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If wz < 2, is z < 1?

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If wz < 2, is z < 1? [#permalink]

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If wz < 2, is z < 1?

(1) w > 2
(2) z < 2
[Reveal] Spoiler: OA
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Re: Is z<1 [#permalink]

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New post 02 Jan 2010, 17:41
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If wz < 2, is z < 1?

(1) w > 2. Subtract \(w > 2\) from \(wz < 2\) (we can do it as the signs are in opposite direction):

\(wz-w<2-2\) --> \(w(z-1)<0\), as w is positive (given w>2), then the product to be negative \(z-1\) must be negative --> \(z-1<0\) --> \(z<1\). Sufficient.

(2) z < 2. If \(z=1.5>1\) and \(w=0\) (wz<2), then the answer is YES but if \(z=0<1\) and \(w=0\) (wz<0), then the answer is NO. Not sufficient.

Answer: A.
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Re: Is z<1 [#permalink]

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New post 02 Jan 2010, 17:46
Thanks Bunuel

Even i got A. However, the OA given in the GMAC paper test is D.

I guess the OA (like many others in the paper tests) is wrong
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Re: If wz < 2, is z < 1? [#permalink]

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New post 22 Feb 2014, 11:58
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Re: If wz < 2, is z < 1? [#permalink]

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New post 14 Apr 2014, 01:31
If wz < 2, is z < 1?

(1) w > 2
(2) z < 2

Sol.
opt.1 => w>2
i.e. z(something>2)<2
=> z<2/(something >2)
in the step above, R.H.S will always be <1. Hence Z<1

opt.2 => z<2
from this we can't say whether z will be less then 1.
Hence, A.
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Re: If wz < 2, is z < 1? [#permalink]

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New post 13 Mar 2016, 06:01
Here the simple way to conclude that statement 1 is sufficient is because w is always positive and greater than 2
for w=2 which it cant be => z=1
so as Z increases z will decrease => Z<1
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Re: If wz < 2, is z < 1?   [#permalink] 13 Mar 2016, 06:01
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