Pipes and Cisterns – Individual times exceed the joint time by fixed minutes: Two pipes A and B fill a tank. When both are opened together, they fill it in T minutes. Pipe A alone takes 16 minutes more than T, and pipe B alone takes 9 minutes more than T. Find T, the time taken when both are opened together.

Introduction / Context:
This is a parameterized pipes problem: each solo time is an additive offset from the unknown joint time T. The physics is the same: work rates add. We set up an equation in T and solve.

Given Data / Assumptions:

• Together: T min ⇒ rate(A + B) = 1/T per min.
• A alone: T + 16 ⇒ rate A = 1/(T + 16).
• B alone: T + 9 ⇒ rate B = 1/(T + 9).

Concept / Approach:
Because rates add, 1/(T + 16) + 1/(T + 9) = 1/T. Clear denominators and solve the resulting quadratic for the positive, realistic T.

Step-by-Step Solution:
1/(T + 16) + 1/(T + 9) = 1/T.Cross-multiply: T(T + 9 + T + 16) = (T + 16)(T + 9).2T^2 + 25T = T^2 + 25T + 144 ⇒ T^2 = 144 ⇒ T = 12 (take positive root).

Verification / Alternative check:
Check rates: 1/28 + 1/21 = (3 + 4)/84 = 7/84 = 1/12, which equals 1/T.

Why Other Options Are Wrong:
8, 10, 14, 15 minutes fail to satisfy the rate equation when substituted into 1/(T + 16) + 1/(T + 9) = 1/T.

Common Pitfalls:
Adding times rather than rates; sign mistakes when clearing denominators.

Final Answer:
12 min