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10 Mar 2019, 08:37
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
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Question Stats:
66% (02:34) correct
34%
(02:19)
wrong
based on 166
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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
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
10 Mar 2019, 09:35
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Bring all the equations to the LHS.
\(\frac{(x-1)}{(x-3)}-\frac{(x-2)}{(x-4)}>0\)
Take the LCM, then we get,
\(\frac{x^2-5x+4-x^2+5x-6}{(x-3)(x-4)}>0\)
Now, x = 3 or x = 4.
Plotting this in wavy curve, we get.
(-∞, 3) to (4,+∞)
So a<x<b can be written as 3<x<4
b-a = 4-3 = 1
D is the answer.
\(\frac{(x-1)}{(x-3)}-\frac{(x-2)}{(x-4)}>0\)
Take the LCM, then we get,
\(\frac{x^2-5x+4-x^2+5x-6}{(x-3)(x-4)}>0\)
Now, x = 3 or x = 4.
Plotting this in wavy curve, we get.
(-∞, 3) to (4,+∞)
So a<x<b can be written as 3<x<4
b-a = 4-3 = 1
D is the answer.
10 Mar 2019, 16:36
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fskilnik
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
Let´s see a "non-standard" (but absolutely rigorous) approach, very interesting for high-level candidates: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
\({{x - 1} \over {x - 3}} > {{x - 2} \over {x - 4}}\,\,\,\left( * \right)\,\,\,\, \Leftrightarrow \,\,\,\,\,a < x < b\)
\(? = b - a\)
\(\left( * \right)\,\,\,\,\,\mathop \Leftrightarrow \limits^{ \cdot \left( {x - 3} \right)\left( {x - 4} \right)} \,\,\,\,\left\{ \matrix{\\
\,\left( {x - 1} \right)\left( {x - 4} \right) > \left( {x - 2} \right)\left( {x - 3} \right)\,\,\,{\rm{if}}\,\,\,\,\left( {x - 3} \right)\left( {x - 4} \right) > 0\,\,\,\left( {\rm{I}} \right) \hfill \cr \\
\left( {x - 1} \right)\left( {x - 4} \right) < \left( {x - 2} \right)\left( {x - 3} \right)\,\,\,{\rm{if}}\,\,\,\,\left( {x - 3} \right)\left( {x - 4} \right) < 0\,\,\,\left( {{\rm{II}}} \right) \hfill \cr} \right.\)
\(\left( {\rm{I}} \right)\,\,\,\,\left( {x - 1} \right)\left( {x - 4} \right) > \left( {x - 2} \right)\left( {x - 3} \right)\,\,\,\,\, \Leftrightarrow \,\,\,\,{x^2} - 5x + 4 > {x^2} - 5x + 6\,\,\,{\rm{impossible!}}\)
\(\left( {{\rm{II}}} \right)\,\,\,\,\left( {x - 1} \right)\left( {x - 4} \right) < \left( {x - 2} \right)\left( {x - 3} \right)\,\,\,\,\, \Leftrightarrow \,\,\,\,{x^2} - 5x + 4 < {x^2} - 5x + 6\,\,\,{\rm{always}}\,\,\left[ {\,{\rm{for}}\,\,\left( {x - 3} \right)\left( {x - 4} \right) < 0\,} \right]\,\,!\,{\rm{!}}\)
\(\left( {x - 3} \right)\left( {x - 4} \right) < 0\,\,\,\,\, \Leftrightarrow \,\,\,\,\,3 < x < 4\)
\(? = 4 - 3 = 1\)
The correct answer is (D).
We follow the notations and rationale taught in the GMATH method.
Regards,
Fabio.
