Question

A tank has four pipes P1, P2, P3, P4. The tank can be filled in 15 minutes by pipes P1, P2, P3 together. It can be filled in 20 minutes by pipes P1, P2, P4 together and it can be filled by pipes P1 and P3 together in 30 minutes. If all the pipes are opened together, then in how much time will the tank be filled?

• A. 10 min 20 sec
• B. 12 min 0 sec
• C. 18 min 20 sec
• D. 11 min 20 sec

Hint:

For a system of linear equations to have a unique solution, you generally need as many independent equations as you have variables. Here we have 4 variables (\(r_1, r_2, r_3, r_4\)) but only 3 independent equations.

Answer 1

The Correct Option is B

Step 1: Define work rates per minute.

Let the rates of P1, P2, P3, P4 be \(a, b, c, d\) respectively (tank/minute).

From the problem conditions: \[a+b+c = \frac{1}{15} \quad (1)\] \[a+b+d = \frac{1}{20} \quad (2)\] \[a+c = \frac{1}{30} \quad (3)\]

Step 2: Subtract equations (1) and (3).

\[(a+b+c) - (a+c) = \frac{1}{15} - \frac{1}{30}\] \[b = \frac{1}{30}\]

Step 3: Determine values of (a+c) and (a+b).

From (3): \[a+c = \frac{1}{30}\] From (1): \[a+b+c = \frac{1}{15}\] Substitute \(b=\frac{1}{30}\): \[a+c + \frac{1}{30} = \frac{1}{15}\] \[a+c = \frac{1}{30} \quad \text{(consistent with (3))}\]

Step 4: Calculate d.

From (2): \[a+b+d = \frac{1}{20}\] Substitute \(b=\frac{1}{30}\): \[a+d = \frac{1}{20} - \frac{1}{30} = \frac{1}{60}\]

Step 5: Find the combined work rate when all pipes are open.

\[a+b+c+d = (a+c) + (b) + (d)\] We know: \[a+c = \frac{1}{30}, \quad b=\frac{1}{30}, \quad a+d=\frac{1}{60}\] So: \[a+b+c+d = \frac{1}{30} + \frac{1}{30} + \frac{1}{60}\] \[= \frac{2}{30} + \frac{1}{60} = \frac{4}{60} + \frac{1}{60} = \frac{5}{60} = \frac{1}{12}\]

Step 6: Determine the total time.

If rate = \(\frac{1}{12}\), then time = 12 minutes. Check the answer options (in minutes and seconds). \[12 \, \text{minutes} = 12 \, \text{min 0 sec}\]

Final Answer:

\[\boxed{12 \, \text{minutes}}\]