Question

A can do a piece of work in 80 days. He works at it for 10 days and then B alone finishes
the work in 42 days. The two together could have completed the work in

Answer 1

Answer:

30 days

Step-by-step explanation:

In 80 days A can do the work, in 1 day he does 1/80 part of the work.

He works for 10 days. In 10 days he does 10x1/80 part=1/8 part of the work.

Remaining part of work left=1-1/8=7/8 part

In 42 days B does 7/8 part of the work. In 1 day he does (1/42)x(7/8)=1/48 part

A and B in 1 day do=1/80+1/48=1/30 part.

They would finish the work in 30 days.

Answer 2

Let total work =1

A's one day work =1/80

Let B take x days to finish work alone,

then B's one day work =1/x

Total Work done by A= (days worked by A)*(A's one day work)=10*(1/80)=1/8

Work left =1−1/8=7/8

Now B completes the left work in 42 days,

(B's one day work)×42=(7/8)

(1/x)×42=7/8

7x=8×42

x=48

(A+B)'s one day work =(1/80)+(1/48)

Total time needed by (A+B) to complete

$work = \frac{1}{ \frac{1}{80} + \frac{1}{48} } = 30$