Question

a can do a piece of work in 14 days while B can do it in 21 days they began together and worked at it for 6 days then a fell ill and b
had to complete the remaining work alone in how many days was the work completed

Answer 1

After days A did 6/14=3/7 piece of work, and B 6/21=2/7 piece. The remaining work:
1 - 3/7-2/7= 7/7-5/7=2/7 worked B in 21 * 2/7 = 6days.
The work was in 6+6 = 12 days completed.

Answer: 12 days

Note: If A had not been ill, it's the work would be finished by x days:
x/14 + x/21 = 1
3x/42 + 2x/42 = 1
5x = 42
x = 8.4 days
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Answer 2

A can do the piece of work in 14 days. So in one day A completes 1/14 part of work.
Similarly B completes 1/21 part of work.
They worked together for 6 days. So the work completed by them- $= \frac {1}{14} + \frac {1}{21}\\ =\frac {14+21}{14 \times21}\\ =\frac {35}{21 \times 14}\\ =\frac {5}{42}$ So now 5/42 part of work is completed by both in one day. In 6 days they would complete $\frac {5}{42}\times 6\\ =\frac {5}{7}$
In 6 days both complete 5/7 part of work.
Now the part of work remaining is $1-\frac {5}{7} \\= \frac {7-5}{7}\\ =\frac{2}{7}$
As A falls ill at this stage ; B has to work alone. B completes 1/21 part of work. So he can complete the remaining 2/7 part of work in:-$\frac {2}{7} \times 21\\ =2\times 3\\ =6 \:days.$
So in all 6+6=12 days were needed to complete the work.