Two pipes A and B can fill a tank respectively in 30 min and 45 min. Both A and B are opened together for some time and then pipe B is turned off. If the tank is filled in 20 min, then find after how many minutes the pipe B is turned off?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 15 min
Let the tank capacity be 1 unit.
Filling rates:
Pipe A: $\frac{1}{30}$ per min
Pipe B: $\frac{1}{45}$ per min
Let pipe B be turned off after $t$ minutes. Both A and B run together for $t$ minutes, then only A runs for remaining $(20 - t)$ minutes.
Total filled:
$t\left(\frac{1}{30} + \frac{1}{45}\right) + (20 - t)\left(\frac{1}{30}\right) = 1$
Compute combined rate: $\frac{1}{30} + \frac{1}{45} = \frac{3+2}{90} = \frac{5}{90} = \frac{1}{18}$
Equation:
$t \cdot \frac{1}{18} + (20 - t) \cdot \frac{1}{30} = 1$
$\frac{t}{18} + \frac{20 - t}{30} = 1$
Multiply through by 90:
$5t + 3(20 - t) = 90$
$5t + 60 - 3t = 90 \Rightarrow 2t + 60 = 90 \Rightarrow 2t = 30 \Rightarrow t = 15$