If a number is 20 % more than the other,how m | vedclass

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(A) Let the second number be $100$.Then,the first number is $100 + 20 = 120$.The difference between the two numbers is $120 - 100 = 20$.To find how mu

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$16 \frac{2}{3} \%$

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(A) Let the second number be $100$.Then,the first number is $100 + 20 = 120$.The difference between the two numbers is $120 - 100 = 20$.To find how much per cent the second number is less than the first,we calculate the percentage decrease relative to the first number:Required percentage $= \left( \frac{	ext{Difference}}{	ext{First Number}} 	imes 100 ight) \%$$= \left( \frac{20}{120} 	imes 100 ight) \%$$= \left( \frac{1}{6} 	imes 100 ight) \% = \frac{100}{6} \% = 16 \frac{2}{3} \%$.

If a number is $20 \%$ more than the other,how much per cent is the second number less than the first?

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