Number of days taken by ‘P’, ‘Q’ and ‘R’ to do a certain work alone is (x + 4) days, (x - 11) days and (x - 1) days, respectively and number of days taken by ‘P’ and ‘Q’ together to do the work is (1/4) days more than the number of days taken by ‘Q’ and ‘R’ together to do the work. Find the number of days taken to complete the work if all three of them decided to do the work alternatively till the work gets completed starting with ‘P’, then ‘Q’ and then ‘R’.

Let the total amount of work be {(x + 4)(x - 11)(x - 1)} units Efficiency of ‘P’ = [{(x + 4)(x - 11)(x - 1)}/(x + 4)] = {(x - 11)(x - 1)} units/day Efficiency of ‘Q’ = [{(x + 4)(x - 11)(x - 1)}/(x - 11)] = {(x + 4)(x - 1)} units/day Efficiency of ‘R’ = [{(x + 4)(x - 11)(x - 1)}/(x - 1)] = {(x + 4)(x - 11)} units/day ATQ; [{(x + 4)(x - 11)(x - 1)}/{(x - 11)(x - 1) + (x + 4)(x - 1)}]
= (1/4) + [{(x + 4)(x - 11)(x - 1)}/{(x + 4)(x - 1) + (x + 4)(x - 11)}] Or, [{(x + 4)(x - 11)}/{2x - 7}] = (1/4) + [{(x - 11)(x - 1)}/{2x - 12}] Or, 8x² - 201x + 1168 = 0 Or, 8x² - 128x - 73x + 1168 = 0 Or, 8x(x - 16) - 73(x - 16) = 0 Or, (8x - 73)(x - 16) = 0 Or, x = 73/8 or x = 16 As number of days taken by ‘Q’ cannot be negative, so value of x cannot be 73/8. So, x = 16 Now, number of days taken by ‘P’ to do the work alone = (16 + 4) = 20 days Number of days taken by ‘Q’ to do the work alone = (16 - 11) = 5 days And, number of days taken by ‘R’ to do the work alone = (16 - 1) = 15 days So, let total work = 60 units
LCM of 20, 5 and 15 = 60 Efficiency of ‘P’ = (60/20) = 3 units/day Efficiency of ‘Q’ = (60/5) = 12 units/day Efficiency of ‘R’ = (60/15) = 4 units/day So, work done by ‘P’, ‘Q’ and ‘R’ in consecutive 3 days, separately = (3 + 12 + 4) = 19 units Similarly, work done by ‘P’, ‘Q’ and ‘R’ in consecutive 9 days, separately = (19 × 3) = 57 units Time taken by ‘P’ to do the rest of the work = (60 - 57)/3 = 1 day So, total number of days taken to complete the whole work = (9 + 1) = 10 days Since, x - 6 = 16 - 6 = 10