Math, asked by kaulinsh3620, 10 months ago

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

Answers

Answered by bhagyashreechowdhury
2

The minimum profit that the merchant could have made on this trade is option (3): Rs. 1580.

Step-by-step explanation:

The C.P. of each article = Rs. 40

∴ The C.P. of 80 articles = 80 × 40 = Rs. 3200

It is given that the merchant sold n articles at n% profit and remaining at (100 - n)% profit, so we have

The profit from n objects = (n% * 40 )* n  

And,

The profit from the remaining objects = [(100 – n)% × 40] × (80 – n)  

Now, as it is asked in the question, we have to find the minimum possible value of  

[(n% * 40 )* n] + [(100 – n)% * 40 * (80 – n)]

⇒ [n²] + [(8000 – 100n – 80n + n²)]  ……. [removing % and 40 for easy calculation will not change the final required value]

⇒ n² + 8000 – 180n + n²

⇒ 2n² + 8000 – 180n

⇒ n² – 90n + 4000 …… [dividing by 2 throughout]

[Since the middle term is 90n so we will now add and subtract 2025 (a square of 45) in order to create an expression that can be expressed as a perfect square. This type of approach can be termed as the “Completion of Squares” approach]  

⇒ n² – 90n + 4000 + 2025 – 2025

⇒ (n² – 90n + 2025) + (4000 - 2025)  

(n – 45)² + 1975This minimum expression is obtained when n = 45

Therefore, we can conclude that,

When n = 45,  the minimum profit made by the merchant will be,

= [45% × 40 × 45] + [55% × 40 × (80-45)]

= [45 * 0.4 * 45] + [55 * 0.4 * 35]  

= 810 + 770

= Rs. 1580

Thus, the minimum profit that the merchant could have made on this trade is Rs. 1580.

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