A merchant buys 80 articles, each at Rs. 40. He sells n of them at a profit of…

2025202420232025

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?

  1. A.

    Rs. 2160

  2. B.

    Rs. 1420

  3. C.

    Rs. 1580

  4. D.

    Rs. 2210

Attempted by 9 students.

Show answer & explanation

Correct answer: C

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 = 1580

The question is "What is the minimum profit the merchant could have made on this trade?"

Hence, the answer is Rs. 1580.

Choice C is the correct answer.

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