Question : The sum of three fractions A, B and C, A > B > C, is $\frac{121}{60}$. When C is divided by B, the resulting fraction is $\frac{9}{10}$, which exceeds A by $\frac{3}{20}$. What is the difference between B and C?
Option 1: $\frac{1}{15}$
Option 2: $\frac{1}{10}$
Option 3: $\frac{3}{10}$
Option 4: $\frac{7}{15}$
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Correct Answer: $\frac{1}{15}$
Solution :
Given: $A + B + C = \frac{121}{60}$
$\frac{C}{B} = \frac{9}{10}$
According to the question,
$\frac{9}{10} = A + \frac{3}{20}$
⇒ $A = \frac{9}{10} - \frac{3}{20} = \frac{18-3}{20} = \frac{15}{20} = \frac{3}{4}$
$B + C = \frac{121}{60} - \frac{3}{4} = \frac{121-45}{60} = \frac{76}{60} = \frac{19}{15}$
Since, C : B = 9 : 10
$C = \frac{19}{15} × \frac{9}{19} = \frac{9}{15} = \frac{3}{5}$
$B = \frac{19}{15}×\frac{10}{19}= \frac{10}{15} = \frac{2}{3}$
Required Difference $=\frac{2}{3} - \frac{3}{5} = \frac{10-9}{15} = \frac{1}{15}$
Hence, the correct answer is $\frac{1}{15}$.
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