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

Let's rephrase the given information and calculations:

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

Given: Pipe A is open from 2 PM to 10 PM (8 hours), and pipe B from 3 PM to 10 PM (7 hours), and the tank gets completely filled. This gives the equation: $8a - 7b = 1$   [1]

In another scenario, pipe A is open from 2 PM to 6 PM (4 hours), and pipe B from 4 PM to 6 PM (2 hours), and the tank also gets filled. So: $4a - 2b = 1$   [2]

Subtracting [1] from twice [2]:

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

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

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

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

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

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