A water tank can be filled by a pipe in 4 minutes and by a smaller pipe in 12 minutes. If both the pipes are opened simultaneously, in how much time will the tank be filled?
The first pipe can fill the tank in 4 minutes.
Rate of filling by the first pipe = V / 4 units per minute.
We can normalize the tank capacity to 1 unit (i.e., the whole tank).
Rate of the first pipe = 1/4 tank per minute.
The smaller pipe can fill the tank in 12 minutes.
Rate of filling by the second pipe = V / 12 units per minute.
Normalized rate of the second pipe = 1/12 tank per minute.
When both pipes are opened simultaneously, their rates of filling are added together.
Combined rate of both pipes = Rate of pipe 1 + Rate of pipe 2
Combined rate = 1/4 + 1/12 tanks per minute.
To add the fractions, find a common denominator, which is 12.
1/4 = 3/12.
Combined rate = 3/12 + 1/12 = 4/12 tanks per minute.
Combined rate = 1/3 tank per minute.
The time taken to fill the tank is the reciprocal of the combined rate (since Rate * Time = Work Done, and Work Done = 1 tank).
Time taken = 1 / (Combined rate)
Time taken = 1 / (1/3) minutes.
Time taken = 3 minutes.
– Rate = 1 / Time (where Time is the time taken to complete the whole work, i.e., fill the tank).
– If pipes A and B take $T_A$ and $T_B$ time respectively to fill a tank, their combined rate is $1/T_A + 1/T_B$, and the time taken together is $1 / (1/T_A + 1/T_B)$.
Using this formula:
$T_{combined} = \frac{4 \times 12}{4 + 12} = \frac{48}{16} = 3$ minutes.
This formula is a shortcut derived from $1/T_1 + 1/T_2 = 1/T_{combined}$.
$(T_2 + T_1) / (T_1 T_2) = 1/T_{combined}$
$T_{combined} = (T_1 T_2) / (T_1 + T_2)$.