A and B together can complete a work in 12 days, B and C together can complete the same work in 8 days and A and C together can complete it in 16 days. In total, how many days do A, B and C together take to complete the same work?

### Concept & Formula / Logic When the work rates of pairs of individuals are given, summing all the pair rates yields exactly twice the combined rate of all individuals working together. $$ 2(A + B + C) = (A + B) + (B + C) + (A + C) $$ ### Step-by-Step Solution 1. **Determine the 1-day work for each pair:** (A + B)'s 1-day work = $\frac{1}{12}$ (B + C)'s 1-day work = $\frac{1}{8}$ (A + C)'s 1-day work = $\frac{1}{16}$ 2. **Sum the 1-day work of all pairs:** $(A + B) + (B + C) + (A + C) = \frac{1}{12} + \frac{1}{8} + \frac{1}{16}$ Find the LCM of 12, 8, and 16, which is 48. $2(A + B + C) = \frac{4}{48} + \frac{6}{48} + \frac{3}{48} = \frac{13}{48}$ 3. **Calculate the combined 1-day work of A, B, and C:** Since $2(A + B + C) = \frac{13}{48}$ $(A + B + C) = \frac{13}{48 \times 2} = \frac{13}{96}$ This means A, B, and C together can do $\frac{13}{96}$ of the work in 1 day. 4. **Find the total time taken:** Total time = Reciprocal of combined 1-day work = $\frac{96}{13}$ days. Convert to a mixed fraction: $\frac{96}{13} = 7 \frac{5}{13}$ days. ### Exam Strategy & Shortcut Use the LCM (Total Work) method. Let total work = LCM(12, 8, 16) = 48 units. Efficiencies: $(A+B) = 4$, $(B+C) = 6$, $(A+C) = 3$. Sum of efficiencies = $4 + 6 + 3 = 13$. This is twice the efficiency of $A+B+C$. Efficiency of $(A+B+C) = \frac{13}{2} = 6.5$ units/day. Time taken = $\frac{48}{6.5} = \frac{96}{13} = 7 \frac{5}{13}$ days. ### Common Pitfall A frequent mistake is forgetting to divide the summed rates by 2. Summing $(A+B)$, $(B+C)$, and $(A+C)$ double-counts each worker's contribution. ### Final Answer Therefore, the correct answer is **$7 \frac{5}{13}$**.