Question 1

Two pipes can fill a tank with water in 15 and 12 hours respectively and a third pipe can empty it in 4 hours. If the pipes be opened in order at 8, 9 and 11 a.m. respectively, the tank will be emptied at (S.S.C., 2005)

Accepted Answer

Concept & Net Emptying Rate: When fill pipes and an emptying pipe operate simultaneously, assign positive efficiencies to filling and negative to emptying. If the net efficiency is negative, the tank will empty. Net Efficiency = \sum Fill Rates - \sum Empty Rates Step-by-Step Solution: 1. Given: Pipe A fills in 15 hrs, Pipe B fills in 12 hrs, Pipe C empties in 4 hrs. 2. Calculation: - Total capacity = LCM(15, 12, 4) = 60 units. - Efficiency of A = +60 / 15 = +4 units/hr. - Efficiency of B = +60 / 12 = +5 units/hr. - Efficiency of C = -60 / 4 = -15 units/hr. 3. Calculate the volume filled before the emptying pipe opens at 11 a.m.: - 8 a.m. to 11 a.m. (3 hrs): Pipe A is open. Work = 3 * 4 = 12 units. - 9 a.m. to 11 a.m. (2 hrs): Pipe B is open. Work = 2 * 5 = 10 units. - Total water in the tank at 11 a.m. = 12 + 10 = 22 units. 4. Calculate emptying time after 11 a.m.: - After 11 a.m., all pipes are open. - Net efficiency = +4 + 5 - 15 = -6 units/hr (tank is emptying at 6 units/hr). - Time required to empty the 22 units = 22 / 6 = 11/3 hours = 3 hours and 40 minutes. 5. Determine final time: - 11:00 a.m. + 3 hours 40 minutes = 2:40 p.m. Exam Strategy & Shortcut: Find the accumulated volume quickly: A works 3 hours, B works 2 hours \rightarrow 3(4) + 2(5) = 22 units. Net rate is -6. Time is 22/6 = 3(2/3) hours. Convert 2/3 hour to 40 mins and add to 11 a.m. Common Pitfall: A common mistake is trying to empty the entire capacity (60 units) rather than just the amount of water that actually collected in the tank by 11 a.m. (22 units). Final Answer: Therefore, the correct answer is 2 : 40 p.m.

Question 2

Tap A fills a tank in 4 hours whereas tap B empties the full tank in 24 hours. A and B are opened alternately for 1 hour each. Every 2 hours the level of water is found to increase by 0.5 m. The depth of the tank is

Accepted Answer

Concept & Proportionality of Volume and Depth: In uniform tanks, the fraction of the tank filled is directly proportional to the fraction of the total depth filled. We analyze the net work done in one complete alternate cycle. Fraction Filled × Total Depth = Observed Depth Increase Step-by-Step Solution: 1. Given: Tap A fills in 4 hours, Tap B empties in 24 hours. They run alternately. 2. Calculation: - In 1 hour, Tap A fills 1/4 of the tank. - In 1 hour, Tap B empties 1/24 of the tank. - Since they operate alternately, a full cycle takes 2 hours. - Net part of the tank filled in 2 hours = (1/4) - (1/24) - Make denominators equal: 1/4 = 6/24. - Net filled = 6/24 - 1/24 = 5/24 of the tank. 3. Relate volume to depth: - We are given that in 2 hours, the water level increases by 0.5 m. - Therefore, 5/24 of the total depth (D) corresponds to 0.5 m. - (5/24) × D = 0.5 4. Solve for Depth (D): - D = 0.5 × (24 / 5) - D = 1/2 × 24 / 5 = 12 / 5 = 2.4 meters. Exam Strategy & Shortcut: Recognize that 2 hours is exactly one cycle. Net fill per cycle is 1/4 - 1/24 = 5/24. If 5/24 of the tank is 0.5 meters, then 1/24 of the tank is 0.1 meters. The full tank (24/24) is 24 × 0.1 = 2.4 meters. Common Pitfall: A common error is confusing the alternate hour setup and accidentally calculating the net rate assuming they run simultaneously for a full 2 hours, which leads to incorrect capacity ratios. Final Answer: Therefore, the correct answer is 2.4 m.

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