Mike, an experienced bricklayer, can build a wall in $3$ hours; his son, who is learning, can do the same job in $6$ hours. How long does it take them to build the wall together?

Their hourly rates: - Mike: $\tfrac{1}{3}$ of the wall per hour. - Son: $\tfrac{1}{6}$ of the wall per hour. Combined rate: $\tfrac{1}{3} + \tfrac{1}{6} = \tfrac{2}{6} + \tfrac{1}{6} = \tfrac{3}{6} = \tfrac{1}{2}$ of the wall per hour. Time to finish one wall: $t = \dfrac{1}{1/2} = 2$ hours. - Trap A ($1.5 = 3 / 2$) takes a simple average of solo times — but the correct combined-time is the **harmonic-mean-style** computation above. - Trap C ($4.5 = (3 + 6) / 2$) is an arithmetic mean — also wrong. - Trap D ($9 = 3 + 6$) adds solo times. Wrong direction entirely. General formula: if two workers alone take $a$ and $b$ hours, together they take $\dfrac{a \cdot b}{a + b}$ hours. Here $\dfrac{3 \cdot 6}{3 + 6} = \dfrac{18}{9} = 2$ ✓.