Question

$A$ and $B$ together can complete a work in $3$ days. They start together but, after $2$ days, $B$ left the work. If the work is completed after $2$ more days, $B$ alone could do the work in

• A. $4$ days
• B. $6$ days
• C. $8$ days
• D. $10$ days

Answer 1

Answer: (B)

$6$ days

Let the total unit of work be $6$ units (LCM of $2$ and $3$)
$(A+B)do=\frac{6}{3}=2units/day$
$(A+B{)}^{\prime }s2day$ work $=2\times 2=4units$
Work left $=6-4=2units$
This remaining work is done by $A$ in $2$ days.
$\therefore A^{\prime }s$ per day work $=\frac{2}{2}=1unit/day$
$\therefore B^{\prime }s$ per unit work $=(A+B{)}^{\prime }swork-A^{\prime }swork$
$=2-1=1unit/day$
$B$ alone can complete the work in $=\frac{6}{1}=6days$.