Question

Solve the following
A takes $6$ days less than $B$ to finish a piece of work. if both $A\&B$ together can finish the work in $4$ days find the time take $B$ to finish the work.

Answer 1

let the work done by $A$ in $1$ day be $x$ and by $B$ be $y$

given $A$ takes $6$ days less than be to finish the work

therefore let the days taken by $A$ alone to finish the work be $k$

then total work $=x.k$

that is also equals to $y.(k+6)$ and $(x+y).4$

$\Rightarrow xk=y(k+6)=(x+y)4$

therefore $x=\frac{4y}{(k-4)}$

substituting it in $ky+6y=4x+4y$ we get

$k+6=4+\frac{16}{(k-4)}$

$k-\frac{16}{(k-4)}+2=0$

$\Rightarrow k^2-4k+2k-8-16=0\Rightarrow k^2-2k-24=0\Rightarrow k=\frac{2\pm \sqrt{4+4.48}}{2}=8$

therefore A alone takes 8 days and B alone takes $8+6=14$ days to finish the work