A man starts climbing a 11 m high wall at 5 pm. In each minute he climbs up 1 m but slips down 50 cm. At what time will he climb the wall?

Explanation:-

1^st Method:
Man climbs 1 m and slips down 50 cm (0.5 m) in one minute
i.e. he climbs (1-0.5 = 0.5 m) in one minute.
But in the last minute he will be climbing 1 m as he gets on the top so no slip.
Time taken to climb 11 meter = [(¹⁰/_0.5) +1] = 21 minutes.

He climbs the wall at 5:21 pm.

2^nd Method (short-cut):

Reqd. time = [(ht. of pole - slipped distance) /(climbed distance - slipped distance)]*t;
$$\eqalign{ & = \left[ {\frac{{\left( {x - z} \right)}}{{y - z}}} \right] \times t \cr & {\text{Required}}\,{\text{time}}, \cr & = \left\{ {\frac{{\left( {11 - 0.5} \right)}}{{\left( {1 - 0.5} \right)}}} \right\} \times 1 \cr & = 21\,{\text{minutes}} \cr}$$