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# If xy < 3, is x < 1?

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Math Expert
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If xy < 3, is x < 1? [#permalink]

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26 Jan 2014, 10:37
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The Official Guide For GMAT® Quantitative Review, 2ND Edition

If xy < 3, is x < 1?

(1) y > 3
(2) x < 3

Data Sufficiency
Question: 56
Category: Algebra Inequalities
Page: 157
Difficulty: 600

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Re: If xy < 3, is x < 1? [#permalink]

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26 Jan 2014, 11:47
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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? [#permalink]

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28 Jan 2014, 03:07
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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.

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Re: If xy < 3, is x < 1? [#permalink]

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26 Jan 2014, 10:38
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SOLUTION

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 contradcit 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.

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if-xy-is-less-than-3-is-x-less-than-61629.html
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Re: If xy < 3, is x < 1? [#permalink]

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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? [#permalink]

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01 Feb 2014, 09:07
SOLUTION

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 contradcit 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.

Similar questions to practice:
if-xy-4-is-x-2-1-y-1-2-y-x-129251.html
if-ab-7-is-b-166192.html
if-xy-is-less-than-3-is-x-less-than-61629.html
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Re: If xy < 3, is x < 1? [#permalink]

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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.

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.

Here, both the positive and negative values of x and y have been considered.

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Re: If xy < 3, is x < 1? [#permalink]

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04 Mar 2016, 10:07
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Re: If xy < 3, is x < 1? [#permalink]

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08 Jun 2017, 14:32
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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If xy < 3, is x < 1? [#permalink]

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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

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If xy < 3, is x < 1?   [#permalink] 08 Jun 2017, 14:47
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