A and B can complete a piece of work in 15 days, B and C in 30 days and C and A in 45 days. In how many days will they complete the work if they start working together?

The correct option is $16\frac{4}{11}$ days.

Step 1: Calculate the work done by each pair in one day.

Let the total work be 1 unit.

Work done by A and B together in 1 day = $\frac{1}{15}$

Work done by B and C together in 1 day = $\frac{1}{30}$

Work done by C and A together in 1 day = $\frac{1}{45}$

Step 2: Calculate the combined work done by A, B, and C together in one day.

Adding the work done by all three pairs in one day:

$2(\text{A}+\text{B}+\text{C})=\frac{1}{15}+\frac{1}{30}+\frac{1}{45}$

Find the least common multiple (LCM) of the denominators 15, 30, and 45, which is 90:

$2(\text{A}+\text{B}+\text{C})=\frac{6+3+2}{90}=\frac{11}{90}$

Therefore, the work done by A, B, and C working together in 1 day is:

$\text{A}+\text{B}+\text{C}=\frac{11}{90\times 2}=\frac{11}{180}$

Step 3: Calculate the total time taken to complete the work together.

The total number of days taken by A, B, and C working together is the reciprocal of their 1-day work:

$\text{Total Days}=\frac{180}{11}$

Converting $\frac{180}{11}$ into a mixed fraction gives:

$180÷11=16$ with a remainder of $4$

So, the total time required is $16\frac{4}{11}$ days.