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?
It is given that X does half as much work as Y in 1/6 of time,
So,
$$\frac{1}{6}\times\frac{1}{X}=\frac{1}{2}\times\frac{1}{Y}$$
or,$$\frac{X}{Y}=\frac{1}{3}.$$
They together do in 10 days.
$$\frac{1}{X}+\frac{1}{Y}=\frac{1}{10}.$$
or,$$\frac{1}{X}+\frac{1}{3X}=\frac{1}{10}.$$
or,$$\frac{4}{3X}=\frac{1}{10}.$$
or,$$X=\frac{40}{3}.$$
So, $$Y=\frac{40}{3}\times3=40.$$
C is correct choice.
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