A is 40\% more efficient than B and both together can complete a work in $9\frac{3}{8}$ days. If A works for the first five days alone and the remaining work is completed by B, then find in how many days total work will be completed?

Question

Accepted Answer

**Solution:**

1. Let the efficiency of B be $5x$ units/day.

2. Since A is 40\% more efficient than B, efficiency of A = $5x 	imes \frac{140}{100} = 7x$ units/day.

3. Both together can complete the work in $9\frac{3}{8} = \frac{75}{8}$ days.

4. Total work = (A's efficiency + B's efficiency) $	imes$ total days
   
   $$	ext{Total work} = (7x + 5x) 	imes \frac{75}{8} = 12x 	imes \frac{75}{8} = \frac{900x}{8} = \frac{225x}{2} 	ext{ units}$$

5. Work done by A in first 5 days:

$$7x 	imes 5 = 35x 	ext{ units}$$

6. Remaining work after A's 5 days:

$$\frac{225x}{2} - 35x = \frac{225x - 70x}{2} = \frac{155x}{2} 	ext{ units}$$

7. B completes the remaining work at $5x$ units/day, so time taken by B:

$$\frac{\frac{155x}{2}}{5x} = \frac{155}{10} = 15.5 	ext{ days}$$

8. Total time to complete the work:

$$5 + 15.5 = 20.5 	ext{ days} = 20 \frac{1}{2} 	ext{ days}$$

**Answer:** $20 \frac{1}{2}$ days

Answer

Answer

Answer

Answer