A can complete a certain work in 30 days. B is 20% more efficient than A. C can complete $$\frac{2}{5}$$​ part of the same work in 8 days. A and B together complete $$\frac{11}{15}$$​ part of the work. The remaining work is completed by A and C together. In how many days was the entire work completed?

Correct option is B

Given:

A can complete the work in 30 days $$\Rightarrow \text{A's 1 day work} = \frac{1}{30}$$

B is 20% more efficient than A $$\Rightarrow \text{B's 1 day work} = \frac{1}{30} \times \frac{120}{100} = \frac{1}{25} \\$$​ 

C can complete 2/5 of the work in 8 days $$\Rightarrow \frac{2}{5} = 8 \times \text{C's 1 day work} \Rightarrow\\$$

C's 1 day work $$= \frac{2}{5 \times 8} = \frac{1}{20} \\$$​​

=> C can complete full work in 20 days

Now, we know:

​$$\begin{aligned}A &= \frac{1}{30} \\B &= \frac{1}{25} \\C &= \frac{1}{20}\end{aligned}$$​​

Solution:

$$\textbf{Step 1: A and B together complete } \frac{11}{15} \text{ of the work} \\\ \\\text{A + B's 1 day work} = \frac{1}{30} + \frac{1}{25} = \frac{5 + 6}{150} = \frac{11}{150} \\\ \\\text{Time taken} = \frac{\frac{11}{15}}{\frac{11}{150}} = \frac{11}{15} \times \frac{150}{11} = 10 \text{ days} \\\ \\\\\textbf{Step 2: Remaining work} = 1 - \frac{11}{15} = \frac{4}{15} \\\ \\\text{A + C's 1 day work} = \frac{1}{30} + \frac{1}{20} = \frac{2 + 3}{60} = \frac{5}{60} = \frac{1}{12} \\\ \\\text{Time taken} = \frac{\frac{4}{15}}{\frac{1}{12}} = \frac{4}{15} \times 12 = \frac{48}{15} = 3\frac{1}{5} \text{ days} \\\ \\\\\textbf{Step 3: Total Time} = 10 + 3\frac{1}{5} = \boxed{13\frac{1}{5} \text{ days}}$$

​Final Answer: (B)