A and B together can do a piece of work in 6 days. If A’s one day’s work is $1\dfrac{1}{2}$ times the one day's work of B, find how many days, each alone can finish the work.

Let A's one day work be x and B's one day work be y.

According to first condition given in the problem,

A works $1\dfrac{1}{2}$ times of B,

$\Rightarrow x = 1\dfrac{1}{2}y$

$\Rightarrow x = \dfrac{3}{2}y$

⇒ 2x = 3y

⇒ 2x - 3y = 0     ……(1)

Also given, A and B together can do a piece of work in 6 days.

$\therefore x + y = \dfrac{1}{6}$

⇒ 6(x + y) = 1

⇒ 6x + 6y = 1     …(2)

Multiplying (1) by 2 we get,

⇒ 2(2x - 3y) = 2 × 0

⇒ 4x - 6y = 0     …(3)

Adding equations (2) and (3) we get,

⇒ 6x + 6y + 4x - 6y = 1 + 0

⇒ 10x = 1

⇒ x = $\dfrac{1}{10}$

Substituting value of x in equation (1), we get :

⇒ 2 × $\dfrac{1}{10}$ - 3y = 0

⇒ $\dfrac{1}{5}$ - 3y = 0

⇒ 3y = $\dfrac{1}{5}$

⇒ y = $\dfrac{1}{15}$

Since, A's one day work is x and B's one day work is y, so A can do complete work in $\dfrac{1}{x}$ and B can do work in $\dfrac{1}{y}$ days.

$\dfrac{1}{x} = \dfrac{1}{\dfrac{1}{10}}$ = 10 days

$\dfrac{1}{y} = \dfrac{1}{\dfrac{1}{15}}$ = 15 days

Hence, A can finish the work in 10 days while B can finish the work in 15 days.