Question : In an examination, B obtained 20% more marks than those obtained by A, and A obtained 10% fewer marks than those obtained by C. D obtained 20% more marks than those obtained by C. By what percentage are the marks obtained by D more than those obtained by A?
Option 1: $33 \frac{1}{3} \%$
Option 2: $13 \frac{1}{3} \%$
Option 3: $43 \frac{1}{3} \%$
Option 4: $23 \frac{1}{3} \%$
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Correct Answer: $33 \frac{1}{3} \%$
Solution : Given: B obtained 20% more marks than A. A obtained 10% less marks than C. D obtained 20% more marks than C Let the marks obtained by C be $x$. Marks obtained by D $=x\times \frac{120}{100}=\frac{6x}{5}$ Marks obtained by A $=x\times \frac{90}{100}=\frac{9x}{10}$ The percentage of marks obtained by D more than by A: $= \frac{\frac{6x}{5}-\frac{9x}{10}}{\frac{9x}{10}}\times 100$ $= \frac{\frac{3x}{10}}{\frac{9x}{10}}\times 100$ $= \frac{1}{3}\times 100$ $= 33\frac{1}{3}\%$ Hence, the correct answer is $33\frac{1}{3}\%$.
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