Question:medium

For all real numbers $ x $, the condition $ |3x - 20| + |3x - 40| = 20 $ necessarily holds if

Updated On: Jan 15, 2026

  • \(6 < x < 11\)

  • \(7 < x < 12\)

  • \(10 < x < 15\)

  • \(9 < x < 14\)

Show Solution

The Correct Option is B

Solution and Explanation

The minimum value of the modulus function is zero. For the expression \(|3x - 20| + |3x - 40| = 20\), the minimum value for \(|3x - 20|\) occurs at \(x = \frac{20}{3}\), and the minimum value for \(|3x - 40|\) occurs at \(x = \frac{40}{3}\).

|3x - 20| + |3x - 40| = 20

Consider the equation:

\[ |3x - 20| + |3x - 40| = 20 \]

Function Behavior:

  • When \( x < \frac{20}{3} \), both \(3x - 20\) and \(3x - 40\) are negative. As \(x\) increases, their absolute values decrease, and thus their sum decreases.
  • When \( x > \frac{40}{3} \), both \(3x - 20\) and \(3x - 40\) are positive. As \(x\) increases, their absolute values increase, and thus their sum increases.
  • In the interval \( \left[ \frac{20}{3}, \frac{40}{3} \right] \), \(3x - 20 \geq 0\) and \(3x - 40 \leq 0\). The expression becomes \((3x - 20) + -(3x - 40) = 3x - 20 - 3x + 40 = 20\).

This implies:

\[ \text{The sum is constant at 20 for } x \in \left[ \frac{20}{3}, \frac{40}{3} \right] \Rightarrow x \in [6.66, 13.33] \]

Correct Option:

The interval that is entirely contained within the solution set \( \left[ \frac{20}{3}, \frac{40}{3} \right] \) is:

\( \boxed{(B)\ 7 < x < 12} \)

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