CAT 2017 Quantitative Ability

CP=80×40
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
n%×40×n+(100−n)%×40×(80−n)
Or, we need to find the minimum possible value of
n2+(100−n)(80−n)
⇒‌Minimum of‌n2+n2−180n+8000
⇒‌Minimum of ‌n2−90n+4000
⇒‌Minimum of ‌n2−90n+2025−2025+4000
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.
⇒‌Minimum of ‌n2−90n+2025+1975
=(n−45)2+1975
This reaches minimum, when n = 45.
When n = 45, the minimum profit made= 45% × 40 × 45 + 55% × 40 × 35 = 810 + 770 = ₹ 1580