A can complete a work in 10 days,B in 12 days | vedclass

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(A) Let the total work be completed in $x$ days.The work done by $A$ in $2$ days is $2 	imes \frac{1}{10} = \frac{1}{5}$.$B$ left $3$ days before the

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$7$

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(A) Let the total work be completed in $x$ days.The work done by $A$ in $2$ days is $2 	imes \frac{1}{10} = \frac{1}{5}$.$B$ left $3$ days before the completion,so $B$ worked for $(x - 3)$ days. The work done by $B$ is $(x - 3) 	imes \frac{1}{12}$.$C$ worked for the entire duration of $x$ days. The work done by $C$ is $x 	imes \frac{1}{15}$.The sum of the work done by $A, B,$ and $C$ equals $1$ (the total work).$\frac{1}{5} + \frac{x-3}{12} + \frac{x}{15} = 1$Multiply the entire equation by the least common multiple of $5, 12,$ and $15$,which is $60$:$12 + 5(x - 3) + 4x = 60$$12 + 5x - 15 + 4x = 60$$9x - 3 = 60$$9x = 63$$x = 7$ days.Thus,the work lasted for $7$ days.

$A$ can complete a work in $10$ days,$B$ in $12$ days,and $C$ in $15$ days. All of them began the work together; but $A$ had to leave the work after $2$ days of the start and $B$ also left $3$ days before the completion of the work. How long did the work last? (in days)

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