A, B, and C together can finish a task in $16\frac{4}{11}$ days, whereas A and B together can complete the same task in 20 days. How many days will C alone take to complete 60\% of the work?

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**Solution:**

1. Convert the mixed fraction $16\frac{4}{11}$ days into an improper fraction:
   $$16\frac{4}{11} = \frac{16 	imes 11 + 4}{11} = \frac{176 + 4}{11} = \frac{180}{11} 	ext{ days}$$

2. Calculate the work rates:
   - Rate of A, B, and C together:
     $$	ext{Rate}_{ABC} = \frac{1}{\frac{180}{11}} = \frac{11}{180} 	ext{ work/day}$$
   - Rate of A and B together:
     $$	ext{Rate}_{AB} = \frac{1}{20} = 0.05 	ext{ work/day}$$

3. Find the rate of C alone:
   $$	ext{Rate}_C = 	ext{Rate}_{ABC} - 	ext{Rate}_{AB} = \frac{11}{180} - \frac{1}{20}$$

Find common denominator (180):
   $$\frac{11}{180} - \frac{9}{180} = \frac{2}{180} = \frac{1}{90} 	ext{ work/day}$$

4. Time taken by C alone to complete the whole work:
   $$	ext{Time}_C = \frac{1}{	ext{Rate}_C} = 90 	ext{ days}$$

5. Time taken by C to complete 60\% of the work:
   $$	ext{Time}_{C, 60\%} = 90 	imes 0.6 = 54 	ext{ days}$$

**Answer:** 54 days

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