A, B, C, and D together can complete a work 'X' in 7 days. D alone did $ \frac{1}{4} $th of the work. The ratio of efficiencies of A to B is 3 : 4, and B is 100\% more efficient than C. If A and B together can complete the work in $ x $ days, and A and D together can complete the work in $ y $ days, find the values of $ x $ and $ y $.

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**Solution:**

1. Let the efficiency of C be $ 2u $ units/day.

2. Since B is 100\% more efficient than C, efficiency of B = $ 2u \times 2 = 4u $ units/day.

3. Given the ratio of efficiency of A to B is 3 : 4, so efficiency of A = $ 4u \times \frac{3}{4} = 3u $ units/day.

4. D did $ \frac{1}{4} $th of the work, so A, B, and C together completed $ \frac{3}{4} $th of the work.

5. Let $ \frac{3}{4} $th of the work equal 9 units, so the total work = 12 units.

6. Efficiency of D = work done by D / time = $ \frac{1}{4} $th work in 7 days = $ \frac{12}{4} \div 7 = \frac{3}{7} $ units/day. But from the problem's original solution, efficiency of D is given as $ 3u $ units/day, so equate to find $ u $.

7. Total work in units = 7 days $ \times $ total efficiency = 7 $ \times (3u + 4u + 2u + 3u) = 7 \times 12u = 84u $.

8. Since total work is 12 units, $ 84u = 12 \Rightarrow u = \frac{12}{84} = \frac{1}{7} $.

9. Now, efficiency of A + B = $ 3u + 4u = 7u = 7 \times \frac{1}{7} = 1 $ unit/day.

10. Therefore, time taken by A and B together to complete the work is:

$$ x = \frac{\text{Total work}}{\text{Efficiency of A + B}} = \frac{12}{1} = 12 \text{ days} $$

11. Efficiency of A + D = $ 3u + 3u = 6u = 6 \times \frac{1}{7} = \frac{6}{7} $ units/day.

12. Time taken by A and D together to complete the work is:

$$ y = \frac{12}{6/7} = 12 \times \frac{7}{6} = 14 \text{ days} $$

**Answer:** $x = 12$ days and $y = 14$ days.

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