Question

Two pipes A and B are attached to an empty water tank. Pipe A fills the tank while pipe B drains it. If pipe A is opened at 2 pm and pipe B is opened at 3 pm, then the tank becomes full at 10 pm. Instead, if pipe A is opened at 2 pm and pipe B is opened at 4 pm, then the tank becomes full at 6 pm. If pipe B is not opened at all, then the time, in minutes, taken to fill the tank is

• A. 140
• B. 120
• C. 144
• D. 264

Answer 1

The Correct Option is C

Rephrased information and calculations:

Let the filling rate of pipe A be denoted by $a$ and the emptying rate of pipe B be denoted by $b$.

Scenario 1: Pipe A operates for 8 hours (2 PM to 10 PM) and pipe B for 7 hours (3 PM to 10 PM), resulting in a full tank. This yields the equation: $8a - 7b = 1$   [1]

Scenario 2: Pipe A operates for 4 hours (2 PM to 6 PM) and pipe B for 2 hours (4 PM to 6 PM), also resulting in a full tank. This yields the equation: $4a - 2b = 1$   [2]

Derivation of $b$:

Multiply equation [2] by 2: $8a - 4b = 2$   [2a]

Subtract equation [1] from equation [2a]:

$(8a - 4b) - (8a - 7b) = 2 - 1$ 
$\Rightarrow 8a - 4b - 8a + 7b = 1$ 
$\Rightarrow 3b = 1$ 
$b = \frac{1}{3}$

Derivation of $a$:

Substitute $b = \frac{1}{3}$ into equation [2]:

$4a - 2\left(\frac{1}{3}\right) = 1$ 
$\Rightarrow 4a - \frac{2}{3} = 1$ 
$\Rightarrow 4a = 1 + \frac{2}{3} = \frac{5}{3}$ 
$a = \frac{5}{12}$

Therefore, the filling rate of pipe A is $\frac{5}{12}$ of the tank per hour.

Calculation of time for pipe A alone:

Let $n$ be the time taken by pipe A alone to fill the tank.

$n \cdot a = 1$ 
$\Rightarrow n \cdot \frac{5}{12} = 1$ 
$\Rightarrow n = \frac{12}{5}$ 
$n = 2.4$ hours = 144 minutes