A and B together can complete a work in 12 days, and A, B, and C together can complete the same work in 6 days. The efficiency of A is twice that of B, and C alone can finish the work in $R + 8$ days. Find the number of days taken by A alone to complete the work.

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Accepted Answer

**Solution:**

1. Let the efficiency of B be $b$ units of work per day. Then efficiency of A is $2b$ units per day.

2. Combined efficiency of A and B:
   $$2b + b = 3b$$
   Since A and B together complete the work in 12 days, their combined daily work is:
   $$\frac{1}{12} = 3b \implies b = \frac{1}{36}$$

3. Therefore, efficiency of A:
   $$2b = 2 	imes \frac{1}{36} = \frac{1}{18}$$

4. Efficiency of C alone is:
   $$\frac{1}{R + 8}$$

5. Efficiency of A, B, and C together is:
   $$\frac{1}{6}$$

6. Sum of efficiencies of A, B, and C:
   $$2b + b + \frac{1}{R + 8} = \frac{1}{6}$$
   Substitute $b = \frac{1}{36}$:
   $$3 	imes \frac{1}{36} + \frac{1}{R + 8} = \frac{1}{6}$$
   $$\frac{1}{12} + \frac{1}{R + 8} = \frac{1}{6}$$

7. Rearranging to find $\frac{1}{R + 8}$:
   $$\frac{1}{R + 8} = \frac{1}{6} - \frac{1}{12} = \frac{1}{12}$$

8. This confirms the given condition for C, so no contradiction.

9. Time taken by A alone to complete the work is the reciprocal of A's efficiency:
   $$	ext{Time by A} = \frac{1}{\frac{1}{18}} = 18$$

10. Expressing 18 in terms of $R$:
    Since $\frac{1}{R + 8} = \frac{1}{12}$, we get $R + 8 = 12 \implies R = 4$

11. Therefore, time taken by A alone:
    $$18 = R + 14$$
    (Because $4 + 14 = 18$)

**Answer:** $R + 14$

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