Question

$A$ takes $6$ days less than the time taken by $B$ to finish a piece of work. If both $A$ And $B$ together can finish it in $4$ days, find the time taken by $B$ to finish the work.

Answer 1

Suppose $B$ alone takes $x$ days to finish the work. Then $A$ alone can finish it in $(x-6)$ days

Now A's one day's work + B's one day's work $=\frac{1}{x}+\frac{1}{x-6}$
and (A+B)'s one day's work $=\frac{1}{4}$

$\therefore \frac{1}{x}+\frac{1}{x-6}=\frac{1}{4}$

$\Rightarrow \frac{x-6+x}{x(x-6)}=\frac{1}{4}\Rightarrow \frac{2x-6}{x^2-6x}=\frac{1}{4}$

$\Rightarrow 8x-24=x^2-6x$

$\Rightarrow x^2-14x+24=0$

$\Rightarrow x^2-12x-2x+24=0$

$\Rightarrow (x-12)(x-2)=0$

$\Rightarrow x=12orx=2$

But $x$ cannot be less than $6$.

So $x=12$

Hence, $B$ also can finish the work in $12$ days.