Two flood gates A and B can be used to fill a reservoir. If both gates A and B are opened together, the reservoir becomes full in 6 hours. Gate A is faster and can fill the reservoir 5 hours sooner than gate B can. How many hours does the faster flood gate take to fill the reservoir when it works alone?

Introduction / Context:
This question deals with the classic pipes and cisterns or inflow problems, which are mathematically similar to time and work problems. Here, two flood gates are filling a reservoir, and we know the combined time and the relationship between the times of the two individual gates. The task is to find the time taken by the faster gate to fill the reservoir alone.

Given Data / Assumptions:

• Gates A and B together can fill the reservoir in 6 hours.
• Gate A is faster and fills the reservoir 5 hours sooner than gate B.
• Let the time taken by gate A be Ta hours.
• Then the time taken by gate B is Ta + 5 hours.
• Total work (reservoir filled once) is considered as 1 unit.

Concept / Approach:
The method is to express the work rates of the gates in terms of Ta. If a gate takes T hours to fill the reservoir, its rate is 1/T units per hour. Since both gates together fill the reservoir in 6 hours, the sum of their rates must equal 1/6. This leads to an equation in Ta, which is then solved to find the time taken by the faster gate.

Step-by-Step Solution:
Step 1: Let Ta be the time taken by gate A alone. Then time taken by gate B alone is Ta + 5. Step 2: Rate of gate A = 1/Ta units per hour. Step 3: Rate of gate B = 1/(Ta + 5) units per hour. Step 4: Together, they fill the reservoir in 6 hours, so combined rate = 1/6. Step 5: Set up the equation: 1/Ta + 1/(Ta + 5) = 1/6. Step 6: Multiply through by 6 * Ta * (Ta + 5) to clear denominators: 6(Ta + 5) + 6Ta = Ta(Ta + 5). Step 7: Simplify the left side: 6Ta + 30 + 6Ta = 12Ta + 30. Step 8: The equation becomes Ta(Ta + 5) = 12Ta + 30. Step 9: Expand and rearrange: Ta^2 + 5Ta - 12Ta - 30 = 0, which simplifies to Ta^2 - 7Ta - 30 = 0. Step 10: Solve the quadratic equation Ta^2 - 7Ta - 30 = 0. The roots are Ta = 10 and Ta = -3. Since time cannot be negative, Ta = 10 hours.

Verification / Alternative check:
If gate A takes 10 hours, then gate B takes 15 hours. Rate of A = 1/10, rate of B = 1/15. Combined rate = 1/10 + 1/15 = (3/30 + 2/30) = 5/30 = 1/6, meaning they fill the reservoir in 6 hours together, which matches the problem statement and confirms the solution.

Why Other Options Are Wrong:

• 7 hours: This would not yield a combined time of 6 hours when paired with a gate that is 5 hours slower.
• 13 hours: Leads to inconsistent combined rate and does not satisfy the equation 1/Ta + 1/(Ta + 5) = 1/6.
• 5 hours: Implies gate B takes 10 hours, and together they would fill much faster than 6 hours.

Common Pitfalls:
Learners often misinterpret the phrase “5 hours faster” and sometimes incorrectly set the time of the slower gate as Ta - 5 instead of Ta + 5. Another frequent error is algebraic manipulation when solving the quadratic equation, such as sign mistakes or missing terms during expansion. Carefully forming and solving the equation is crucial to arriving at the correct answer.

Final Answer:
The faster flood gate alone will fill the reservoir in 10 hours.