Number of days taken by ‘P’, ‘Q’ and ‘R’ to

Solution

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