Question

A and B did a piece of work in 10 days b and c in 15 days and c and a in 20 days they all worked together for 6 days and then A leaves,and b and c goes on for 4 days more .if b leaves how long will c take to complete the work​

Answer 1

Answer:

$10$ $days$

Step-by-step explanation:

A and B do a piece of work in $10$ $days$

⇒Work Done By A and B in One Day $(A+B)$ = $\frac{1}{10}$      ----- $(1)$

B and C do a piece of work in $15$ $days$

⇒Work Done By B and C in One Day $(B+C)$ = $\frac{1}{15}$       ------$(2)$

A and C can do a piece of work in $20$ $days$

⇒Work Done By A and C in One Day $(A+C)$ = $\frac{1}{20}$                     ------$(3)$

Work Done by A, B, and C in One Day = $(A+B+C)$

Adding the three equations $(1)+(2)+(3)$

⇒$A+B+B+C+C+A$ = $\frac{1}{10} + \frac{1}{15} +\frac{1}{20}$

⇒$2A+2B+2C = \frac{6 + 4 + 3}{60}$

⇒$2(A+B+C) = \frac{13}{60}$

⇒$A+B+C = \frac{13}{120}$

$A$, $B$, and $C$ worked together for 6 days

⇒Work Done By A, B, and C in 6 days = $\frac{13}{120} *6$ = $\frac{13}{20}$

Work Left After A Leaves = $1 - \frac{13}{20}$ = $\frac{7}{20}$

$B$, and $C$ Work together for 4 days

⇒Work Done by B and C in 4 days = $\frac{1}{15} *4$ = $\frac{4}{15}$

Work Left after B leaves = $\frac{7}{20} - \frac{4}{15}$ = $\frac{21-16}{60}$ = $\frac{5}{60} = \frac{1}{12}$

Work done by C in One Day = $C$

We know that $A+B+C=\frac{13}{120}$ and $A+B = \frac{1}{10}$

Substituting the values

⇒$\frac{1}{10} +C=\frac{13}{120}$

⇒$C = \frac{13-12}{120}$

⇒$C = \frac{1}{120}$

⇒Work done by C in One Day = $\frac{1}{120}$

No Of days taken to finish the job by $C$ = $x$

⇒$\frac{1}{120} * x = \frac{1}{12}$

⇒$x = 10$ $days$

∴ $C$ takes $10$ $days$ to finish the job