IBPS PO Mains 5 Nov 2023 Paper

Concept:The given conditions produce no valid integer solution, so the quantities cannot be determined.Explanation:Let total balls in bag A = 48 and in bag B = 37.Green balls are half of red balls, so red balls must be an even integer.One bag has yellow balls, so that bag has 4 colours and the other bag has 3 colours.Case 1: Bag A has 3 colours (R, B, G).$R_A + B_A + \frac{R_A}{2} = 48$ gives $1.5R_A + B_A = 48$.Using $|R_A - B_A| = 6$:If $B_A = R_A + 6$, then $2.5R_A = 42$, so $R_A = 16.8$ (not an integer).If $B_A = R_A - 6$, then $2.5R_A = 54$, so $R_A = 21.6$ (not an integer).Thus, Case 1 is invalid.Case 2: Bag A has 4 colours (R, B, G, Y).Here $Y_A = \frac{R_A}{2} + 3$.Total: $R_A + B_A + \frac{R_A}{2} + \frac{R_A}{2} + 3 = 48$ gives $2R_A + B_A = 45$.Using $|R_A - B_A| = 6$:If $B_A = R_A + 6$, then $3R_A = 39$, so $R_A = 13$ (odd, invalid).If $B_A = R_A - 6$, then $3R_A = 51$, so $R_A = 17$ (odd, invalid).Thus, Case 2 is also invalid.Since no set of integer values satisfies all conditions, the values of Quantity I and Quantity II cannot be found.Therefore, the relationship cannot be established.Answer:Option E: $Quantity\ I = Quantity\ II$ or relationship cannot be established.