Two pipes A and B can fill a tank in 15 minutes and 20 minutes, respectively. Both the pipes are opened togeth

Question

Two pipes A and B can fill a tank in 15 minutes and 20 minutes, respectively. Both the pipes are opened together but after 4 minutes, pipe A is turned off. What is the total time required to fill the tank?

Examoogle · Accepted Answer

The correct answer is Option C: 14 min 40 sec. Calculate the work done by B alone after A is closed using the remaining capacity: $(Remaining Work / B's rate) + 4$ minutes.

Step-by-Step Solution & Concepts:
Shortcut Trick: Calculate the work done by B alone after A is closed using the remaining capacity: $(Remaining Work / B's rate) + 4$ minutes.

Formula Used: $$TotalWork = LCM(15, 20) = 60, WorkRate_A = \frac{60}{15} = 4, WorkRate_B = \frac{60}{20} = 3$$

Given: Pipe A time = 15 min, Pipe B time = 20 min. Both open for 4 min, then A is closed.

Steps: Calculate work rates: Assuming total capacity is the LCM of 15 and 20, which is 60 units. Pipe A rate = $60/15 = 4$ units/min. Pipe B rate = $60/20 = 3$ units/min.. Work done in first 4 minutes: Both pipes A and B work together for 4 minutes at a combined rate of $4+3=7$ units/min.. Remaining work for pipe B: After pipe A is turned off, pipe B fills the remaining $60 - 28 = 32$ units.. Calculate total time: Convert $10 \frac{2}{3}$ minutes to $10$ minutes and $40$ seconds. Add the initial 4 minutes: $4 + 10$ min $40$ sec $= 14$ min $40$ sec.

Answer: C is correct because the total time is the sum of the initial 4 minutes of joint work and the 10 minutes and 40 seconds B takes to finish the remaining capacity.

Exam Tip: This concept of 'shared work followed by individual completion' is frequently applied in pipe-cistern problems and can also be solved using the linear equation $x/15 + (x-4)/20 = 1$.

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