Three workers working together need 1 hour to construct a wall. The first worker, working alone, can construct the wall twice as fast as the third worker, and can complete the task an hour sooner than the second worker. Then, the average time in hours taken by the three workers, when working alone, to construct the wall is

Let the time taken for the first worker to construct the wall alone be A hours. Let the time taken for the second worker to build the wall alone be B hours. Let the time taken for the third worker to construct the wall alone be C hours.

The first worker, working alone, can construct the wall twice as fast as the third worker.

=> $$\dfrac{1}{A}=2\left(\dfrac{1}{C}\right)$$ => $$\dfrac{1}{C}=\dfrac{1}{2A}$$

The first worker can complete the task an hour sooner than the second worker.

=> $$A=B-1$$ => $$B=A+1$$

The three workers working together need 1 hour to construct a wall.

=> $$\dfrac{1}{A}+\dfrac{1}{B}+\dfrac{1}{C}=\frac{1}{1}$$

=> $$\dfrac{1}{A}+\dfrac{1}{A+1}+\dfrac{1}{2A}=1$$

=> $$\dfrac{2A+2+2A+A+1}{2A\left(A+1\right)}=1$$

=> $$5A+3=2A^2+2A$$

=> $$2A^2-3A-3=0$$

=> $$A=\dfrac{3\pm\sqrt{9-4\left(-3\right)\left(2\right)}}{2\left(2\right)}$$

=> $$A=\dfrac{3+\sqrt{33}}{4}$$          [Since the value of A cannot be negative]

The average time in hours taken by the three workers, when working alone, to construct the wall is = $$\dfrac{A+B+C}{3}$$

$$=\dfrac{\frac{3+\sqrt{33}}{4}+\frac{3+\sqrt{33}}{4}+1+2\left(\frac{3+\sqrt{33}}{4}\right)}{3}$$

$$=\dfrac{3+\sqrt{33}+1}{3}$$

$$=\dfrac{4+\sqrt{33}}{3}$$