If frac{3}{5} of a number is 23 more than 50 | vedclass

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(D) Let the number be $x$.According to the problem,$\frac{3}{5} x = 23 + 50 \% 	ext{ of } x$.Since $50 \% = \frac{50}{100} = \frac{1}{2}$,the equatio

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$184$

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(D) Let the number be $x$.According to the problem,$\frac{3}{5} x = 23 + 50 \% 	ext{ of } x$.Since $50 \% = \frac{50}{100} = \frac{1}{2}$,the equation becomes $\frac{3}{5} x = 23 + \frac{1}{2} x$.Subtracting $\frac{1}{2} x$ from both sides: $\frac{3}{5} x - \frac{1}{2} x = 23$.Finding a common denominator $(10)$: $\frac{6x - 5x}{10} = 23 \Rightarrow \frac{x}{10} = 23 \Rightarrow x = 230$.Now,we need to find $80 \% 	ext{ of } x$: $\frac{80}{100} 	imes 230 = 0.8 	imes 230 = 184$.

If $\frac{3}{5}$ of a number is $23$ more than $50 \%$ of the same number,then what will be $80 \%$ of the number?

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