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 1/3 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 order.
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.
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