P and Q together can do a piece of work in 15 days, Q and R together can do it in 18 days. P starts the work and works on it for 5 days, then Q takes it up and works for 15 days. Finally, R finishes the work in 14 days. In how many days can R do the work when doing it separately (in days)?

Solution

(P + Q)’s 1 day work = 1/15 And (Q + R)’s 1 day work = 1/10 Also, P’s 5 days work + Q’s 15 days work + R’s 14 days work = 1 work ⇒ P’s 5 days work + Q’s 5 days work + Q’s 10 days work + R’s 10 day work + R’s 4 day work = 1 work ⇒ (P + Q)’s 5 days work + (Q + R)’s 10 days work + R’s 4 days work = 1 work ⇒ 5/15 + 10/18 + R’s 4 days work = 1 work ⇒ 1/3 + 5/9 + R’s 4 days work = 1 work ⇒ R’s 4 days work = 1 – ( 1/3 + 5/9 ) work ⇒ R’s 4 days work = 1 – ((3 + 5)/9) = 1 - 8/9 = 1 - 8/9 ∴ R’s 4 days work = 1/9 work ∴ R will complete the work in 4 × 9 = 36 days Alternate method: P+Q can do a work in 15 days & Q+R can do a same work in 18 days Let total workdone = LCM of 15 & 18 = 90 units Hence P+ Q’s one day work = 90/15 = 6 units Q+ R’s one day work = 90/18 = 5 units So P worked for 5 days, Q for 15 days & R for 14 days Or P worked for 5 days, Q for 5 days and Q worked for 10 days, R worked for 10 days & R for 4 days Or P+Q’s 5 days together work = 6×5 = 30 units & Q+R’s 10 days together work = 10×5 = 50 units So remaining work = 90 – 30 – 50 = 10 units This remaining work is done by R in 4 days So R ‘s 1 day work = 10/4 = 5/2 So R can do alone complete job in = 90 /(5/2) = 90×2 /5= 180/5 = 36 days