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

$3\frac{3}{5}$Minute

Total work = efficiency × time ⇒ Efficiency = \(\frac{Total\;work}{time}\)

Let air in tyre (total work) = 54 unit (i.e. lcm of 6 and 9)

Efficiency of 1st puncture = \(\frac{54}{6}\) = 9 unit/min

Efficiency of 2nd puncture = \(\frac{54}{9}\) = 6 unit/min

If both work together, then efficiency = 9 + 6 = 15 unit/min

Time taken by both together = \(\frac{54}{15}\) = $3\frac{3}{5}$Minute