[Rs. 1580] A merchant buys 80 articles, each at Rs. 40. He sells n of them at a profit of n% and the remaining

For Competitive Exams

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

Rs. 2160

Rs. 1420

Rs. 1580

Rs. 2210

Explanation:

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 n^2 + (100 – n) (80 – n).
Minimum of n^2 + n^2 – 180n + 8000
Minimum of n^2 – 90n + 4000
Minimum of n^2 – 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.

Minimum of n^2 – 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 = 1580

The question is "What is the minimum profit the merchant could have made on this trade?"
Hence, the answer is Rs. 1580.

Additional Questions

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