A spider climbed 62 frac{1}{2} % of the heigh | vedclass

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(D) The total height of the pole is $192 \, m$.In the first hour,the spider climbs $62 \frac{1}{2} \% = 62.5 \% = \frac{125}{200} = \frac{5}{8}$ of th

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$9$

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(D) The total height of the pole is $192 \, m$.In the first hour,the spider climbs $62 \frac{1}{2} \% = 62.5 \% = \frac{125}{200} = \frac{5}{8}$ of the total height.Remaining height after the first hour $= 192 	imes (1 - \frac{5}{8}) = 192 	imes \frac{3}{8} = 72 \, m$.In the second hour,the spider climbs $12 \frac{1}{2} \% = 12.5 \% = \frac{25}{200} = \frac{1}{8}$ of the remaining height.Distance climbed in the second hour $= \frac{1}{8} 	imes 72 = 9 \, m$.

$A$ spider climbed $62 \frac{1}{2} \%$ of the height of the pole in one hour and in the next hour it covered $12 \frac{1}{2} \%$ of the remaining height. If the height of the pole is $192 \, m$,then the distance climbed in the second hour is:

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