Question

Suppose $A$ can do a piece of work in $14$ days while $B$ can do it in $21$ days. They begin together and worked at it for $6$ days. Then $A$ fell ill and $B$ had to complete the work alone. In how many days was the work completed?

Answer 1

$A$ can do a piece of work in $14$ days.

Therefore $A^{\prime }s$ one days work will be $\frac{1}{14}$

$B$ can do a piece of work in $21$ days.

Therefore $B^{\prime }s$ one days work will be $\frac{1}{21}$

If $A$ and $B$ together work then their one days work will be

$\Rightarrow \frac{1}{A}+\frac{1}{B}=\frac{1}{14}+\frac{1}{21}$

$\Rightarrow \frac{1}{A}+\frac{1}{B}=\frac{3+2}{42}$ taking LCM

$\Rightarrow \frac{1}{A}+\frac{1}{B}=\frac{5}{42}$

$A$ and $B$ worked for $6$ days

Therefore their $6$ days work is $\frac{5}{42}\times 6=\frac{5}{7}$

Amount of work completed is $\frac{5}{7}$

Work to be completed is $1-\frac{5}{7}=\frac{7-5}{7}=\frac{2}{7}$

Therefore $\frac{2}{7}$ work has to be completed by $B$ only.

$=\frac{2}{7}\times 21$

$=2\times 3=6days$

Work will be completed in $6$ days