Pipes M, N and S can fill a tank in 25, 50 and 100 minutes, respectively. Initially, pipes N and S are kept open for 10 minutes, and then pipe N is shut while pipe M is opened. Pipe S is closed 15 minutes before the tank overflows. How much time (in minutes) will it take to fill the tank if the three pipes work in this pattern?

Question

Accepted Answer

Let the total capacity of the tank be 100 units (for ease of calculation).

Pipe M fills in 25 min → 4 units/min
Pipe N fills in 50 min → 2 units/min
Pipe S fills in 100 min → 1 unit/min

Step 1: First 10 minutes, only N and S are open.
Combined rate = 2 + 1 = 3 units/min
Water filled in 10 min = 3 × 10 = 30 units

Step 2: After 10 min, N is closed, M is opened, S continues.
Now, M and S are open. Combined rate = 4 + 1 = 5 units/min
Let the total time taken to fill the tank be T minutes.

Pipe S is closed 15 minutes before the tank overflows. So, for the last 15 minutes, only M is open (rate = 4 units/min).

Let the total time taken to fill the tank be T minutes.

- First 10 min: N and S open → 30 units filled
- Next (T - 25) min: M and S open (since S is closed 15 min before end, so S is open for (T - 10 - 15) = (T - 25) min after the first 10 min)
- Last 15 min: Only M open

So,
- First 10 min: 30 units
- Next (T - 25) min: M and S open → 5 units/min × (T - 25) min = 5(T - 25) units
- Last 15 min: Only M open → 4 units/min × 15 = 60 units

Total water filled = 30 + 5(T - 25) + 60 = 100 units

Let's solve for T:

30 + 5(T - 25) + 60 = 100
30 + 5T - 125 + 60 = 100
(30 + 60) - 125 + 5T = 100
90 - 125 + 5T = 100
-35 + 5T = 100
5T = 135
T = 27

Let's check the breakdown carefully.

Let’s clarify the time intervals:
- First 10 min: N and S (3 units/min)
- Next (T - 10 - 15) min: M and S (5 units/min)
- Last 15 min: Only M (4 units/min)

Total time: T = 10 + (T - 10 - 15) + 15 = T

So, after the first 10 min, S continues for (T - 10 - 15) min, i.e., S is open for (T - 25) min after the first 10 min.

So, total water:
- First 10 min: 3 × 10 = 30 units
- Next (T - 25) min: 5 × (T - 25) units
- Last 15 min: 4 × 15 = 60 units

Total = 30 + 5(T - 25) + 60 = 100

30 + 5T - 125 + 60 = 100
90 - 125 + 5T = 100
-35 + 5T = 100
5T = 135
T = 27

But 27 is an option. Let's check the logic again.

Let’s try with a variable for the time S is open after the first 10 min.
Let x = time (in min) for which M and S are open together after the first 10 min.

So,
- First 10 min: N and S → 3 × 10 = 30 units
- Next x min: M and S → 5 × x units
- Last 15 min: Only M → 4 × 15 = 60 units

Total time: 10 + x + 15 = T

Total water: 30 + 5x + 60 = 100
5x = 10
x = 2

So, total time = 10 + 2 + 15 = 27 min

But this matches the calculation above. However, let's check if the pattern is interpreted correctly.

After the first 10 min, N is closed, M is opened, S continues. S is closed 15 min before the tank is full. So, S is open for (T - 15) min in total.

So,
- S is open for first 10 min + next (T - 10 - 15) min = (T - 15) min

Let’s calculate the total water filled by each pipe:
- N: open for 10 min → 2 × 10 = 20 units
- S: open for (T - 15) min → 1 × (T - 15) = (T - 15) units
- M: open for (T - 10) min (since after 10 min, M is opened and remains open till the end) → 4 × (T - 10) units

Total water = 20 (N) + (T - 15) (S) + 4(T - 10) (M) = 100

20 + (T - 15) + 4(T - 10) = 100
20 + T - 15 + 4T - 40 = 100
(20 - 15 - 40) + T + 4T = 100
(-35) + 5T = 100
5T = 135
T = 27

So, the correct answer is 27 minutes.

Therefore, the correct option is 27.

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