A and B together can complete a piece of work in 25 days, B and C together can complete the same piece of work in 36 days, while C and A together can complete it in 30 days. If A, B, C, and D together can complete this piece of work in 18 days, then in how many days can D alone complete this piece of work?

The correct answer is 200.

Step 1: Determine the rate of work done by pairs of individuals.
Let the total work to be completed be 1 unit.

Work done by A and B together in 1 day = $\frac{1}{25}$

Work done by B and C together in 1 day = $\frac{1}{36}$

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

Step 2: Calculate the combined 1-day work of A, B, and C together.
Add the 1-day work of all three pairs:

$2\times \left(\text{Work done by A, B, and C together in 1 day}\right)=\frac{1}{25}+\frac{1}{36}+\frac{1}{30}$

To sum these fractions, find the Least Common Multiple (LCM) of the denominators 25, 36, and 30, which is 900.

$2\times \left(\text{Work done by A, B, and C together in 1 day}\right)=\frac{36+25+30}{900}=\frac{91}{900}$

Now, divide by 2 to get the combined 1-day work of A, B, and C:

$\text{Work done by A, B, and C together in 1 day}=\frac{91}{2\times 900}=\frac{91}{1800}$

Step 3: Calculate the 1-day work of D alone.
It is given that A, B, C, and D together can complete the work in 18 days.

Work done by A, B, C, and D together in 1 day = $\frac{1}{18}$

The work done by D alone in 1 day is the total 1-day work of all four minus the 1-day work of A, B, and C together:

$\text{Work done by D in 1 day}=\frac{1}{18}-\frac{91}{1800}$

Expressing $\frac{1}{18}$ with a denominator of 1800:

$\text{Work done by D in 1 day}=\frac{100}{1800}-\frac{91}{1800}=\frac{100-91}{1800}=\frac{9}{1800}=\frac{1}{200}$

Step 4: Find the total days required by D alone.
Since D completes $\frac{1}{200}$ of the work in 1 day, the total time required for D alone to complete the entire work is:

$\text{Days taken by D}=200\text{days}$

Therefore, D alone can complete the piece of work in 200 days.