A can do a piece of work in 10 days working 8 hours per day. If B is two-thirds as efficient as A, then in how many days can B alone do the same piece of work, working 5 hours per day?

### Concept & Multi-Variable Work Equations When dealing with varying days, hours per day, and efficiency, convert everything into a standard unit (like total hours) and apply the formula linking these variables. $$ \text{Days}_1 \times \text{Hours}_1 \times \text{Efficiency}_1 = \text{Days}_2 \times \text{Hours}_2 \times \text{Efficiency}_2 $$ ### Step-by-Step Solution 1. **Determine Total Hours for A:** A takes 10 days, working 8 hours a day. Total hours taken by A = $10 \times 8 = 80$ hours. 2. **Establish the Efficiency Ratio:** B is $\frac{2}{3}$ as efficient as A. If A's efficiency is 3 units/hour, then B's efficiency is 2 units/hour. 3. **Calculate Total Work:** Total Work = A's Total Hours $\times$ A's Efficiency Total Work = $80 \times 3 = 240$ units. 4. **Calculate B's Time:** B needs to complete 240 units of work. B's total hours needed = $\frac{\text{Total Work}}{\text{B's Efficiency}} = \frac{240}{2} = 120$ hours. B works 5 hours per day. Number of days required by B = $\frac{120}{5} = 24$ days. ### Exam Strategy & Shortcut Use the direct inverse relationship equation: $M_1 D_1 H_1 E_1 = M_2 D_2 H_2 E_2$. Here, workers are just 1 person (so $M_1 = M_2 = 1$). $D_1 = 10$, $H_1 = 8$, $E_1 = 3$ (since B is 2/3 of A, A=3, B=2). $D_2 = x$, $H_2 = 5$, $E_2 = 2$. $10 \times 8 \times 3 = x \times 5 \times 2$ $240 = 10x \implies x = 24$. ### Common Pitfall ### Final Answer Therefore, the correct answer is **24**.