There are $2$ people $A$ and $B$. $A$ requires $a\;$ days to complete certain amount of work and $B$ requires $b\;$ days to complete the same amount of work. If $A$ begins the work a day before $B$ begins, then in which of the following conditions will the work get completed quicker than in the case where $B$ begins a day before $A?$
$1.\;n$ is a positive integer and $n(1/a+1/b)=1$
$2.\;b>a$
In my analysis I found that only condition $2$ is necessary to get the work done quicker, but the answer given is only $1$ is necessary. Can someone please explain why $2$ is not sufficient to prove the case?
I take it that fractional days will be taken into account in assessing the difference.
Your answer is the correct one.
ADDED:
If "certain amount of work" and "same amount of work" mean the full work, all that condition $1$ means is that working together, $A$ and $B$ complete the full work in $n$ days, and if instead they mean some fraction (may be improper fraction) of the full work, they complete that fraction of the work in $n$ days.
Either way, it has no bearing on the question asked.