Children’s Day distribution: Sweets were to be equally distributed among 300 children. On the day, 50 children were absent, so each attending child received exactly one extra sweet. How many sweets were prepared for distribution?

Introduction / Context:
This is a ratio and difference-per-recipient problem. A fixed total of sweets is split among different headcounts, leading to a per-child change of exactly one sweet. Setting up a simple equation in terms of total sweets solves it cleanly.

Given Data / Assumptions:

• Originally planned recipients = 300 children.
• Actual recipients = 250 children (since 50 were absent).
• Each of the 250 received 1 more sweet than originally planned.

Concept / Approach:
Let total sweets be S. Then planned share per child is S/300, and actual share per child is S/250. The given condition S/250 = S/300 + 1 yields a simple linear equation for S.

Step-by-Step Solution:
Set S/250 = S/300 + 1.Multiply through by 1500 (LCM of 250 and 300): 6S = 5S + 1500.Solve: S = 1500.Therefore, 1500 sweets were prepared.

Verification / Alternative check:
Original per-child share: 1500/300 = 5. Actual per-child share: 1500/250 = 6. The difference is exactly 1 sweet, as required.

Why Other Options Are Wrong:
1450, 1650, 1700, and 1800 do not satisfy S/250 − S/300 = 1.

Common Pitfalls:
Using 300 − 50 = 200 by mistake; mixing up “one extra per child” as “one extra total.”

Final Answer:
1500