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 _________.

• A. $24days$
• B. $25days$
• C. $30days$
• D. $35days$

Answer 1

Answer: (C)

$30days$

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)$\times 42=\left(7/8\right)$

$\left(1/x\right)\times 42=7/8$

$7x=8\times 42$

$x=48$

(A+B)'s one day work $=\left(1/80\right)+\left(1/48\right)$

Total time needed by (A+B) to complete work =$\frac{1}{\frac{1}{80}+\frac{1}{48}}=$ $30$