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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