Question

A film show lasted of $3\frac{2}{3}$ hours. Out of this time $1\frac{1}{2}$ hours was spent on advertisements. What was the actual duration of the film?

Answer 1

Actual duration of the film = $\left(3\frac{2}{3}-1\frac{1}{2}\right)\mathrm{hours}$

= $\left(\frac{11}{3}-\frac{3}{2}\right)\mathrm{hours}$

= $\left(\frac{22-9}{6}\right)\mathrm{hours}$ [∵ LCM of 3 and 2 = 6]

= $\frac{13}{6}\mathrm{hours}=2\frac{1}{6}\mathrm{hours}$

Hence, the actual duration of the film was $2\frac{1}{6}\mathrm{hours}$.