Question 59

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


Correct Answer: 407

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 $$\ \frac{\ 1000m+1125n}{m+n}$$. It is given thatĀ 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 =Ā $$\left(\ \frac{\ 1000m+1125n}{m+n}\right)\times\ \frac{5}{4}\times\ \frac{9}{10}=\frac{9}{8}\left(\ \frac{\ 1000m+1125n}{m+n}\right)$$

The average profit of the shirts =Ā $$\frac{9}{8}\left(\ \frac{\ 1000m+1125n}{m+n}\right)-\frac{\ 1000m+1125n}{m+n}=\frac{1}{8}\left(\frac{\ 1000m+1125n}{m+n}\right)$$

The total profit of the shirts =Ā $$\frac{1}{8}\left(\frac{\ 1000m+1125n}{m+n}\right)\times\ \left(m+n\right)\ =\ \frac{1}{8}\left(1000m+1125n\right)$$

Now,Ā $$=>\frac{1}{8}\left(1000m+1125n\right)=51000$$

$$=>1000m+1125n=51000\times\ 8=408000$$

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 beĀ zero. Therefore, m has to be maximum.

$$m\ =\ \ \frac{\ 408000-1125n}{1000}$$

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

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