CAT 2023 Slot 2 · Quant · Question 15

Let the total number of shirts be n and the average cost per shirt be A, so the total cost is nA.

The marked price is 25% above the average cost, that is 1.25A for every shirt.

Selling at a 10% discount gives a selling price of 0.9\times 1.25A=1.125A per shirt.

The total revenue is 1.125An, so the profit is 1.125An-An=0.125An.

Since An is the total cost, the profit is simply one-eighth of the total cost.

Setting \frac{1}{8}\times \text{total cost}=51000 gives a total cost of 408000.

With w white shirts at ₹1000 and b blue at ₹1125, 1000w+1125b=408000.

Dividing throughout by 125, 8w+9b=3264.

To maximise w+b, favour the cheaper shirt, so make b as small as possible.

Since 3264 is divisible by 8, the term 9b must also leave no remainder modulo 8, which forces b to be a multiple of 8.

He bought both colours, so the smallest allowed value is b=8.

Then 8w=3264-72=3192, giving w=399.

The maximum total is 399+8=407 shirts.

Shortcut. The discount and markup combine into a single factor: 1.25\times 0.9=1.125, so the profit is always exactly \frac{1}{8} of the cost, whatever the mix. After that it is one linear equation, and cheaper shirts mean more shirts — so minimise the blue count subject to divisibility.