One quick way to approach this would be to
pick the rate of work for A and B instead of going into algebra
Evaluating statement 2 first since it seems easier
Since b > a, implies that
a is fasterLets say A does 2 units of work per day and B does 1 unit of work per day.
Let total work be 6 unit ---> if a and b are working on alternate days then it
wont matter who starts work will finish in 4 days i.e. 2, 1, 2 ,1 units of work done by a starting or 1 , 2 , 1 , 2 done by B starting
However, if the total work is say 5 units
Then if A starts ---> 2 , 1 , 2 ---> full work done in 3 days
But if B starts 1 , 2 , 1 ---> only 4 units done in 3 days hence work will spill over to the 4th day
Hence B is not sufficient
Now coming to statement 1
Notice that this tells you that n is the days since days x units per day = total work
So if the total work is 1, then rates would be 1/a and 1/b and since they are multiplied to give you 1, that means that n is number of days.
But when you look closely what statement 1 tells you. ---> It implies that that when the rates are added together they divide evenly into the total work i.e The total work equals an integer number of complete AB cycles.---> which again means that it wont matter who starts cause the alternate pairs will complete the work in integer integer number of days and there would be no spill over.
Taking our example n(2+1) = 6 ---> this means that n *3 = 6 and we know that n is an integer hence
A is sufficient.