A, B and C. each working alone, can finish a piece of work in 27, 33 and 45 days respectively. A starts by working alone for 12 days, then…

Solution
A's 1 day's work = $\frac{1}{27}$
=> A's 12 days work = $\frac{1}{27} \times 12 = \frac{4}{9}$
Remaining work = $1 - \frac{4}{9} = \frac{5}{9}$
Similarly, B's 11 days work = $\frac{1}{33} \times 11 = \frac{1}{3}$
Remaining work = $\frac{5}{9} - \frac{1}{3} = \frac{2}{9}$
Now, days taken by C to complete the work = 45 days
=> Days taken to complete $\frac{2}{9}$ th of the work = $\frac{2}{9} \times 45 = 10$ days
$\therefore$ Days in which the whole work is completed = 12 + 11 + 10 = 33 days

Solution
A's 1 day's work = $\frac{1}{27}$
=> A's 12 days work = $\frac{1}{27} \times 12 = \frac{4}{9}$
Remaining work = $1 - \frac{4}{9} = \frac{5}{9}$
Similarly, B's 11 days work = $\frac{1}{33} \times 11 = \frac{1}{3}$
Remaining work = $\frac{5}{9} - \frac{1}{3} = \frac{2}{9}$
Now, days taken by C to complete the work = 45 days
=> Days taken to complete $\frac{2}{9}$ th of the work = $\frac{2}{9} \times 45 = 10$ days
$\therefore$ Days in which the whole work is completed = 12 + 11 + 10 = 33 days