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For a real number $x$ the condition $|3x – 20| + |3x – 40| = 20$ necessarily holds if

  1. $9 < x < 14$
  2. $6 < x < 11$
  3. $7 < x < 12$
  4. $10 < x < 15$
in Quantitative Aptitude 2.7k points 5 79 342
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1 Answer

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Given that, $|3x-20| + |3x-40| = 20 ; x \in \mathbb{R} \quad \longrightarrow (1)$

We know that $,|x| = \left\{\begin{matrix} x\;;&x\geq 0 \\ -x\;; &x<0 \end{matrix}\right.$

We can open mod as positive and negative. There are four such cases.

$\textbf{Case 1:}\;\text{ Positive, Positive}$

$\Rightarrow 3x – 20 + 3x – 40 = 20$

$\Rightarrow 6x – 60 = 20$

$\Rightarrow 6x = 80$

$\Rightarrow \boxed{x = \frac{40}{3} = 13.33}$

$\textbf{Case 2:}\;\text{ Positive, Negative}$

$\Rightarrow 3x – 20 – (3x – 40) = 20$

$\Rightarrow 3x – 20 – 3x + 40 = 20$

$\Rightarrow \boxed{20 = 20\; {\color{Green} {\text{(True)}}}}$

$\textbf{Case 3:}\;\text{ Negative, Positive}$

$\Rightarrow \;– (3x – 20) + 3x – 40 = 20$

$\Rightarrow\; – 3x + 20 + 3x – 40 = 20$

$\Rightarrow \boxed{- 20 = 20\;\color{Red}{\text{(False)}}}$

$\textbf{Case 4:}\;\text{ Negative, Negative}$

$\Rightarrow \;– (3x – 20) – (3x – 40) = 20$

$\Rightarrow \;– 3x + 20 – 3x + 40 = 20$

$\Rightarrow \;– 6x =\; – 40$

$\Rightarrow \boxed{x = \frac{20}{3} = 6.66}$

$\therefore$ $\boxed{7 < x < 12}$

Correct Answer $: \text{C}$
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