A tyre has two punctures. The first puncture alone would have made the tyre flat in 9 minutes and the second alone would have done it in 6 minutes. If air leaks out at a constant rate, how long does it take both the punctures together to make it flat ?

### Concept & Combined Rates (Pipes and Cisterns Logic) This problem is identical in logic to "Pipes and Cisterns" or "Time and Work". A puncture acts like a pipe emptying a tank. If one empties in $A$ minutes and another in $B$ minutes, together they empty $\frac{1}{A} + \frac{1}{B}$ of the capacity per minute. ### Step-by-Step Solution * **Given:** First puncture time = 9 minutes. Second puncture time = 6 minutes. * **Calculation:** Rate of deflation by first puncture = $\frac{1}{9}$ of the tyre per minute. Rate of deflation by second puncture = $\frac{1}{6}$ of the tyre per minute. * Combined rate of deflation: $= \frac{1}{9} + \frac{1}{6}$ LCM of 9 and 6 is 18. $= \frac{2}{18} + \frac{3}{18} = \frac{5}{18}$ of the tyre per minute. * Total time taken to make the tyre flat: $= \text{Reciprocal of combined rate} = \frac{18}{5}$ minutes. * Converting to a mixed fraction: $18 \div 5 = 3$ with a remainder of 3. Time $= 3\frac{3}{5}$ minutes. ### Exam Strategy & Shortcut For two entities working together, you can directly use the formula: $$ \text{Total Time} = \frac{xy}{x + y} $$ where $x$ and $y$ are the individual times. Here, $x = 9$ and $y = 6$. Time $= \frac{9 \times 6}{9 + 6} = \frac{54}{15} = \frac{18}{5} = 3\frac{3}{5}$ minutes. ### Common Pitfall A common error is to subtract the rates if treating one as an inlet and one as an outlet. Remember both punctures are "emptying" the tyre, so their rates must be added. ### Final Answer Therefore, the correct answer is **$3\frac{3}{5}$ minutes**.