S does half as much work as T in 1/7 of the time taken by T. If together they take 21 days to complete a work, then how may days shall S take to complete that work alone?

Let time taken by T to complete the work = $$7x$$ days

=> S can do 1/2 work in = $$\frac{1}{7}\times7x=x$$ days

So, time taken by S to complete the work = $$2x$$ days

Let total work to be done = L.C.M $$(7x,2x)=14x$$ units

=> T's efficiency = $$\frac{14x}{7x}=2$$ units/day

Similarly, S's efficiency = $$\frac{14x}{2x}=7$$ units/day

Now, total time taken by both S and T together = $$21(2+7)=14x$$

=> $$3\times9=2x$$

=> $$x=\frac{27}{2}=13.5$$

$$\therefore$$ Time taken by S to complete work alone = $$2\times13.5=27$$ days

=> Ans - (B)