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A and B together can do a piece of work in 30 days, B and C together can do it in 20 days, A starts the work and works on it for 5 days, B takes up and work for 15 days. Finally C finishes the work in 18 days. The number of days in which C alone can do the work where doing it separately is = ?

Correct Answer: 24 days
L.C.M. of Total Work =60
One day work of A + B = $$\frac{{60}}{{30}}$$ = 2 unit/day
One day work of B + C = $$\frac{{60}}{{20}}$$ = 3 unit/day
$$\eqalign{ & {\text{According to the question,}} \cr & \Rightarrow \left( {{\text{A}} + {\text{B}}} \right).....{\text{30 days}} \cr & \Rightarrow \left( {{\text{B}} + {\text{C}}{\text{.}}} \right).....{\text{20 days}} \cr & \Rightarrow {\text{A}}.....{\text{5}} \cr & \Rightarrow {\text{B}}.....{\text{5}} + {\text{10}} \cr & \Rightarrow {\text{C}}.....{\text{10}} + {\text{8 days}} \cr} $$
Work done by (A +B) in 5 days
= 2 × 5 = 10 units
⇒ Work done by (B + C) in 10 days
= 10 × 3 = 30 units
⇒ Total work (finished till now).....40 units
⇒ Remaining work
= 60 - 40
= 20 units
⇒ Here we find that C does remaining 20 units in 8 days
⇒ C's efficiency = $$\frac{{{\text{Work}}}}{{{\text{Day}}}}$$
$$ = \frac{{{\text{20 Work}}}}{{{\text{8 Days}}}} = \frac{5}{2}$$
⇒ Therefore, time taken by C alone to complete the work
$$ = \frac{{60}}{5} \times 2 = 24{\text{ days}}$$

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