B is 20\% more efficient than A. If B is 60\% more efficient than A, then B could complete the work in 18 days less than the number of days taken by A. What fraction of work would be left after 12 days, if A and B work together?

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**Solution:**

1. Let the time taken by A alone to complete the work be $24x$ days.

2. Since B is 20\% more efficient than A, time taken by B alone is:
   $$\frac{24x}{1.2} = 20x 	ext{ days}$$

3. When B is 60\% more efficient than A, time taken by B alone is:
   $$\frac{24x}{1.6} = 15x 	ext{ days}$$

4. Given that the difference in days between A and B (at 60\% efficiency) is 18 days:
   $$24x - 15x = 18 \implies 9x = 18 \implies x = 2$$

5. Therefore, original time taken by A and B alone are:
   - A: $24x = 24 	imes 2 = 48$ days
   - B: $20x = 20 	imes 2 = 40$ days

6. Let total work be 240 units (LCM of 48 and 40 for convenience).

7. Efficiency of A:
   $$\frac{240}{48} = 5 	ext{ units/day}$$

8. Efficiency of B:
   $$\frac{240}{40} = 6 	ext{ units/day}$$

9. Work done by A and B together in 12 days:
   $$12 	imes (5 + 6) = 12 	imes 11 = 132 	ext{ units}$$

10. Fraction of work left:
    $$\frac{240 - 132}{240} = \frac{108}{240} = \frac{9}{20}$$

**Answer:** $\frac{9}{20}$

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