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If c and d are real numbers, is the value of c less than the

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If c and d are real numbers, is the value of c less than the [#permalink]  29 Nov 2012, 08:43
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If c and d are real numbers, is the value of c less than the value of d?

(1) cd < 1
(2) (c - 1)/( d - 2) < 1

Last edited by Bunuel on 29 Nov 2012, 08:46, edited 1 time in total.
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Re: If c and d are real numbers, is the value of c less than the [#permalink]  29 Nov 2012, 09:18
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If c and d are real numbers, is the value of c less than the value of d?

(1) cd < 1 --> if $$d=0$$, then $$cd=0<1$$, thus $$c$$ can take any value, less as well as more than $$d$$. Not sufficient.

(2) (c - 1)/( d - 2) < 1. Consider the same case: if $$d=0$$, then we have that $$\frac{c-1}{-2}<1$$ --> $$c>-1$$, thus $$c$$ again can be less (for example -1/2) as well as more than $$d$$ (for example 1). Not sufficient.

(1)+(2) Examples from (2) are naturally valid for (1), thus $$c$$ can be less as well as more than $$d$$. Not sufficient.

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Re: If c and d are real numbers, is the value of c less than the [#permalink]  19 Dec 2012, 22:42
Bunuel wrote:
If c and d are real numbers, is the value of c less than the value of d?

(1) cd < 1 --> if $$d=0$$, then $$cd=0<1$$, thus $$c$$ can take any value, less as well as more than $$d$$. Not sufficient.

(2) (c - 1)/( d - 2) < 1. Consider the same case: if $$d=0$$, then we have that $$\frac{c-1}{-2}<1$$ --> $$c>-1$$, thus $$c$$ again can be less (for example -1/2) as well as more than $$d$$ (for example 1). Not sufficient.

(1)+(2) Examples from (2) are naturally valid for (1), thus $$c$$ can be less as well as more than $$d$$. Not sufficient.

Hi Bunuel,

St 2 is given as (c-1)/(d-2) < 1, This implies (c-1) < (d-2) or (c-1)> (d-2) depending upon the sign of (d-2).

Combining we get cd<1, c-1<d-2 or c-1> d-2. From these combinations we cannot make out whether c>d or d>c. Hence E

Is this explanation correct allegbrically ?

Thanks
Mridul
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Re: If c and d are real numbers, is the value of c less than the   [#permalink] 19 Dec 2012, 22:42
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