A can complete a piece of work in 10 days, B in 15 days and C in 20 days. A and C together for 2 days and A was replaced by B. In how many days, altogether, was the work complete ?

Solution(By Examveda Team)

$$\eqalign{ & \left( {{\text{A}} + {\text{C}}} \right){\text{'s 1 day's work}} \cr & = \left( {\frac{1}{{10}} + \frac{1}{{20}}} \right) \cr & = \frac{3}{{20}} \cr & \left( {{\text{A}} + {\text{C}}} \right){\text{'s 2 day's work}} \cr & = \left( {\frac{3}{{20}} \times 2} \right) \cr & = \frac{3}{{10}} \cr & {\text{Remaining work }} \cr & = \left( {1 - \frac{3}{{10}}} \right) \cr & = \frac{7}{{10}}{\text{ }} \cr & \left( {{\text{B}} + {\text{C}}} \right){\text{'s 1 day's work}} \cr & = \left( {\frac{1}{{15}} + \frac{1}{{20}}} \right) \cr & = \frac{7}{{60}} \cr} $$
$$\frac{7}{{60}}$$ work is done by B and C in 1 day
∴ $$\frac{7}{{10}}$$ work is done by B and C in
$$\eqalign{ & = \left( {\frac{{60}}{7} \times \frac{7}{{10}}} \right) \cr & = 6{\text{ days}}{\text{. }} \cr & {\text{Hence, total time taken }} \cr & = \left( {2 + 6} \right){\text{days}} \cr & = 8{\text{ days}} \cr} $$