On a school's Annual Day sweets were to be equally distributed amongst 112 children. But on that particular day, 32 children were absent. Thus the remaining children got 6 extra sweets. How many sweets was each child originally supposed to get?

## Concept & Logic This problem relies on equating the total volume of items based on two different distribution scenarios. The extra sweets received by the present children are exactly the sweets that would have belonged to the absent children. $$ Sweets\ of\ Absent\ Children = Extra\ Sweets\ given\ to\ Present\ Children $$ ## Step-by-Step Solution * **Given:** Original children = 112. Absent = 32. Present = $112 - 32 = 80$. Extra sweets per present child = 6. * **Deduction:** Let the original number of sweets per child be $x$. The total number of sweets originally meant for the 32 absent children is $32 \times x$. * **Calculation:** Because these children are absent, their sweets ($32x$) are redistributed among the 80 present children. * **Calculation:** The problem states each present child gets 6 extra sweets. This means the redistributed sweets amounted to $80 \times 6 = 480$ sweets. * **Substitution:** Set the two amounts equal to each other: $32x = 480$ * **Calculation:** Solve for $x$: $x = \frac{480}{32} = 15$. * **Conclusion:** Each child was originally supposed to get 15 sweets. ## Exam Strategy & Shortcut Instead of algebra, use pure logic based on totals. 80 children each got 6 extra sweets. That's $80 \times 6 = 480$ total extra sweets distributed. Where did these 480 sweets come from? They came from the shares of the 32 absent children. So, 32 children were supposed to get a total of 480 sweets. Originally, 1 child was supposed to get $\frac{480}{32} = 15$ sweets. Done in 10 seconds. ## Common Pitfall Setting up a complex fractional equation like $\frac{Total}{80} - \frac{Total}{112} = 6$ is correct but mathematically tedious, requiring large LCM calculations ($560$) and increasing the chance of arithmetic errors under pressure. The logical "transfer" method shown above is vastly safer. ## Final Answer Therefore, the correct answer is **15**.