A can do a certain work in 30 days. B is 25% more efficient than A, and C is 20% more efficient than B. A and B work together for 10 days. C alone completes the remaining work in x days. The value of x is:

Let total work = 30 units (LCM of 30). A's efficiency = 30/30 = 1 unit/day. B is 25% more efficient than A, so B's efficiency = 1.25 units/day. C is 20% more efficient than B, so C's efficiency = 1.25 × 1.2 = 1.5 units/day. Work done by A and B in 10 days = (1 + 1.25) × 10 = 22.5 units. Remaining work = 30 - 22.5 = 7.5 units. Time for C alone = 7.5/1.5 = 5 days. This matches option D.