IBPS Clerk Mains 29 Nov 2025 Paper

Concept:Use the given relationships between boys and girls in schools A, B, and C to find their exact numbers, then compute the required difference after increasing the girls in school B by 50%.Explanation:Let $\mathrm{A_b}$ and $\mathrm{A_g}$ be boys and girls in school A, $\mathrm{B_b}$ and $\mathrm{B_g}$ in school B, and $\mathrm{C_b}$ and $\mathrm{C_g}$ in school C.From the first condition, $\mathrm{A_g}-5 = \frac{1}{5}(\mathrm{A_b}+5)$.This simplifies to $\mathrm{A_b} = 5\mathrm{A_g} - 30$.The total students in school B is $\mathrm{T_B} = 2\mathrm{A_b} - 10$.Also, $\mathrm{B_b} = 60\%$ of $\mathrm{T_B}$, and $\mathrm{T_C} = \frac{\mathrm{T_B}}{2}$.For school C, $\mathrm{C_b} = \frac{\mathrm{B_b}}{3}$ and $\mathrm{C_g} = \mathrm{A_b} - 9$.Since $\mathrm{T_C} = \mathrm{C_b} + \mathrm{C_g}$:$\frac{\mathrm{T_B}}{2} = \frac{\mathrm{B_b}}{3} + \mathrm{A_b} - 9$.Substitute $\mathrm{B_b} = 0.6\,\mathrm{T_B}$:$0.5\,\mathrm{T_B} = 0.2\,\mathrm{T_B} + \mathrm{A_b} - 9$, giving $0.3\,\mathrm{T_B} = \mathrm{A_b} - 9$.Using $\mathrm{T_B} = 2\mathrm{A_b} - 10$:$0.3(2\mathrm{A_b} - 10) = \mathrm{A_b} - 9$.So $0.6\mathrm{A_b} - 3 = \mathrm{A_b} - 9$, which gives $\mathrm{A_b} = 15$.Now $\mathrm{A_g} = 9$, so total students in A: $15 + 9 = 24$.Total students in B: $2(15) - 10 = 20$.Boys in B: $60\%$ of $20 = 12$, so girls in B: $20 - 12 = 8$.Total students in C: $\frac{20}{2} = 10$.Increase in girls of B: $50\%$ of $8 = 4$, so new girls in B: $8 + 4 = 12$.Sum of total students in A and C: $24 + 10 = 34$.Required difference: $34 - 12 = 22$.Answer:$22$ (Option B)