P and Q together can do a piece of work in 30 days, Q and R together can do it in 20 days. P starts the work and works on it for 5 days, then Q takes it up and works for 15 days. Then, R finishes the remaining work in 18 days. If R works alone, how many days he will take to complete the work?

Solution

Given, (P + Q)’s 1 day work = 1/30 And (Q + R)’s 1 day work = 1/20 Also, P’s 5 days w + Q’s 15 days w + R’s 18 days w = 1 work ⇒P’s 5 days w + Q’s 5 days w + Q’s 10 days w + R’s 10 day w + R’s 8 day w = 1 work ⇒ (P + Q)’s 5 days w + (Q+ R)’s 10 days w + R’s 8 days work = 1 work ⇒ (5/30) + (10/20) + R’s 8 days work = 1 work ⇒ (1/6) + (1/2) + R’s 8 days work = 1 work ⇒R’s 8 days work = 1 – ( 1/6 + 1/2 ) work ⇒R’s 8 days work = 1 – ((1+3)/6) = 1 - (4/6) = 1 - (2/3) R’s 8 days work = 1/3 work R will complete the work in 8 × 3 = 24 days Alternate Method: P+Q Q+R Time 30 20 Let work = LCM = 60 units Eff 2 : 3 P`xx` 5+Q`xx` 15+R`xx` 18 = 60 P`xx` 5+Q`xx` 5+Q`xx` 10+R`xx` 10+R`xx` 8=60 (P+Q)`xx` 5 + (Q+R)`xx` 10+8R = 60 (2)`xx` 5 + (3)`xx` 10+8R = 60