Two pipes A and B can fill the cistern in 37.5 minutes and 45 minutes respectively. Both pipes are opened. The cistern will be filled in just half an hour, if the B is turned off after:
1. Define the Goal and Roles - The cistern must be filled in exactly 30 minutes (half an hour).
- This means Pipe A runs for the entire 30 minutes.
- Pipe B runs for an unknown time, T, before it is turned off.
2. Define the Rates (per minute) - Pipe A's time = 37.5 minutes = 75/2 minutes. So, its rate is 2/75 of the cistern per minute.
- Pipe B's time = 45 minutes. So, its rate is 1/45 of the cistern per minute.
3. Calculate the Work Done by Pipe A - In its 30 minutes of operation, the amount of the cistern filled by Pipe A is:
- Work = Rate × Time = (2/75) × 30 = 60/75 = 4/5 of the cistern.
4. Find the Work and Time for Pipe B - The remaining work must have been done by Pipe B before it was turned off.
- Remaining Work = 1 - 4/5 = 1/5 of the cistern.
- To find the time Pipe B was open (T), we use: Time = Work ÷ Rate.
- T = (1/5) ÷ (1/45) = (1/5) × 45 = 9 minutes.
Final Answer: Pipe B must be turned off after 9 minutes.
Reinforce your conceptual grasp with related questions from NIMCET 2020:
Which of the four options should fill the missing cell?
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