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.

Answer: E.

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 ?

Please confirm

Thanks

Mridul

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