A does half as much work as B in one-sixth of the time. If together they take 10 days to complete a work, how much time shall B alone take to do it? (S.S.C., 2005)

### Concept & Efficiency Ratio The core concept involves relating the amount of work done to the time taken to establish an efficiency ratio between two workers. $$ \text{Efficiency} \propto \frac{\text{Work}}{\text{Time}} $$ ### Step-by-Step Solution * **Given:** A does $1/2$ work in $1/6$ time of B. A and B together complete the work in 10 days. * **Calculation:** Let's assume B takes $x$ days to complete 1 unit of work. * Thus, B's 1-day work is $1/x$. * A does $1/2$ of the work in $(1/6)x$ days. * To do the full work (1 unit), A will take $2 \times (1/6)x = x/3$ days. * So, A's 1-day work is $3/x$. * Together, their 1-day work is $(1/x) + (3/x) = 4/x$. * We are given that together they take 10 days, so their combined 1-day work is $1/10$. * Equating the two: $4/x = 1/10 \implies x = 40$. * Therefore, B alone takes 40 days to complete the work. ### Exam Strategy & Shortcut Instead of fractions, use efficiency ratios. If A does $1/2$ work in $1/6$ time, for the same time (1 full unit of time), A would do $1/2 \times 6 = 3$ times the work of B. So, Efficiency of A : B = 3 : 1. Together, they have an efficiency of 4 units/day. In 10 days, total work = 40 units. Time for B alone = Total Work / B's Efficiency = $40 / 1 = 40$ days. ### Common Pitfall Misinterpreting "half as much work in one-sixth of the time". Students often directly write the efficiency ratio as 1:2 or 1:6, forgetting to mathematically combine both the work and time modifiers. ### Final Answer Therefore, the correct answer is **40 days**.