CAT 2023 Slot 2 Solved Paper

Let the number of white shirts be $\mathit{m}$, and the number of blue shirts be $\mathit{n}$. Hence, the total cost of the shirts = $\left(1000\mathit{m}+1125\mathit{n}\right)$, and the number of shirts is $\left(\mathit{m}+\mathit{n}\right)$
The average price of the shirts is $\frac{1000\mathit{m}+1125\mathit{n}}{\mathit{m}+\mathit{n}}$. It is given that he set a fixed market price which was $25\%$ higher than the average cost of all the shirts. He sold all the shirts at a discount of $10\%$.
Hence, the average selling price of the shirts The average profit of the shirts
The total profit of the shirts Now, $\Rightarrow \frac{1}{8}\left(1000\mathit{m}+1125\mathit{n}\right)=51000$
$\Rightarrow 1000\mathit{m}+1125\mathit{n}=51000\times 8=408000$
Now to get the maximum number of shirts, we need to minimize $\mathit{n}$ (since the coefficient of $\mathit{n}$ is greater than the coefficient of $\mathit{m}$ ), but it can't be zero. Therefore, $\mathit{m}$ has to be maximum.
$\mathit{m}=\frac{408000-1125\mathit{n}}{1000}$
The maximum value of $\mathrm{m}$ such that $\mathrm{m}$, and both are integers is $\mathrm{m}=399$, and $\mathrm{n}=8$ (by inspection)
Hence, the maximum number of shirts $=\mathit{m}+\mathit{n}=399+8=407$