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Bunuel
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Nidzo
Rearranging the inequality: \(\frac{a}{f}*\frac{bc}{de}>\frac{bc}{de}\)

divide both sides by \(\frac{bc}{de}\)

\(\frac{a}{f}>1\)

\(a>f\)

Now the question is asking merely if a is greater than f.
You are implying that \(\frac{bc}{de}\) is positive. If \(\frac{bc}{de}\) is negative, you have to flip the inequality.
We do not know if \(\frac{bc}{de}\) is zero, positive, or negative. So your rearrangement is not correct.
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Bunuel
Is \((\frac{a}{d})(\frac{b}{e})(\frac{c}{f}) > (\frac{b}{d})(\frac{c}{e})\) ?


(1) \(bc > de\)

(2) \(a > f\)


Can you please explain why is it not B?
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Bunuel
Is \((\frac{a}{d})(\frac{b}{e})(\frac{c}{f}) > (\frac{b}{d})(\frac{c}{e})\) ?

(1) \(bc > de\)
(2) \(a > f\)

Solution:
Pre Analysis:
  • We are asked if \((\frac{a}{d})(\frac{b}{e})(\frac{c}{f}) > (\frac{b}{d})(\frac{c}{e})\) or not
    \(⇒\frac{abc}{def}-\frac{bc}{de}>0\)
    \(⇒\frac{bc}{de}(\frac{a}{f}-1)>0\)
  • This is possible in 2 cases:
    • Case 1: \(\frac{bc}{de}>0\) and \(\frac{a}{f}>1\)
    • Case 2: \(\frac{bc}{de}<0\) and \(\frac{a}{f}<1\)

Statement 1: \(bc > de\)
  • We can have both the cases here
    • Case 1: \(\frac{bc}{de}>0\) when \(bc=4\) and \(de=2\)
    • Case 2: \(\frac{bc}{de}<0\) when \(bc=4\) and \(de=-2\)
  • Thus, statement 1 alone is not sufficient and we can eliminate options A and D

Statement 2: \(a > f\)
  • We can have both the cases here
    • Case 1: \(\frac{a}{f}>1\) when \(a=4\) and \(f=2\)
    • Case 2: \(\frac{a}{f}<1\) when \(a=4\) and \(f=-2\)
  • Thus, statement 2 alone is also not sufficient

Combining:
  • Still both the cases are possible

Hence the right answer is Option E


NOTE: The most important thing in this question is the ability to handle inequalities well. For example \(bc > de\) doesnt mean that we can say \(\frac{bc}{de}>1\). We cannot say that without knowing the positive negative nature of \(de\)
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