Question:
A tank is emptied everyday at a fixed time point. Immediately thereafter, either pump A or pump B or both start working until the tank is full. On Monday, A alone completed ?lling the tank at 8 pm. On Tuesday, B alone completed filling the tank at 6 pm. On Wednesday, A alone worked till 5 pm, and then B worked alone from 5 pm to 7 pm, to fill the tank. At what time was the tank ?lled on Thursday if both pumps were used simultaneously all along?
A tank is emptied everyday at a fixed time point. Immediately thereafter, either pump A or pump B or both start working until the tank is full. On Monday, A alone completed ?lling the tank at 8 pm. On Tuesday, B alone completed filling the tank at 6 pm. On Wednesday, A alone worked till 5 pm, and then B worked alone from 5 pm to 7 pm, to fill the tank. At what time was the tank ?lled on Thursday if both pumps were used simultaneously all along?
Updated On: Jul 29, 2025
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- 4:12 pm
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Verified By Collegedunia
The Correct Option is C
Solution and Explanation
Step 1: Define the Variable
Let \( x \) be the time (on a 24-hour clock) at which the tank gets empty.
- Pipe A starts at 20:00 (8 PM) → fills for \( (20 - x) \) hours
- Pipe B starts at 18:00 (6 PM) → fills for \( (18 - x) \) hours
On another day, the schedule changes:
- Pipe A starts at 15:00 (3 PM) → fills for \( (15 - x) \) hours
- Pipe B runs for 2 hours before the tank fills
Step 2: Let \( A \) and \( B \) be the Rates of Work
Then we write:
\[ (20 - x)A = (18 - x)B = (17 - x)A + 2B \]
Step 3: Express \( A \) in Terms of \( B \)
From \( (20 - x)A = (18 - x)B \), divide both sides by \( A \) and \( B \):
\[ \frac{A}{B} = \frac{18 - x}{20 - x} \]
From the third equation: substitute into:
\[ (20 - x)A = (17 - x)A + 2B \Rightarrow A = 2B \]
So: \[ \frac{A}{B} = \frac{2}{1} \Rightarrow \frac{18 - x}{20 - x} = \frac{2}{1} \Rightarrow 18 - x = 2(20 - x) = 40 - 2x \Rightarrow x = 22 \]
Step 4: Find Individual Rates
From earlier:
\[ (20 - x)A = (20 - 22)A = (-2)A = 1 \Rightarrow A = -\frac{1}{2} \]
This leads to a contradiction. But from: \[ (20 - x)A = 1 \Rightarrow 20 - x = \frac{1}{A} \]
Let us try \( x = 14 \). Then:
- \( (20 - 14)A = 6A \)
- \( (18 - 14)B = 4B \)
- \( (17 - 14)A + 2B = 3A + 2B \)
Let’s assume: \[ 6A = 4B \Rightarrow \frac{A}{B} = \frac{2}{3} \Rightarrow A = \frac{2}{3}B \] Substitute into: \[ 3A + 2B = 3 \cdot \frac{2}{3}B + 2B = 2B + 2B = 4B \Rightarrow \text{Verified} \]
Step 5: Determine Time Taken if Both Work Together
We know from earlier: \[ (20 - x)A = 6A = 1 \Rightarrow A = \frac{1}{6} \Rightarrow B = \frac{1}{4} \]
Total rate when both work together:
\[ A + B = \frac{1}{6} + \frac{1}{4} = \frac{5}{12} \Rightarrow \text{Time to fill} = \frac{1}{5/12} = \frac{12}{5} = 2.4 \text{ hours} \]
2.4 hours = 2 hours 24 minutes
Step 6: Final Answer Time
Both pipes started at 14:00 (2 PM), so the tank is full at:
\[ 14:00 + 2 \text{ hr } 24 \text{ min } = \boxed{16:24} \text{ or } \boxed{4:24 \text{ PM}} \]
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