CAT 2017 Quantitative Ability

$\mathit{C}\mathit{P}=80\times 40$
Profit from the n objects = n% × 40 × n.
Profit from the remaining objects= (100 - n)% x 40 x (80 - n)
We need to find the minimum possible value of n
$\mathit{n}\%\times 40\times \mathit{n}+\left(100-\mathit{n}\right)\%\times 40\times \left(80-\mathit{n}\right)$
Or, we need to find the minimum possible value of
${\mathit{n}}^2+\left(100-\mathit{n}\right)\left(80-\mathit{n}\right)$



We add and subtract 2025 to this expression in order to create an expression that can be expressed as a perfect square This approach is termed as the “Completion of Squares" approach.

$={\left(\mathit{n}-45\right)}^2+1975$
This reaches minimum, when n = 45.
When n = 45, the minimum profit made= 45% × 40 × 45 + 55% × 40 × 35 = 810 + 770 = ₹ 1580