CAT 2023 Slot 3 · Quant · Question 14

Let the total stock be T, so the mangoes number 0.4T and the bananas and apples together number 0.6T.

Write A for the apples; the bananas are then 0.6T-A.

Half the mangoes sold is 0.2T, and 40% of the apples sold is 0.4A.

Half the entire stock is sold, so 0.2T+96+0.4A=0.5T.

Rearranging, 0.4A=0.3T-96, so A=0.75T-240.

The bananas are then 0.6T-\left( 0.75T-240 \right)=240-0.15T.

For at least one apple, 0.75T-240\ge 1, which forces T\ge 321.33.

For 96 bananas to be available, 240-0.15T\ge 96, which caps T at 960 — easily satisfied here.

Divisibility pins the rest down: 0.4T whole needs T to be a multiple of 5, 0.75T whole needs T to be a multiple of 4, and 0.15T whole needs T to be a multiple of 20.

So T must be a multiple of 20, and the smallest multiple of 20 beyond 321.33 is 340.

Verifying at T=340, the mangoes are 0.4\times 340=136, the apples are 0.75\times 340-240=15, and the bananas are 340-136-15=189.

Sold that day: 68+96+6=170, which is exactly half of 340.

The smallest possible total is 340.

Shortcut. Every percentage in the question has denominator 20 once simplified, so T must be a multiple of 20 before anything else is checked. The single inequality A\ge 1 then gives T>320, and the next multiple of 20 is the answer.