In an examination in which full marks were 80 | vedclass

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(C) Let the marks obtained by $D$ be $x.$Given that $C$ gets $15 \%$ less than $D,$ so marks of $C = x - 0.15x = 0.85x.$Given that $B$ gets $20 \%$ mo

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$58.8$

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(C) Let the marks obtained by $D$ be $x.$Given that $C$ gets $15 \%$ less than $D,$ so marks of $C = x - 0.15x = 0.85x.$Given that $B$ gets $20 \%$ more than $C,$ so marks of $B = 1.20 	imes 0.85x = 1.02x.$Given that $A$ gets $20 \%$ more than $B,$ so marks of $A = 1.20 	imes 1.02x = 1.224x.$We are given that $A = 576.$Therefore,$1.224x = 576 \implies x = \frac{576}{1.224} \approx 470.588.$Now,the percentage of full marks obtained by $D$ is $\frac{x}{800} 	imes 100 = \frac{470.588}{800} 	imes 100 = \frac{470.588}{8} \approx 58.82 \%.$

In an examination in which full marks were $800,$ $A$ gets $20 \%$ more than $B,$ $B$ gets $20 \%$ more than $C,$ and $C$ gets $15 \%$ less than $D.$ If $A$ got $576,$ what percentage of full marks did $D$ get (approximately) (in $\%$)?

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