If A's height is 40 % less than that of B,how | vedclass

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(D) Let $B$'s height be $100 	ext{ units}$.Since $A$'s height is $40 \%$ less than $B$'s height,$A$'s height $= 100 - 40 = 60 	ext{ units}$.We need

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$66 \frac{2}{3} \%$

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(D) Let $B$'s height be $100 	ext{ units}$.Since $A$'s height is $40 \%$ less than $B$'s height,$A$'s height $= 100 - 40 = 60 	ext{ units}$.We need to find how much percent $B$'s height is more than $A$'s height.Difference in height $= 100 - 60 = 40 	ext{ units}$.Percentage more $= \left( \frac{	ext{Difference}}{	ext{A's height}} ight) 	imes 100 = \left( \frac{40}{60} ight) 	imes 100$.Percentage more $= \left( \frac{2}{3} ight) 	imes 100 = \frac{200}{3} = 66 \frac{2}{3} \%$.

If $A$'s height is $40 \%$ less than that of $B$,how much percent is $B$'s height more than that of $A$?

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