8 men and 10 women can complete a work in 15 days, while 10 men and 18 women can finish the same work in 10 days. If 4 men and 5 women start the work and after 10 days all the men leave, how many additional women are needed to complete the remaining work within a total of 21 days?

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**Solution:**

1. Let the work done by one man in one day be *m* and by one woman in one day be *w*.

2. From the first condition:
   - 8 men and 10 women complete the work in 15 days.
   - Total work = 1 unit.
   - Work done per day by 8 men and 10 women = $ \frac{1}{15} $

So, 
   $$8m + 10w = \frac{1}{15} \quad \Rightarrow \quad 8m + 10w = \frac{1}{15}$$

3. From the second condition:
   - 10 men and 18 women complete the work in 10 days.
   - Work done per day by 10 men and 18 women = $ \frac{1}{10} $

So, 
   $$10m + 18w = \frac{1}{10}$$

4. Solve these two equations:

Multiply the first equation by 10:
   $$80m + 100w = \frac{10}{15} = \frac{2}{3}$$

Multiply the second equation by 8:
   $$80m + 144w = \frac{8}{10} = \frac{4}{5}$$

5. Subtract the first from the second:
   $$ (80m + 144w) - (80m + 100w) = \frac{4}{5} - \frac{2}{3} $$
   $$44w = \frac{12}{15} - \frac{10}{15} = \frac{2}{15}$$

Note: $ \frac{4}{5} = \frac{12}{15} $ and $ \frac{2}{3} = \frac{10}{15} $

So,
   $$44w = \frac{2}{15} \Rightarrow w = \frac{2}{15 \times 44} = \frac{1}{330}$$

6. Substitute $ w = \frac{1}{330} $ into the first equation:
   $$8m + 10 \times \frac{1}{330} = \frac{1}{15}$$
   $$8m + \frac{10}{330} = \frac{1}{15}$$
   $$8m = \frac{1}{15} - \frac{1}{33} = \frac{11 - 5}{165} = \frac{6}{165} = \frac{2}{55}$$
   $$m = \frac{2}{55 \times 8} = \frac{1}{220}$$

7. Work done by 4 men and 5 women per day:
   $$4m + 5w = 4 \times \frac{1}{220} + 5 \times \frac{1}{330} = \frac{4}{220} + \frac{5}{330}$$
   Find common denominator 660:
   $$\frac{4}{220} = \frac{12}{660}, \quad \frac{5}{330} = \frac{10}{660}$$
   So,
   $$4m + 5w = \frac{12 + 10}{660} = \frac{22}{660} = \frac{1}{30}$$

8. Work done by 4 men and 5 women in 10 days:
   $$10 \times \frac{1}{30} = \frac{10}{30} = \frac{1}{3}$$

9. Remaining work:
   $$1 - \frac{1}{3} = \frac{2}{3}$$

10. After 10 days, men leave. The remaining work $ \frac{2}{3} $ must be done by women alone in the next $21 - 10 = 11$ days.

11. Let the number of additional women required be $x$.

Total women working now = $5 + x$.

Work done per day by these women:
    $$ (5 + x) \times w = (5 + x) \times \frac{1}{330} = \frac{5 + x}{330}$$

12. Work done in 11 days:
    $$11 \times \frac{5 + x}{330} = \frac{2}{3}$$

13. Solve for $x$:
    $$\frac{11(5 + x)}{330} = \frac{2}{3}$$
    $$11(5 + x) = \frac{2}{3} \times 330 = 220$$
    $$5 + x = \frac{220}{11} = 20$$
    $$x = 20 - 5 = 15$$

**Answer:** 15

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