Two pipes, P and Q, together can fill a tank in 5 hours. If pipe P alone can fill the tank in 10 hours, then how much time (in hours) will pipe Q alone take to fill the tank?

Correct option is B

Given:
Time taken by (P + Q) = 5 hours
Time taken by P alone = 10 hours
Concept Used:
​$$\text{Time} = \frac{\text{Total Work}}{\text{Efficiency}}$$​​
Solution:
Let the total capacity of the tank be the LCM of 5 and 10.
​$$\text{Total Capacity} = 10 \text{ units}$$​​
Calculate the efficiency of the pipes:
​$$\text{Efficiency of (P + Q)} = \frac{10}{5} = 2 \text{ units/hour} \\\text{Efficiency of P} = \frac{10}{10} = 1 \text{ unit/hour}$$​​
Find the efficiency of pipe Q:
​$$\text{Efficiency of Q} = \text{Efficiency of (P + Q)} - \text{Efficiency of P} \\\text{Efficiency of Q} = 2 - 1 = 1 \text{ unit/hour}$$​​
Calculate the time taken by Q:
​$$\text{Time}_Q = \frac{\text{Total Capacity}}{\text{Efficiency of Q}} = \frac{10}{1} = 10 \text{ hours}$$​​
Exam-Hall Method:
Product upon Difference formula:
​$$T_Q = \frac{P \times T_{total}}{P - T_{total}} = \frac{10 \times 5}{10 - 5} = \frac{50}{5} = 10 \text{ hours}$$​​
So the correct answer is (b)