IBPS PO Mains 30 Nov 2024 Shift 2 Paper

Concept:Convert all comparison clues into a single score order, then apply the average condition to find the highest feasible score of F from the given options.Explanation:From the clues: $D > B > A$ and $B > E$, so we get $D > B > E$.Also, $E > F > C$, and since C scores more than at least one person, C must be above A.Thus, the complete order of scores is: $D > B > E > F > C > A$.The average of A, C and E is 17, so their total is $17 \times 3 = 51$.Hence, $A + C + E = 51$.All scores are distinct whole numbers, and we know $A < C < E$.With $A \geq 1$ and $C \geq 2$, the minimum value of $A + C$ is $1 + 2 = 3$.So, the maximum possible score of E is $51 - 3 = 48$.Since E must score more than F, F cannot be 49, 51 or 54.Between the remaining options, 35 and 18, the higher feasible value is 35.This is possible, for example: $A = 1$, $C = 2$, $E = 48$, $F = 35$, $B = 49$, $D = 50$.This set satisfies the full order $D > B > E > F > C > A$ and the total $A + C + E = 51$.Therefore, the highest possible score of F from the options is 35.Answer:The highest possible score of F is 35, i.e., option A.