Two numbers are 90 % and 75 % lesser than a t | vedclass

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Question

(D) Let the third number be $100$.Since the first number is $90 \%$ less than the third number,the first number $= 100 - (90 \% 	ext{ of } 100) = 100

Vedclass · Accepted Answer

$150$

Vedclass · Answer

(D) Let the third number be $100$.Since the first number is $90 \%$ less than the third number,the first number $= 100 - (90 \% 	ext{ of } 100) = 100 - 90 = 10$.Since the second number is $75 \%$ less than the third number,the second number $= 100 - (75 \% 	ext{ of } 100) = 100 - 75 = 25$.We need to increase the first number $(10)$ to make it equal to the second number $(25)$.The required increase $= 25 - 10 = 15$.The percentage increase required $= \left( \frac{	ext{Increase}}{	ext{Original Number}} ight) 	imes 100 = \left( \frac{15}{10} ight) 	imes 100 = 150 \%$.

Two numbers are $90 \%$ and $75 \%$ lesser than a third number. By what percentage should the first number be increased so that it becomes equal to the second number?

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