A and B, working separately can do a piece of work in 12 days and 20 days respectively. If they work alternately, beginning with A, then in how many days can the work be completed?

A can complete the work in 12 days, so A's work in 1 day $= \dfrac{1}{12}$

B can complete the work in 20 days, so B's work in 1 day $= \dfrac{1}{20}$

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They work alternately, with A starting first.

In one complete cycle of 2 days:

Day 1: A works and completes $\dfrac{1}{12}$ of the work

Day 2: B works and completes $\dfrac{1}{20}$ of the work

Work completed in 2 days $= \dfrac{1}{12} + \dfrac{1}{20}$

$= \dfrac{5}{60} + \dfrac{3}{60}$

$= \dfrac{8}{60}$

$= \dfrac{2}{15}$

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Number of complete 2-day cycles needed:

In 2 days: $\dfrac{2}{15}$ work completed

In 14 days (7 cycles): $7 \times \dfrac{2}{15} = \dfrac{14}{15}$ work completed

Remaining work $= 1 - \dfrac{14}{15} = \dfrac{1}{15}$

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After 14 days, it is A's turn to work on day 15.

A completes $\dfrac{1}{12}$ work in 1 day.

Since $\dfrac{1}{12} > \dfrac{1}{15}$, A can finish the remaining work in less than a day.

Time required by A to complete remaining work $= \dfrac{1/15}{1/12}$

$= \dfrac{1}{15} \times 12$

$= \dfrac{12}{15}$

$= \dfrac{4}{5}$ day

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Total time to complete the work $= 14 + \dfrac{4}{5} = 14\dfrac{4}{5}$ days

Since the work is completed during the 15th day, the answer is 15 days.