X does half as much work as Y in 1/6 of time. If together they can complete a job in 10 days, how many days will Y take to complete it, when he works alone?

Worker X does half as much work as Y in $$\frac{1}{6}$$​​ of the time.

Together, X and Y can complete the job in 10 days. 

Formula Used: 

​$$\text{Total Work} = \text{Efficiency}\times \text{Times}$$ 

Solution: 

X does half as much work as Y in $$\frac{1}{6}$$​ of the time. 

If Y's efficiency is 1 unit , then,

X"s efficiency is 3.

Now, Time taken together X + Y = 10 days

Total work = $$(1+3)\times 10 = 40 \,unit$$ 

Time taken by Y to work alone,

Time = $$\frac{40}{1} = \bf 40 \, days$$ ​