Equal distribution with absentees — Sweets were to be equally distributed among 300 children, but 50 were absent. Each present child then received one extra sweet. How many sweets were distributed in total?

Introduction / Context:
This is a classic allocation problem using the idea that the total quantity remains constant, but the per-person share changes because fewer recipients are present. The key is to set up the relation between original planned share and the increased share when some children are absent.

Given Data / Assumptions:

• Planned recipients: 300 children.
• Actual recipients: 300 - 50 = 250 children.
• Each present child received one more sweet than originally planned.
• Total sweets distributed remained the same.

Concept / Approach:
Let the originally planned per-child share be x. Then the actual per-child share is x + 1 (since fewer children shared the same total). Equate the two expressions for total sweets and solve for x, then compute the total number of sweets.

Step-by-Step Solution:
Let original per-child share = x.Original total = 300 * x.Actual per-child share = x + 1 with 250 children → Actual total = 250 * (x + 1).Equate totals: 300x = 250(x + 1) → 300x = 250x + 250 → 50x = 250 → x = 5.Total sweets = 300 * x = 300 * 5 = 1500.

Verification / Alternative check:
With 250 children, each gets x + 1 = 6 sweets: 250 * 6 = 1500, same total—consistent.

Why Other Options Are Wrong:

• 1450 / 1700 / 1650 / 1800: These totals do not satisfy both equations simultaneously; they arise from misreading the “one extra” condition or miscomputing the absent count.

Common Pitfalls:
Assuming the total changes; mixing up who got “one extra”; arithmetic slips when equating totals.

Final Answer:
1500