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:

We designate the filling rate of pipe A as 'a' and the emptying rate of tank by pipe B as 'b'.

Given the provided data, when pipe A operates from 2 PM to 10 PM (8 hours) and pipe B operates from 3 PM to 10 PM (7 hours), the tank becomes completely filled. Thus, we have:

8a - 7b = 1 [1]

Also, according to the second scenario, when pipe A is open from 2 PM to 6 PM (4 hours) and pipe B operates for 2 hours (from 4 PM to 6 PM), the tank gets filled. This leads to:

4a - 2b = 1 [2]

By multiplying equation [2] and subtracting [1] from it, we obtain:

8a - 4b - (8a - 7b) = 2 - 1

This simplifies to:

7b - 4b = 1

Which means: b = \(\frac{1}{3}\)

Substituting the value of b into equation [2], we get:

\(4a - 2(\frac{1}{3}) = 1\)

This simplifies to:

\(4a -\frac{ 2}{3} = 1\)

So: \(4a = 1 + \frac{2}{3}\)

Which leads to: \(a = \frac{5}{12}\)

Hence, the filling rate is \(\frac{5}{12}.\)

If only pipe A is operational, and the tank gets filled in 'n' hours, we have the equation:

\(n \times a = 1\)

Substituting the value of a:

\(n \times (\frac{5}{12}) = 1\)

This simplifies to: \(n = \frac{12}{5}\)

So, n = 2.4 hours or 144 minutes.