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?

1^st Method:
Man climbs 1m and slips down 50 cm (0.5m) 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 1m as he gets on the top so no slip.
Time taken to climb 11 meter = $$ {\frac{{10}}{{0.5}} + 1} $$  = 21 minutes.
He climbs the wall at 5:21 pm

2^nd Method (short-cut):
$$ = \frac{{{\text{hight}}\,\,{\text{of}}\,\,{\text{pole}} - {\text{slipped}}\,\,{\text{distance}}}}{{{\text{climbed}}\,\,{\text{distance}} - {\text{slipped}}\,\,{\text{distance}}}}$$
$$\eqalign{ & = {\frac{{x - z}}{{y - z}}} \times t \cr & = {\frac{{11 - 0.5}}{{1 - 0.5}}} \times 1 \cr & = 21\,\,{\text{minutes}} \cr} $$