P can do a work in 15 days. Q can do a work in 25 days. Let P left after x days. Q worked for (x+5) days. ∴ 1/15 × x + 1/25 (x+5) = 1 x/15 + x/25 + 1/5 = 1 (5x+3x)/75 = 4/5 8x/75 = 4/5 x = 15/2 days Alternate method: Let total work = LCM(15, 25) = 75 So P’s 1 day work = 5 & Q’s 1 day work =3 So Q’s 5 days work in last = 3×5 = 15 So remaining work = 75 – 15 = 60 So this 60 units of work is completed by both in = 60/(5+3) = 60/8 =7.5 days
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