Percentage Practice Test 2 with solutions

Let the total weight = x gm, then weight of the container $=\frac{\mathit{x}}{4}$ gm
Now weight of the total filled fluid = $\mathit{x}-\frac{\mathit{x}}{4}$ gm $=\frac{3}{4}\mathit{x}$ gm
Latter weight of the remaining fluid with container $=\frac{50}{100}\times \mathit{x}=\frac{\mathit{x}}{2}\mathrm{gm}$
Now weight of the remaining fluid $=\frac{\mathit{x}}{2}-\frac{\mathit{x}}{4}=\frac{\mathit{x}}{4}\mathrm{gm}$
Hence, required fractional part $=\frac{\frac{3}{4}\mathit{x}-\frac{\mathit{x}}{4}}{\frac{3}{4}\mathit{x}}=\frac{\mathit{x}}{2}\times \frac{4}{3\mathit{x}}=\frac{2}{3}$