Question

A spider climbed $62\frac{1}{2}$ $\%$ of the height of the pole in one hour and in the next hour 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

• A. $3$ m
• B. $5$ m
• C. $7$ m
• D. $9$ m

Answer 1

Answer: (D)

$9$ m

Distance climbed by the spider in one hour
$=62.5\%$ of $192$ m $=\frac{62.5}{100}\times 192$ m $=120$ m
$\therefore$ Remaining distance $=192$ m $-120$ m $=72$ m
$\therefore$ Distance climbed in the second hour
$=12.5\%$ of $72$ m $=$ $\frac{12.5}{100}\times 72$ m
$=\frac{125}{1000}\times 72$ m $=9$ m