**Q5. 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 (TITA) **

Solution:

Let the number of white shirts be m, and the number of blue shirts be n. Hence, the total cost of the shirts = (1000m+1125n), and the number of shirts is (m+n)

The average price of the shirts is (1000m + 1125n)/(m+n)

It is given that he 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%.

Hence, the average selling price of the shirts = [(1000m + 1125n)/(m+n)] * 5/4 * 9/10 = 9/8 * (1000m + 1125n)/(m+n)

The average profit of the shirts = 1/8 * (1000m + 1125n)/(m+n) → 1/8 *(1000m + 1125n) = 51000

1000*m *+ 1125*n *= 51000 × 8 = 408000

Now to get the maximum number of shirts, we need to minimize n (since the coefficient of n is greater than the coefficient of m), but it can't be zero. Therefore, m has to be maximum.

m = (408000 – 1125n)/1000

The maximum value of m such that m, and both are integers is m = 399, and n = 8 (by inspection)

Hence, the maximum number of shirts = m + n = 399 + 8 = **407**