A sum of money is equally divided among a number of children. Had there been 16 children more, each would have received ₹ 2 less and had there been 16 fewer, each would have received ₹ 3 more. Find the sum of the money distributed.

## Concept & Logic This problem involves two variables: the number of children and the original share per child. The total money remains constant. We establish a system of two equations based on the variations in group size and share amount. $$ Total = Number\ of\ Children \times Share\ per\ Child $$ ## Step-by-Step Solution * **Given:** Let the number of children be $N$ and the original share per child be $S$. Total Money $M = N \times S$. Condition 1: If $N+16$, share is $S-2$. Condition 2: If $N-16$, share is $S+3$. * **Calculation:** Set up the equations equating to the constant Total Money ($NS$): Eq 1: $(N + 16)(S - 2) = NS$ Eq 2: $(N - 16)(S + 3) = NS$ * **Expansion:** Expand Eq 1: $NS - 2N + 16S - 32 = NS$ $-2N + 16S = 32 \implies 16S - 2N = 32$ Divide by 2: $8S - N = 16 \implies N = 8S - 16$ * **Expansion:** Expand Eq 2: $NS + 3N - 16S - 48 = NS$ $3N - 16S = 48$ * **Substitution:** Substitute $N = 8S - 16$ into the simplified Eq 2: $3(8S - 16) - 16S = 48$ $24S - 48 - 16S = 48$ $8S = 96 \implies S = 12$ * **Calculation:** Find $N$: $N = 8(12) - 16 = 96 - 16 = 80$. * **Conclusion:** Total Money distributed $M = N \times S = 80 \times 12 = 960$. ## Exam Strategy & Shortcut For equations of the form $(n \pm a)(s \mp b) = ns$, the simplifications always result in linear equations: $as - bn \pm ab = 0$. A rapid formulaic approach when the change in people ($a$) is symmetric (both are 16): $\frac{Number}{16} = \frac{2 \times 3}{3 - 2} \times 2$ is false, but analyzing the shifts gives rapid ratio checks. Alternatively, test the options. We need $M$ such that $\frac{M}{N}$ is an integer, and $\frac{M}{N+16}$ is an integer. Take Option (d) 960. $960 \div 80 = 12$. If 16 more (96): $960 \div 96 = 10$. ($12 - 2 = 10$, matches!). If 16 fewer (64): $960 \div 64 = 15$. ($12 + 3 = 15$, matches!). Option testing is often faster than algebra for this specific question type. ## Common Pitfall Getting lost in the algebraic expansion and dropping a negative sign, leading to incorrect values for $N$ or $S$. Ensure you cancel out the $NS$ terms carefully on both sides of the equations. ## Final Answer Therefore, the correct answer is **₹ 960**.