Question

$A$ and $B$ together can do a piece of work in $10$ days, but $A$ alone can do it in $15$ days.How many days would $B$ alone take to do the same piece of work.

Answer 1

Let $B$ alone take $x$ days to do the work. Since $A$ alone can complete the work in $15$ days.

$\therefore$ in one day $A$ completes ${\frac{1}{15}}^{th}$ part of the work.

Similarly, in one day $B$ completes ${\frac{1}{x}}^{th}$ part of the work.

$A$ and $B$ together completes ${\frac{1}{10}}^{th}$ part of the work in one day.

Now $A^{\prime }s$ one day's, work$+B^{\prime }s$ one day's work$={(A+B)}^{\prime }s$ one day's work

$\Rightarrow \frac{1}{15}+\frac{1}{x}=\frac{1}{10}$

$\Rightarrow \frac{1}{x}=\frac{1}{10}-\frac{1}{15}=\frac{15-10}{150}=\frac{1}{30}$

$\Rightarrow x=30$ days

$B$ can alone do the work in $30$ days.