A and B can complete a task together in 12 days, while B and C can finish it in 16 days. After A works on it for 5 days and B for 7 days, C takes the remaining 13 days to finish the work. How many days would it take for C to complete the work alone?
24 days
Let the total work be the LCM of 12 and 16, i.e. 48 units.
Step 1 – Combined rates from given pairs:
\((A + B) = \dfrac{48}{12} = 4 \text{ units/day}\)
\((B + C) = \dfrac{48}{16} = 3 \text{ units/day}\)
So \(A = 4 - B\) and \(C = 3 - B\).
Step 2 – Translate the work-distribution condition:
\(5A + 7B + 13C = 48\)
\(5(4 - B) + 7B + 13(3 - B) = 48\)
\(20 - 5B + 7B + 39 - 13B = 48\)
\(59 - 11B = 48 \;\Rightarrow\; B = 1 \text{ unit/day}\)
Step 3 – Find C's rate and time alone:
\(C = 3 - B = 3 - 1 = 2 \text{ units/day}\)
\(\text{Time for C alone} = \dfrac{48}{2} = 24 \text{ days}\)
Hence C alone would take 24 days.
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