Question

Pipe A, B and C are kept open and together fill a tank in t minutes. Pipe A is kept open throughout, pipe B is kept open for the first 10 minutes and then closed. Two minutes after pipe B is closed, pipe C is opened and is kept open till the tank is full. Each pipe fills an equal share of the tank. Furthermore, it is known that if pipe A and B are kept open continuously, the tank would be filled completely in t minutes. How long will it take C alone to fill the tank?

• A. 20 minutes
• B. 24 minutes
• C. 32 minutes
• D. 48 minutes
• E. 18 minutes

Hint:

When a problem states "each unit does an equal share of the work", immediately relate their active working time to that fraction (e.g., $1/3$) to find the time it would take them to do the whole job independently.

Answer 1

The Correct Option is B

Step 1: Determine individual filling times from their shares.
The total filling time is $t$ minutes. Since each pipe fills exactly $1/3$ of the tank:
Pipe A is open for $t$ mins and fills $1/3 \Rightarrow$ A's full tank time = $3t$ mins.
Pipe B is open for 10 mins and fills $1/3 \Rightarrow$ B's full tank time = 30 mins.
Pipe C is opened at 12 mins and closes at $t$ mins. It is open for $(t - 12)$ mins and fills $1/3 \Rightarrow$ C's full tank time = $3(t - 12)$ mins.

Step 2: Use the combined rate of A and B.
If A and B are continuously open, they fill the tank in $t$ mins.
$\frac{1}{A} + \frac{1}{B} = \frac{1}{t} \Rightarrow \frac{1}{3t} + \frac{1}{30} = \frac{1}{t}$

Step 3: Solve for $t$ and C's time.
$\frac{1}{30} = \frac{1}{t} - \frac{1}{3t} = \frac{2}{3t} \Rightarrow 3t = 60 \Rightarrow t = 20$ minutes.
C's full time = $3(t - 12) = 3(20 - 12) = 3(8) = 24$ minutes.