Question : Let $x=\frac{5 \frac{3}{4}-\frac{3}{7} \times 15 \frac{3}{4}+2 \frac{2}{35} \div 1 \frac{11}{25}}{\frac{3}{4} \div 5 \frac{1}{4}+5 \frac{3}{5} \div 3 \frac{4}{15}}$. When $y$ is added to $x$, the result is $\frac{7}{13}$. What is the value of $y$?
Option 1: $\frac{1}{13}$
Option 2: $\frac{9}{13}$
Option 3: $\frac{2}{13}$
Option 4: $\frac{4}{13}$
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Correct Answer: $\frac{4}{13}$
Solution :
$x=\frac{5 \frac{3}{4}-\frac{3}{7} \times 15 \frac{3}{4}+2 \frac{2}{35} \div 1 \frac{11}{25}}{\frac{3}{4} \div 5 \frac{1}{4}+5 \frac{3}{5} \div 3 \frac{4}{15}}$
$⇒x=\frac{\frac{23}{4}-\frac{3}{7} \times \frac{63}{4}+\frac{72}{35} \div \frac{36}{25}}{\frac{3}{4} \div \frac{21}{4}+\frac{28}{5} \div \frac{49}{15}}$
$⇒x=\frac{\frac{23}{4}-\frac{3}{7} \times \frac{63}{4}+\frac{72}{35} \times \frac{25}{36}}{\frac{3}{4} \times \frac{4}{21} +\frac{28}{5} \times \frac{15}{49}}$
$⇒x=\frac{\frac{23}{4}- \frac{27}{4}+\frac{10}{7}}{\frac{1}{7}+\frac{12}{7}}$
$⇒x=\frac{\frac{3}{7}}{\frac{13}{7}}$
$⇒x=\frac{3}{13}$
Given that $x + y = \frac{7}{13}$,
$⇒y = \frac{7}{13} - x = \frac{7}{13} - \frac{3}{13} = \frac{4}{13}$
Hence, the correct answer is $ \frac{4}{13}$.
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