A is 50\% more efficient than B, and B takes twice as much time as C to finish a task. If A completes the work in *x* days and C completes it in *(x - 15)* days, find the number of days in which A, B, and C together can complete the work.

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**Solution:**

- Given B takes double the time of C, so:
  $$ $$\text{Time}_B = 2 \times \text{Time}_C $$$$

- Let time taken by C = *t* days.

- Then time taken by B = 2*t* days.

- Efficiency is inverse of time, so:
  $$ $$\text{Efficiency}_B = \frac{1}{2t}, \quad \text{Efficiency}_C = \frac{1}{t} $$$$

- A is 50\% more efficient than B:
  $$ $$\text{Efficiency}_A = 1.5 \times \text{Efficiency}_B = 1.5 \times \frac{1}{2t} = \frac{3}{4t} $$$$

- Time taken by A:
  $$ $$\text{Time}_A = \frac{1}{\text{Efficiency}_A} = \frac{4t}{3} $$$$

- Given time taken by A is *x* days and time taken by C is $ x - 15 $ days:
  $$ $$x = \frac{4t}{3} \quad \text{and} \quad t = x - 15$$ $$

- Substitute $ t = x - 15 $ into $ $$x = \frac{4t}{3} $$$:
  $$$$ x = \frac{4}{3} (x - 15) $$$$

- Multiply both sides by 3:
  $$ 3x = 4x - 60 $$

- Rearranged:
  $$ 4x - 3x = 60 $$
  $$ x = 60 $$

- So, time taken by A = 60 days.

- Time taken by C = $ x - 15 = 60 - 15 = 45 $ days.

- Time taken by B = $$$ 2 \times 45 = 90$$ $ days.

- Calculate efficiencies:
  $$ $$\text{Efficiency}_A = \frac{1}{60}, \quad \text{Efficiency}_B = \frac{1}{90}, \quad \text{Efficiency}_C = \frac{1}{45} $$$$

- Combined efficiency:
  $$ $$\frac{1}{60} + \frac{1}{90} + \frac{1}{45} = \frac{3}{180} + \frac{2}{180} + \frac{4}{180} = \frac{9}{180} = \frac{1}{20}$$ $$

- Hence, all three together can complete the work in 20 days.

**Answer:** 20 days

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