Question:

<p><strong>Step 1:</strong> From mean = 5: $x_1 + x_2 + x_3 + x_4 + x_5 = 25$</p><p>With three observations as 3, 4, 11: $3 + 4 + 11 + x_4 + x_5 = 25$</p><p>Therefore: $x_4 + x_5 = 7$</p><p><strong>Step 2:</strong> From variance = 124: $\frac{1}{5}\sum_{i=1}^{5}(x_i - 5)^2 = 124$</p><p>So: $\sum_{i=1}^{5}(x_i - 5)^2 = 620$</p><p><strong>Step 3:</strong> Calculate known terms: $(3-5)^2 + (4-5)^2 + (11-5)^2 = 4 + 1 + 36 = 41$</p><p>Therefore: $(x_4 - 5)^2 + (x_5 - 5)^2 = 579$</p><p><strong>Step 4:</strong> Let $x_4 - 5 = a$ and $x_5 - 5 = b$. Then $a + b = -3$ (since $x_4 + x_5 = 7$) and $a^2 + b^2 = 579$</p><p>From $(a+b)^2 = a^2 + 2ab + b^2$: $9 = 579 + 2ab$, so $ab = -285$</p><p><strong>Step 5:</strong> The values $a, b$ satisfy $t^2 + 3t - 285 = 0$, giving $t = \frac{-3 \pm \sqrt{9+1140}}{2} = \frac{-3 \pm 33}{2}$</p><p>So $a = 15, b = -18$ (or vice versa), meaning $x_4 = 20, x_5 = -13$ (or vice versa)</p><p><strong>Step 6:</strong> $|3-5| + |4-5| + |11-5| + |20-5| + |-13-5| = 2 + 1 + 6 + 15 + 18 = 42$</p><p>∴ Answer: C</p>