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Is xy < 12? (1) x < 4 and y < 3 (2) x^2 < 121 and y^2 < 4

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Is xy < 12? (1) x < 4 and y < 3 (2) x^2 < 121 and y^2 < 4  [#permalink]

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New post Updated on: 06 Jul 2018, 07:27
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  35% (medium)

Question Stats:

71% (01:02) correct 29% (00:39) wrong based on 24 sessions

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Is xy < 12?

(1) x < 4 and y < 3
(2) x^2 < 121 and y^2 < 4

Originally posted by neiln413b on 06 Jul 2018, 07:21.
Last edited by Bunuel on 06 Jul 2018, 07:27, edited 1 time in total.
Renamed the topic, edited the question and added the OA.
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Re: Is xy < 12? (1) x < 4 and y < 3 (2) x^2 < 121 and y^2 < 4  [#permalink]

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New post 06 Jul 2018, 07:30
Is xy < 12?


(1) x < 4 and y < 3.

If x = y = 0, then xy = 0 < 12. So, in this case we have an YES answer.
If x = y = -10, then xy = 100 > 12. So, in this case we have a NO answer.

Not sufficient.


(2) x^2 < 121 and y^2 < 4:

-11 < x < 11 and -2 < y < 2.

If x = y = 0, then xy = 0 < 12. So, in this case we have an YES answer.
If x = -10 and y = -1.5 then xy = 15 > 12. So, in this case we have a NO answer.

Not sufficient.


(1)+(2) -11 < x < 4 and -2 < y < 2:

If x = y = 0, then xy = 0 < 12. So, in this case we have an YES answer.
If x = -10 and y = -1.5 then xy = 15 > 12. So, in this case we have a NO answer.

Not sufficient.


Answer: E.

Hope it's clear.

P.S. Please provide the OA's when posting questions. Thank you.
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Re: Is xy < 12? (1) x < 4 and y < 3 (2) x^2 < 121 and y^2 < 4  [#permalink]

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New post 06 Jul 2018, 07:32
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Re: Is xy < 12? (1) x < 4 and y < 3 (2) x^2 < 121 and y^2 < 4  [#permalink]

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New post 07 Jul 2018, 08:40
The key to getting this question right is to consider that x and y could be negative. While combining statements being careful with inequalities and testing appropriate cases as which give a "yes" and a "no" answer so that we can conclude "E" is the right answer.
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Re: Is xy < 12? (1) x < 4 and y < 3 (2) x^2 < 121 and y^2 < 4 &nbs [#permalink] 07 Jul 2018, 08:40
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