Question

14. A monkey standing on the ground wants to climb to the top of a vertical pole of \( 15 m \). He climbs \( 5 m \) in \( 1 s \) and slips downward \( 3 m \) in next second. He again climbs \( 5 m \) in \( 1 s \) and slips by \( 3 m \) in the nexd second and so on. The time taken by the monkey to climb at the top of pole is a. \( 2 s \) b. \( 3 s \) c. \( 9 s \) d. \( 11 s \)

Answer 1

In first 1 s, monkey climbs 5 meters,

In second 1 s, it slips down 3 meters.

So it climbs 2 meters every 2 s, except the last second. In the last second, it would cover all 5 meters and escape.

So keeping 1 s aside for last 5 meters.

Total distance to be covered is 15m, in which we subtract the last 5 m for the last climb.

So, to cover 10 m:

Time required=10/2 * 2 = 10 s

Adding that last 1 s, it takes 11 seconds.