Work done by A in one day is half of the work done by B in one day. Work done by B is half of the work done by C in one day. If C alone can complete the work in 7 days, in how many days can A, B and C together complete the work?

### Concept & Efficiency This problem compares "work done in one day", which is exactly the definition of efficiency. We can trace the efficiencies relative to the known worker (C) to find the combined working rate. ### Step-by-Step Solution * **Calculate C's 1-day work:** C completes the work in $7$ days. Work done by C in $1$ day = $\frac{1}{7}$. * **Calculate B's 1-day work:** Work done by B in $1$ day is half of C's 1-day work. B's $1$ day work = $\frac{1}{2} \times \frac{1}{7} = \frac{1}{14}$. (This means B alone takes $14$ days). * **Calculate A's 1-day work:** Work done by A in $1$ day is half of B's 1-day work. A's $1$ day work = $\frac{1}{2} \times \frac{1}{14} = \frac{1}{28}$. (This means A alone takes $28$ days). * **Calculate Combined 1-day work:** (A + B + C)'s 1-day work = $\frac{1}{28} + \frac{1}{14} + \frac{1}{7}$ LCM of $28$, $14$, $7$ is $28$. Combined 1-day work = $\frac{1 + 2 + 4}{28} = \frac{7}{28} = \frac{1}{4}$. * **Total Time:** Since they complete $\frac{1}{4}$ of the work in one day, they will finish the entire work in $4$ days. ### Exam Strategy & Shortcut Work backwards from C's efficiency using ratios. C's efficiency = $4x$. B's efficiency = $2x$. A's efficiency = $x$. Total efficiency = $7x$. Since C ($4x$ efficiency) takes $7$ days, Total Work = $4x \times 7 = 28x$. Time taken by all together = $\frac{\text{Total Work}}{\text{Total Efficiency}} = \frac{28x}{7x} = 4$ days. ### Common Pitfall Mixing up the ratios by assuming A is the fastest. "Work done by A is half" means A is less efficient and slower, not faster. ### Final Answer Therefore, the correct answer is **4**.