tcs - 4 computers namely p q r and s in a shop p and q running together can finish the task in 10 days if r works twice as p and s works 13 as much as q then the same order of work can be finished in 6 days find the time taken by p alone to complete the same o - skillgun

P and Q together complete the work in 10 days.
Then (P+Q)'s 1 day work = 1/10 i.e P+Q=1/10
Suppose, R works twice as P then R's 1 day work 2P.
And, S works 1/3 as much as Q then S's 1 day work be Q/3.
Now, the time taken to complete the order by (R+S) = 6 days
so solve P+Q=1/10 and 2P+Q/3 = 1/6
then we get P = 4/50.
Thus, P's 1 day work = 4/50.
so P's total work in 50/4 days = 12.5 days.