Question

Jayant bought a certain number of white shirts at the rate of Rs 1000 per piece and a certain number of blue shirts at the rate of Rs 1125 per piece. For each shirt, he then set a fixed market price which was 25% higher than the average cost of all the shirts. He sold all the shirts at a discount of 10% and made a total profit of Rs 51000. If he bought both colors of shirts, then the maximum possible total number of shirts that he could have bought is

• A. 395
• B. 407
• C. 413
• D. None of Above

Answer 1

The Correct Option is B

Given the following quantities:

• Blue shirts: \( n \)
• White shirts: \( m \)
• Total shirts: \( m + n \)

The total cost of the shirts is calculated as:

\(1000m + 1125n\)

The average cost per shirt is:

\(\frac{1000m + 1125n}{m + n}\)

The market price is established at 25% above the average cost. Thus, the selling price is:

\(\frac{1000m + 1125n}{m + n} \times \frac{5}{4} \times \frac{9}{10}\)

Upon simplification, the average selling price per shirt is:

\(\frac{9}{8} \times \frac{1000m + 1125n}{m + n}\)

The average profit per shirt is determined by:

\(\frac{1}{8} \times \frac{1000m + 1125n}{m + n}\)

The total profit for all shirts is:

\(\frac{1}{8} \times \frac{1000m + 1125n}{m + n} \times (m + n)\)

This simplifies to:

\(\frac{1}{8} (1000m + 1125n)\)

Given that the total profit equals 51,000:

\(\frac{1}{8} (1000m + 1125n) = 51000\)

Multiplying both sides by 8 yields:

\(1000m + 1125n = 408000\)

To maximize the total number of shirts (\( m + n \)), we must minimize \( n \), with the constraint that \( n \) cannot be zero. Consequently, \( m \) must be maximized.

The equation is rearranged to solve for \( m \):

\(m = \frac{408000 - 1125n}{1000}\)

By minimizing \( n \) and maximizing \( m \), inspection reveals the maximum value for \( m \) to be 399 and for \( n \) to be 8.

The total number of shirts is then:

\(m + n = 399 + 8 = 407\)

Conclusion:

The maximum number of shirts achievable is \( \boxed{407} \).

The correct option is (B): 407.