IBPS PO Mains 12 Oct 2025 Shift 1 Paper

Concept:Let the quantities of apples be consecutive multiples of $5$, with Type C having the smallest quantity. Then solve for the total weight equation.Explanation:Let the quantity of Type C apples be $5n$.Since the quantities are consecutive multiples of $5$, the other two quantities are $5n + 5$ and $5n + 10$.Case 1: Type A has $5n + 5$ apples and Type B has $5n + 10$ apples.Total weight is:$25(5n) + 40(5n+5) + 60(5n+10) = 2050$$125n + 200n + 200 + 300n + 600 = 2050$$625n + 800 = 2050$$625n = 1250$$n = 2$So Type B quantity $= 5(2) + 10 = 20$ apples.Case 2: Type B has $5n + 5$ apples and Type A has $5n + 10$ apples.Total weight would be $625n + 700 = 2050$, giving $n = 2.16$, not an integer. So this case is invalid.All quantities are within the available stock: A $= 15$, B $= 20$, C $= 10$.Answer:The shopkeeper selected $20$ Type B apples.Correct option: C. $20$