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

Answer: (B)

407

Answer 2

Let's assume the following:
number of blue shirts be n
number of white shirts be m
the number of shirts is $(m+n)$
Therefore, the total cost of the shirts $= (1000m+1125n)$

Now, average price of the shirt: $\frac{1000m+1125n}{m+n}$

It is mentioned that he set a fixed market price which was 25% higher than the average cost of all the shirts.

And he sold all the shirts at a discount of 10%.
So, the average selling price of the shirts :
$=(\frac{1000m+1125n}{m+n})\times\frac{5}{4}\times\frac{9}{10}$

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

Hence, the average profit of the shirts :
$=\frac{1}{8}(\frac{1000m+1125n}{m+n})-\frac{1000m+1125n}{m+n}$
$=\frac{1}{8}(\frac{1000m+1125n}{m+n})$

Now, the total profit of the shirts :
$=\frac{1}{8}(\frac{1000m+1125n}{m+n})\times(m+n)$

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

Now , $=\frac{1}{8}(1000m+1125n)=51000$
$⇒ 1000m + 1125n = 51000 × 8 = 408000$

So, to get the maximum number of shirts , we need to minimize the value of n, which can't be zero. Hence, m has to be maximum.
$m=\frac{408000-1125n}{1000}$

Now, the maximum value of m such that m, and both the integers is m = 399 and n = 8 (which is taken by inspection)
Therefore, the maximum number of shirts :
= m + n = 399 + 8
= 407
So, the correct option is (B) : 407.