A and B can finish a work in 20 days while B and C can do it in 36 days. A started the work, worked for 5 days, then B worked for 10 days and the remaining work was finished by C in 15 days. In how many days could C alone have finished the whole work?

Correct option is A

Given:
A + B can finish in 20 days
B + C can finish in 36 days
A worked 5 days, B worked 10 days, C worked 15 days and finished the work
Solution:
Total work = LCM of 20 and 36 = 180 units
A + B = 180 ÷ 20 = 9 units/day
B + C = 180 ÷ 36 = 5 units/day
 (A + B) − (B + C) = A − C = 9 − 5 = 4 units/day
 A = C + 4
Work done: 5A + 10B + 15C = 180 units
But B = (B + C) − C = 5 − C
 5(C + 4) + 10(5 − C) + 15C = 180
 5C + 20 + 50 − 10C + 15C = 180
 (5 − 10 + 15)C + 70 = 180
 10C = 110
C = 11 units/day
Time for C alone = Total work ÷ C = $$\frac{180}{ 11}$$​ days = $$16 \frac{4}{11} \text{ days}$$​ days