Question

an insect crawls up5 cm every secondon a60 cm vertical rodandfalls down 2 cm over the next second. how many seconds will it take to climb the rod for that insect ??

Answer 1

$$\begin{gathered} \mathrm{In}\mathrm{first}\mathrm{second}\mathrm{insect}\mathrm{crawls}\mathrm{up}\mathrm{distance}=5\mathrm{cm}. \\ \mathrm{In}\mathrm{next}\mathrm{second}\mathrm{insects}\mathrm{falls}\mathrm{down}\mathrm{distance}=2\mathrm{cm}. \\ \mathrm{Total}\mathrm{distance}\mathrm{covered}\mathrm{in}2\mathrm{seconds}=5-2=3\mathrm{cm}. \\ \mathrm{Distance}\mathrm{covered}\mathrm{in}1\mathrm{second}=\frac{3}{2}\mathrm{cm}. \\ \mathrm{Now},\mathrm{after}38\mathrm{seconds}\mathrm{distance}\mathrm{covered}=\frac{3}{2}\times 38=57\mathrm{cm}. \\ \mathrm{Remaining}3\mathrm{cm}\mathrm{will}\mathrm{be}\mathrm{covered}\mathrm{in}\mathrm{last}\mathrm{second} \\ 3\mathrm{cm}\mathrm{will}\mathrm{be}\mathrm{covered}\mathrm{in}\mathrm{time}=\frac{3}{5}\mathrm{seconds}. \\ \mathrm{Total}\mathrm{time}\mathrm{taken}=38\frac{3}{5}\mathrm{seconds}=\frac{193}{5}\mathrm{seconds}=38.60\mathrm{seconds}.\mathrm{ANS}... \end{gathered}$$