A and B can do a work in 9 days and 12 days respectively. If they work on alternate days starting with A, then in how many days will the work be completed?

Question

Accepted Answer

A's one-day work = $$ \frac{1}{9} $$  
B's one-day work = $$ \frac{1}{12} $$

In two alternate days (one day A and one day B), the work done is:  
$$
	ext{Work in two days} = \frac{1}{9} + \frac{1}{12}
$$

Take LCM of 9 and 12:  
$$
	ext{Work in two days} = \frac{4}{36} + \frac{3}{36} = \frac{7}{36}
$$

Thus, in two days, they complete $$ \frac{7}{36} $$ of the total work.

To find how many such cycles (of two days) are needed to complete most of the work:  
$$
	ext{Number of cycles} = \left\lfloor \frac{36}{7} ightfloor = 5 	ext{ cycles (10 days)}.
$$

In these 5 cycles (10 days), the total work completed is:  
$$
5 	imes \frac{7}{36} = \frac{35}{36}.
$$

The remaining work is:  
$$
1 - \frac{35}{36} = \frac{1}{36}.
$$

On the next day (11th day), A will work and can complete $$ \frac{1}{9} $$ of the work in a day. To complete $$ \frac{1}{36} $$, A will take:  
$$
	ext{Time taken by A} = \frac{\frac{1}{36}}{\frac{1}{9}} = \frac{1}{4} 	ext{ day}.
$$

Thus, the total time taken to complete the work is:  
$$
10 + \frac{1}{4} = 10 \frac{1}{4} 	ext{ days}.
$$

Hence, the total time required is **10 1/4 days**.

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