A group of workers promise to complete a piece of work in 10 days, but five of them do not report for work. If it took the remaining workers 12 days to complete the work, then the number of workers originally hired was

### Concept & Constant Work Equation The total amount of work is equivalent to the number of workers multiplied by the number of days they work. This total work remains constant regardless of how the group size changes. $$ M_1 \times D_1 = M_2 \times D_2 $$ ### Step-by-Step Solution * Let the original number of workers hired be $x$. * Original plan: $x$ workers finishing in 10 days. Total work = $10x$. * Actual scenario: $(x - 5)$ workers report for work and finish in 12 days. Total work = $12(x - 5)$. * Equate the total work from both scenarios: $10x = 12(x - 5)$. * Expand the right side: $10x = 12x - 60$. * Rearrange to solve for $x$: $12x - 10x = 60$. * $2x = 60 \Rightarrow x = 30$. * The original number of workers was 30. ### Exam Strategy & Shortcut Use inverse ratio of days. Time ratio is $10 : 12 = 5 : 6$. Therefore, the worker ratio must be $6 : 5$. The difference in parts is $6 - 5 = 1$ part. We are given this 1 part difference equals 5 workers. Original workers (6 parts) = $6 \times 5 = 30$. ### Common Pitfall Setting up the equation as $10x = 12x + 5$ or failing to distribute the 12 to both $x$ and $-5$ in the expression $12(x-5)$. ### Final Answer Therefore, the correct answer is **30**.