Question

Two pipes A, B can fill a tank in 24 min and 32 min respectively. If both pipes are opened simultaneously, after how much time should B be closed so that the tank is full in 18 min?

• A. 8 min
• B. 12 min
• C. 15 min
• D. 20 min

Hint:

In variable-time pipe problems, keep one pipe running the full duration and treat the other for \(t\) minutes. Add their work (rate \(\times\) time) to equal 1 full tank.

Answer 1

The Correct Option is A

Step 1: Write filling rates. 
A's rate \(=\frac{1}{24}\) tank/min; 

B's rate \(=\frac{1}{32}\) tank/min. 

Step 2: Set up the total-fill equation. 
Let B be closed after \(t\) minutes. 
A works for all \(18\) minutes; B works for \(t\) minutes. 
\[ 18\cdot\frac{1}{24} + t\cdot\frac{1}{32} = 1. \] 

Step 3: Solve for \(t\). 
\[ \frac{18}{24} + \frac{t}{32} = 1 \Rightarrow \frac{3}{4} + \frac{t}{32} = 1 \Rightarrow \frac{t}{32} = \frac{1}{4} \Rightarrow t = 8 \text{ min}. \] \[ \boxed{8\ \text{minutes}} \]

Answer 2

Working with whole tank-units instead of fractions makes each option easy to check directly.

1. 8 min: Taking the tank as 96 units (LCM of 24 and 32), A fills \(96/24=4\) units/min and B fills \(96/32=3\) units/min. Over the full 18 minutes, A alone contributes \(18\times4=72\) units; if B runs for 8 minutes, B contributes \(8\times3=24\) units, for a total of \(72+24=96\) units, exactly filling the tank.
2. 12 min: B running for 12 minutes contributes \(12\times3=36\) units, giving a total of \(72+36=108\) units, which overfills the 96-unit tank.
3. 15 min: B running for 15 minutes contributes \(15\times3=45\) units, giving \(72+45=117\) units, again overfilling the tank.
4. 20 min: B running for 20 minutes contributes \(20\times3=60\) units, giving \(72+60=132\) units, far exceeding the 96-unit capacity.

Only 8 minutes of B's running gives exactly 96 units of work, filling the tank precisely in the 18 minutes allowed.

So the correct answer is 8 min.