xy < 3, is x < 1? &nbs
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14 Sep 2018, 17:37






