GMATH practice exercise (Quant Class 11)

If \(\,{{x - 1} \over {x - 3}}\,>\,{{x - 2} \over {x - 4}}\,\) is equivalent to \(\,a < x < b\,\), the value of \(b-a\) is:

(A) 0.25
(B) 0.5
(C) 0.75
(D) 1
(E) 1.25
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Let´s see a "non-standard" (but absolutely rigorous) approach, very interesting for high-level candidates:

\({{x - 1} \over {x - 3}} > {{x - 2} \over {x - 4}}\,\,\,\left( * \right)\,\,\,\, \Leftrightarrow \,\,\,\,\,a < x < b\)

\(? = b - a\)


\(\left( * \right)\,\,\,\,\,\mathop \Leftrightarrow \limits^{ \cdot \left( {x - 3} \right)\left( {x - 4} \right)} \,\,\,\,\left\{ \matrix{
\,\left( {x - 1} \right)\left( {x - 4} \right) > \left( {x - 2} \right)\left( {x - 3} \right)\,\,\,{\rm{if}}\,\,\,\,\left( {x - 3} \right)\left( {x - 4} \right) > 0\,\,\,\left( {\rm{I}} \right) \hfill \cr
\left( {x - 1} \right)\left( {x - 4} \right) < \left( {x - 2} \right)\left( {x - 3} \right)\,\,\,{\rm{if}}\,\,\,\,\left( {x - 3} \right)\left( {x - 4} \right) < 0\,\,\,\left( {{\rm{II}}} \right) \hfill \cr} \right.\)

\(\left( {\rm{I}} \right)\,\,\,\,\left( {x - 1} \right)\left( {x - 4} \right) > \left( {x - 2} \right)\left( {x - 3} \right)\,\,\,\,\, \Leftrightarrow \,\,\,\,{x^2} - 5x + 4 > {x^2} - 5x + 6\,\,\,{\rm{impossible!}}\)


\(\left( {{\rm{II}}} \right)\,\,\,\,\left( {x - 1} \right)\left( {x - 4} \right) < \left( {x - 2} \right)\left( {x - 3} \right)\,\,\,\,\, \Leftrightarrow \,\,\,\,{x^2} - 5x + 4 < {x^2} - 5x + 6\,\,\,{\rm{always}}\,\,\left[ {\,{\rm{for}}\,\,\left( {x - 3} \right)\left( {x - 4} \right) < 0\,} \right]\,\,!\,{\rm{!}}\)

\(\left( {x - 3} \right)\left( {x - 4} \right) < 0\,\,\,\,\, \Leftrightarrow \,\,\,\,\,3 < x < 4\)


\(? = 4 - 3 = 1\)


The correct answer is (D).


We follow the notations and rationale taught in the GMATH method.

Regards,
Fabio.
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