Shortcut Trick Time for P to fill tank = 20 × (9/2) = 90 hours; Time for Q = 18 × (5/2) = 45 hours. Let Total Work = 360 units. Efficiency: P = 4 units/hr, Q = 8 units/hr. Work done by P and Q in 3 hours = (4 + 8) × 3 = 36 units. R empties 36 units in 4 hours ⇒ R's efficiency = −36 ÷ 4 = −9 units/hr. Time taken by P, Q, and R together = 360 ÷ (4 + 8 − 9) = 360 ÷ 3 = 120 hours. ∴ The correct answer is 120 hours. Alternate Method Given: ⇒ Pipe P fills 2/9 of the tank in 20 hours. ⇒ Time taken by P to fill the full tank = 20 × (9/2) = 90 hours. ⇒ Pipe Q fills 2/5 of the tank in 18 hours. ⇒ Time taken by Q to fill the full tank = 18 × (5/2) = 45 hours. Calculations: +------+-------+------------+------------+ | Pipe | Hours | Total Work | Efficiency | +------+-------+------------+------------+ | P | 90 | 360 | 4 | | Q | 45 | 8 | | +------+-------+------------+------------+ ⇒ Combined efficiency of P and Q = 4 + 8 = 12 units/hour. ⇒ Tank filled by P and Q in 3 hours = 12 × 3 = 36 units. ⇒ Pipe R alone empties these 36 units in 4 hours. ⇒ Efficiency of Pipe R = −36 ÷ 4 = −9 units/hour. ⇒ Combined efficiency of Pipes P, Q, and R = 4 + 8 − 9 = 3 units/hour. ⇒ Time taken by P, Q, and R together = Total Work ÷ Combined Efficiency = 360 ÷ 3 = 120 hours. ∴ The correct answer is 120 hours. Related Concepts: Pipes and Cisterns (LCM Method) Total capacity is assumed as the LCM of time taken by individual pipes. Efficiency = Total Capacity ÷ Time. Negative Work A pipe that empties a tank performs negative work. Its efficiency is subtracted from the efficiency of the filling pipes. Combined Work Rule If multiple pipes operate simultaneously, their combined efficiency is the algebraic sum of their individual efficiencies.