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Question

(A) Let the third number be $100$.Since the first number is $20 \%$ more than the third,the first number $= 100 + 20 = 120$.Since the second number is

Vedclass · Accepted Answer

$9 \frac{1}{11} \%$

Vedclass · Answer

(A) Let the third number be $100$.Since the first number is $20 \%$ more than the third,the first number $= 100 + 20 = 120$.Since the second number is $10 \%$ more than the third,the second number $= 100 + 10 = 110$.We need to find how much percent the first number is more than the second.Difference $= 120 - 110 = 10$.Percentage more $= \left( \frac{	ext{Difference}}{	ext{Second number}} 	imes 100 ight) \%$.Percentage more $= \left( \frac{10}{110} 	imes 100 ight) \% = \frac{100}{11} \% = 9 \frac{1}{11} \%$.Thus,the first number is $9 \frac{1}{11} \%$ more than the second.

Two numbers are respectively $20 \%$ and $10 \%$ more than a third number. How much per cent is the first number more than the second?

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