Question

In a book on arithmetic an example was printed thus:
Add $\frac{1}{14\frac{2}{3}},\frac{1}{19\frac{1}{4}},\frac{1}{?},\frac{1}{13\frac{3}{4}}$ together, the denominator of the fraction being left out. The answer given at the end of the book to this question was $\frac{11}{28}$. Find the missing denominator:

Answer 1

Let the number be x

$\frac{1}{14\frac{2}{3}}+\frac{1}{19\frac{1}{4}}+\frac{1}{x}+\frac{1}{13\frac{3}{4}}=\frac{11}{28}$

$\frac{1}{\frac{44}{3}}+\frac{1}{\frac{77}{4}}+\frac{1}{x}+\frac{1}{\frac{55}{4}}=\frac{11}{28}$

$\frac{3}{44}+\frac{4}{77}+\frac{1}{x}+\frac{4}{55}=\frac{11}{28}$

$\frac{3\times 35}{44\times 35}+\frac{4\times 20}{77\times 20}+\frac{1}{x}+\frac{4\times 28}{55\times 28}=\frac{11}{28}$

$\frac{105}{1540}+\frac{80}{1540}+\frac{112}{1540}+\frac{1}{x}=\frac{11}{28}$

$\frac{297}{1540}+\frac{1}{x}=\frac{11\times 55}{28\times 55}$

$\frac{1}{x}=\frac{605-297}{1540}$

$\frac{1}{x}=\frac{308}{1540}$

$x=5$