Question

A can do a piece of work in 100 days, b and c together can do the same work in 20 days. if b can do the work in same time as that of c and a together then how long c alone can do the same work?

Answer 1

1/b=1/a+1/c means 1/b-1/c=1/100
1/b+1/c=1/20
equating this will get i/c=1/50
so c will complete the work in 50 days if he work alone

Answer 2

Answer:

C alone can do the same work in 50 days.

Step-by-step explanation:

It is given that A can do a piece of work in 100 days. The part of work done by  A in 1 day is $\frac{1}{100}$.

The part of work done by B and C in one day is $\frac{1}{b}$ and $\frac{1}{c}$ respectively.

It is given that B and C together can do the same work in 20 days.

$\frac{1}{b}+\frac{1}{c}=\frac{1}{20}$          ..... (1)

B can do the work in same time as that of C and A together.

$\frac{1}{b}=\frac{1}{a}+\frac{1}{c}$

$\frac{1}{b}-\frac{1}{c}=\frac{1}{100}$         ..... (2)

Subtract equation (2) from (1).

$\frac{1}{b}+\frac{1}{c}-(\frac{1}{b}-\frac{1}{c})=\frac{1}{20}-\frac{1}{100}$

$2(\frac{1}{c})=\frac{4}{100}$

$2(\frac{1}{c})=\frac{1}{25}$

$\frac{1}{c}=\frac{1}{50}$

$c=50$

Since c=50, therefore c alone can do the same work in 50 days.