The time taken by A to complete a work is 20 days. B takes $2x\%$ more days than A to complete the same work, and C takes $x\%$ more days than B to complete the work. The time taken by C alone to complete the work is equal to the sum of the time taken by A and B alone to complete the work. Find the time taken by all three working together to complete 11 times the work.

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**Solution:**

1. Let the time taken by A to complete the work be $20$ days.

2. B takes $2x\%$ more days than A, so time taken by B is:
   $$20 	imes \left(1 + \frac{2x}{100}ight)$$

3. C takes $x\%$ more days than B, so time taken by C is:
   $$20 	imes \left(1 + \frac{2x}{100}ight) 	imes \left(1 + \frac{x}{100}ight)$$

4. Given that time taken by C equals the sum of times taken by A and B:
   $$20 	imes \left(1 + \frac{2x}{100}ight) 	imes \left(1 + \frac{x}{100}ight) = 20 + 20 	imes \left(1 + \frac{2x}{100}ight)$$

5. Dividing both sides by 20:
   $$\left(1 + \frac{2x}{100}ight) 	imes \left(1 + \frac{x}{100}ight) = 1 + \left(1 + \frac{2x}{100}ight)$$

6. Simplify the right side:
   $$1 + 1 + \frac{2x}{100} = 2 + \frac{2x}{100}$$

7. Expand the left side:
   $$1 + \frac{x}{100} + \frac{2x}{100} + \frac{2x^2}{10000} = 2 + \frac{2x}{100}$$

8. Combine like terms on the left:
   $$1 + \frac{3x}{100} + \frac{2x^2}{10000} = 2 + \frac{2x}{100}$$

9. Rearranged:
   $$1 + \frac{3x}{100} + \frac{2x^2}{10000} - 2 - \frac{2x}{100} = 0$$

10. Simplify:
    $$-1 + \frac{x}{100} + \frac{2x^2}{10000} = 0$$

11. Multiply entire equation by 10000 to clear denominators:
    $$-10000 + 100x + 2x^2 = 0$$

12. Rearranged quadratic:
    $$2x^2 + 100x - 10000 = 0$$

13. Divide entire equation by 2:
    $$x^2 + 50x - 5000 = 0$$

14. Solve quadratic using factorization or formula:
    - Factors of $-5000$ that sum to $50$ are $100$ and $-50$.

15. So,
    $$(x + 100)(x - 50) = 0$$

16. Hence, $x = 50$ (reject negative root).

17. Calculate times:
    - Time taken by B:
      $$20 	imes \left(1 + \frac{2 	imes 50}{100}ight) = 20 	imes (1 + 1) = 40 	ext{ days}$$
    - Time taken by C:
      $$40 	imes \left(1 + \frac{50}{100}ight) = 40 	imes 1.5 = 60 	ext{ days}$$

18. Find the least common multiple (LCM) of times 20, 40, and 60:
    $$	ext{LCM} = 120$$

19. Calculate efficiencies (work units per day):
    - Efficiency of A: $\frac{120}{20} = 6$
    - Efficiency of B: $\frac{120}{40} = 3$
    - Efficiency of C: $\frac{120}{60} = 2$

20. Combined efficiency:
    $$6 + 3 + 2 = 11 	ext{ units per day}$$

21. Total work for one unit is 120 units, so for 11 times the work:
    $$11 	imes 120 = 1320 	ext{ units}$$

22. Time taken by all together to complete 11 times the work:
    $$\frac{1320}{11} = 120 	ext{ days}$$

**Answer:** $120$ days

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