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Sohil Rajput

· started a discussion

· 1 Months ago

NS S (Question: 336943.0 )

CP = 80 × 40
Profit from the n objects = n% × 40 × n.
Profit from the remaining objects = (100 – n)% × 40 × (80 – n).
We need to find the minimum possible value of n% × 40 × n + (100 – n)% × 40 × (80 – n).
Or, we need to find the minimum possible value of n2 + (100 – n) (80 – n).
Minimum of n2 + 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 crate an expression that can be expressed as a perfect square. This approach is termed as the “Completion of Squares” approach. We keep revisiting this in multiple chapters.

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
18 × 45 + 22 × 35 = 810 + 770
Rs. 1580. Answer choice (C)

Correct Answer: Rs. 1580

Question:

A merchant buys 80 articles, each at Rs.40. He sells x of them at a profit of x% and the remaining at a profit of (100 – x)%. What is the minimum profit the merchant could have made on this trade?  

Options:
A)

Rs.2160

B)

Rs.1420

C)

Rs.1580

D)

Rs.2210

Solution:

Ans: (c)


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