A millionaire bought a lot of hats $$\frac{1}{4}$$ of which were brown. The millionaire sold $$\frac{2}{3}$$ of the including $$\frac{4}{5}$$ of the brown hats. What fraction of the unsold hats were brown ?
Options:
A . $$\frac{1}{{60}}$$
B . $$\frac{1}{{15}}$$
C . $$\frac{3}{{20}}$$
D . $$\frac{3}{5}$$
E . $$\frac{3}{4}$$
Answer: Option C Let the number of hats purchase be x, $$\eqalign{ & {\text{Then, number of brown hats}} \cr & {\text{ = }}\frac{x}{4} \cr & {\text{Number of hats sold}} \cr & {\text{ = }}\frac{{2x}}{3} \cr & {\text{Number of hats left unsold}} \cr & {\text{ = }}\left( {x - \frac{{2x}}{3}} \right) = \frac{x}{3} \cr & {\text{Number of brown hats sold}} \cr & {\text{ = }}\frac{4}{5}{\text{ of }}\frac{x}{4} = \frac{x}{5} \cr & {\text{Number of brown hats left unsold}} \cr & {\text{ = }}\left( {\frac{x}{4} - \frac{x}{5}} \right) = \frac{x}{{20}} \cr & \therefore {\text{Required fraction}} \cr & {\text{ = }}\frac{{\left( {\frac{x}{{20}}} \right)}}{{\left( {\frac{x}{3}} \right)}} \cr & = \frac{x}{{20}} \times \frac{3}{x} \cr & = \frac{3}{{20}} \cr} $$
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