A and B can complete a job together in $12.5$ days. B and C can complete the same job together in $18.75$ days, while C and A can complete the same job together in $15$ days. In how many days will A, B, and C together be able to complete the same job alongside D, who is only $40\%$ as efficient as C?

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Accepted Answer

**Solution:**

1. Let the efficiencies (work done per day) of A, B, and C be $a$, $b$, and $c$ respectively.

2. Given:
   - A and B together complete the job in $12.5$ days, so their combined efficiency:
     $$a + b = \frac{1}{12.5} = \frac{1}{\frac{25}{2}} = \frac{2}{25} = 0.08$$

- B and C together complete the job in $18.75$ days, so:
     $$b + c = \frac{1}{18.75} = \frac{1}{\frac{75}{4}} = \frac{4}{75} \approx 0.0533$$

- C and A together complete the job in $15$ days, so:
     $$c + a = \frac{1}{15} = 0.0667$$

3. Add all three equations:
   $$ (a + b) + (b + c) + (c + a) = 0.08 + 0.0533 + 0.0667 = 0.2 $$

This simplifies to:
   $$ 2(a + b + c) = 0.2 \implies a + b + c = 0.1 $$

4. Now, find individual efficiencies:
   - From $a + b = 0.08$, and $a + b + c = 0.1$, we get:
     $$ c = 0.1 - 0.08 = 0.02 $$

5. Efficiency of D is $40\%$ of C:
   $$ d = 0.4 	imes 0.02 = 0.008 $$

6. Total efficiency of A, B, C, and D together:
   $$ a + b + c + d = 0.1 + 0.008 = 0.108 $$

7. Time taken by all four together to complete the job:
   $$ 	ext{Time} = \frac{1}{0.108} = \frac{1000}{108} = \frac{250}{27} = 9 \frac{7}{27} 	ext{ days} $$

**Answer:** $9 \frac{7}{27}$ days

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