Question

A water tank has steps inside it. A monkey is sitting on the topmost step (i.e., the first step). The water level is at the $9^{th}$ step.

(i) He jumps $3$ steps down and then jumps back $2$ steps up. In how many jumps will he reach the water level?

Answer 1

Given:

Monkey jumps $3$ steps down and then jumps back $2$ steps up.

Now, starting from first,

$\begin{array}{ccc}\text{Jump}& \text{Total steps moved}& \text{Current position}\\ \text{Start}& & 1^{st}\text{step}\\ 1^{st}\text{Jump}& 3\text{steps down}=-3& 1+3=4^{th}\text{step}\\ 2^{nd}\text{Jump}& 2\text{steps up}=-3+2=-1& 4-2=2^{nd}\text{step}\\ 3^{rd}\text{Jump}& 3\text{steps down}=-1-3=-4& 2+3=5^{th}\text{step}\\ 4^{th}\text{Jump}& 2\text{steps up}=-4+2=-2& 5-2=3^{rd}\text{step}\\ 5^{th}\text{Jump}& 3\text{steps down}=-2-3=-5& 3+3=6^{th}\text{step}\\ 6^{th}\text{Jump}& 2\text{steps up}=-5+2=-3& 6-2=4^{th}\text{step}\\ 7^{th}\text{Jump}& 3\text{steps down}=-3-3=-6& 4+3=7^{th}\text{step}\\ 8^{th}\text{Jump}& 2\text{steps up}=-6+2=-4& 7-2=5^{th}\text{step}\\ 9^{th}\text{Jump}& 3\text{steps down}=-4-3=-7& 5+3=8^{th}\text{step}\\ {10}^{th}\text{Jump}& 2\text{steps up}=-7+2=-5& 8-2=6^{th}\text{step}\\ {11}^{th}\text{Jump}& 3\text{steps down}=-5-3=-8& 6+3=9^{th}\text{step}\end{array}$

Since, the water level is at $9^{th}$ step.
So, the monkey will be at water level at $9^{th}$ step after $11$ jumps.