Maximum number of shirts bought

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

Accepted Answer

407. Let $w$ = number of white shirts, $b$ = number of blue shirts ($w, b \ge 1$). Total cost $= 1000w + 1125b$. Average cost per shirt $= \frac{1000w + 1125b}{w+b}$. Marked price per shirt $= 1.25 	imes 	ext{Avg Cost}$. Selling price per shirt $= 0.90 	imes 1.25 	imes 	ext{Avg Cost} = 1.125 	imes 	ext{Avg Cost}$. Total revenue $= (w+b) 	imes 1.125 	imes 	ext{Avg Cost} = 1.125(1000w + 1125b)$. Total profit $= 0.125(1000w + 1125b) = 51000$. $$1000w + 1125b = 51000 	imes 8 = 408000$$ Dividing by 125: $$8w + 9b = 3264$$ Total shirts $= w + b = \frac{3264 - 9b}{8} + b = 408 - \frac{b}{8}$. To maximize $w + b$, we must minimize $b$. Since $8w + 9b = 3264$, $9b \equiv 3264 \equiv 0 \pmod 8 \implies b$ is a multiple of $8$. Since $b \ge 1$, the minimum positive integer value for $b$ is $8$. When $b = 8$, $8w = 3264 - 72 = 3192 \implies w = 399$. Maximum total shirts $= 399 + 8 = 407$.

Answer

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Maximum number of shirts bought

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