P, Q and R are 3 small pumps fitted to a tank. S is a large pump fitted to the tank. Q is 50% more efficient than P. R is `33(1)/(3)%` m ore efficient than Q. S is 50% more efficient than R. All of the pipes are used to fill the tank.
If pump S starts emptying the tank instead of filling it, then find the ratio of time taken by all pumps fill the tank and pipe S emptying to the time taken by all pumps to fill the tank together?

Options

A. `7:5`
B. `5:1`
C. `1:5`
D. `2:3`

Answer

To solve the problem step by step, we will first determine the efficiencies of the pumps P, Q, R, and S, and then calculate the time taken for the different scenarios.

### Step 1: Determine the efficiencies of the pumps

1. Let the efficiency of pump P be \( E_P = 2 \) (arbitrary units).
2. Since Q is 50% more efficient than P, we calculate:
\[ E_Q = E_P + 0.5 \times E_P = 2 + 1 = 3 \]
3. R is \( 33\frac{1}{3}\% \) more efficient than Q. This percentage can be converted to a fraction:
\[ 33\frac{1}{3}\% = \frac{1}{3} \implies E_R = E_Q + \frac{1}{3} \times E_Q = 3 + 1 = 4 \]
4. S is 50% more efficient than R:
\[ E_S = E_R + 0.5 \times E_R = 4 + 2 = 6 \]

### Step 2: Calculate the total efficiency when all pumps are filling the tank

- The total efficiency when all pumps are filling the tank is:
\[ E_{total\_filling} = E_P + E_Q + E_R + E_S = 2 + 3 + 4 + 6 = 15 \]

### Step 3: Calculate the efficiency when S is emptying the tank

- When S is emptying the tank, we need to subtract its efficiency from the total:
\[ E_{total\_emptying} = E_P + E_Q + E_R - E_S = 2 + 3 + 4 - 6 = 3 \]

### Step 4: Calculate the time taken for both scenarios

1. **Time taken when all pumps are filling the tank**:
- Let the total work (capacity of the tank) be \( W = 12 \) (arbitrary units).
- Time taken to fill the tank:
\[ T_{filling} = \frac{W}{E_{total\_filling}} = \frac{12}{15} = \frac{4}{5} \text{ time units} \]

2. **Time taken when all pumps are filling and S is emptying**:
- Time taken to fill the tank when S is emptying:
\[ T_{emptying} = \frac{W}{E_{total\_emptying}} = \frac{12}{3} = 4 \text{ time units} \]

### Step 5: Calculate the ratio of the times

- The ratio of the time taken by all pumps filling the tank with S emptying to the time taken by all pumps filling the tank together is:
\[ \text{Ratio} = \frac{T_{emptying}}{T_{filling}} = \frac{4}{\frac{4}{5}} = 4 \times \frac{5}{4} = 5 \]

### Final Answer

Thus, the ratio of time taken by all pumps filling the tank with S emptying to the time taken by all pumps filling the tank together is:
\[ \text{Ratio} = 5:1 \]