NIMCET 2019 Question Paper with solutions

Let the CP of the article$=\mathit{₹}\mathit{x}$
Then, the SP of the article $=\mathit{x}\times \frac{\left(100+20\right)}{100}=\mathit{₹}1.2\mathit{x}$
∵ He incurs loss because, he allows 16 articles to 12 articles,
loss% $=\frac{16-12}{16}\times 100=\frac{4}{4}\times 25=25\%$
∴ His $\mathrm{SP}=\mathrm{SP}\times \frac{\left(100-25\right)}{100}=\mathit{₹}0.75\mathrm{SP}$
Hence, 0.75 SP$=1.2\mathit{x}$
$\Rightarrow \mathrm{SP}=\mathit{₹}1.6\mathit{x}$
This SP is come after giving a discount of 20% on MP
$\therefore \mathrm{MP}\times \frac{\left(100-80\right)}{100}=\mathrm{SP}$
$\Rightarrow \mathrm{MP}\times \frac{20}{100}=1.6\mathit{x}$
$\therefore \mathrm{MP}=2\mathit{x}$
It means that the article has been marked 100% above thecost price.