Question

A monkey standing on the ground wants to climb to the top of the vertical pole of $13m$ in tall. He climbs $5m$ in $1s$ and then slips downwards $3m$ in the next second. He again climbs $5m$ in $1s$ and slips by $3m$ in the next second as so on. How much time will the monkey take to reach the top of the pole when he moves like this?
$\begin{align}
  & A.7s \\
 & B.9s \\
 & C.11s \\
 & D.13s \\
\end{align}$

Answer 1

Hint: The monkey’s journey is like $5m$ upwards and then $3m$downwards. So the effective distance it travelled will be only $2m$. But the total time taken will be the sum of the time taken to climb up and climb down. These will help you to solve this question.

Complete answer:
As per the question says, the monkey is climbing the pole for $5m$ first. It will take $1s$ to climb up this distance. After that it will descend down $3m$ in next $1s$. This continues for a while. Therefore we can say that the effective distance travelled by the monkey will be $2m$. The time taken for one cycle will be the sum of the time taken to complete the climbing up and down. Therefore the time taken to climb the effective distance will be,
$t=1+1=2s$
Hence we can say that, after the ten cycles of ascending and descending, the monkey will be at a height of $10m$. On the $11s$, the monkey will reach the top of the pole of height $13m$.
So the monkey will take $11s$ to climb up the pole.

So, the correct answer is “Option C”.

Note:
Here the effective length taken should be taken care of well. The chance of making a mistake is very high. And also the time taken to complete one cycle of the journey is the total time as the time is always positive. As per the question it is needed only to find the time at which it just reaches the top, after the climbing it may descend as per the cycle. However we can take the minimum time to reach the top.